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How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high.
Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
Cr2+ is strong reducing agent than Fe2+.
Reason: d4 → d5 occurs in case of Cr2+ to Cr3+.
But d6 → d5 occurs in case of Fe2+ to Fe3+. In a medium (like water) d3 is more stable as compared to d5.
Cu+ in aqueous solution undergoes disproportionation i.e.,
2Cu+(aq) → Cu2+(aq) + Cu(s)
The E°, value for this is favourable.
The outer electronic configuration of copper is 3d10 4s1, yet it is considered transition element. Why?
Though copper, silver and gold have completely filled sets of d-orbitals yet they are considered as transition metals. Why?
These metals in their common oxidation states have incompletely filled d-orbitals e.g., Cu2+ has 3d10 and Au3+ has 5d8 configuration.
copper displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (3d9). Hence, it is a transition element. wherein it will have incompletely filled d-orbitals (4d), hence a transition metal.
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There is a very less energy difference between (n – 1) d and ns orbitals because of which there exists exceptions to the general electronic configuration.
Why do the transition elements exhibit higher enthalpies of atomisation?
Why does Zr and Hf exhibit similar properties?
Zr and Hf exhibit similar properties due to lanthanoid contractin. Electrons present in f subshell didn't do good shielding due to which with the increasing atomic number or increasing effective nuclear charge size gets constricted and size of Halfenium and Zirconium becomes almost equal.
2(40) = [Kr]364d2 5s2, Hf(72) = [Xe]54 4f14 5d2 6s2
Why is the following electronic configuration not correct for the ground state of Mo atom? (at. no. = 42).
1s2 2s2 2p6 3s23p63d10 4s2 4p6 4d4 5s2.
Half filled orbitals have extra stability. Therefore, the correct configuration is:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 4d5
Name a transition element which does not exhibit variable oxidation states.
Which element shows the highest oxidation state among the d-block elements?
What is the most stable oxidation state of Mn (Z = 25)?
The electronic configuration of an element is 3d5 4s1. Write its (i) most stable oxidation state and (ii) most oxidising state.
Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as 3t2g configuration, which is a more stable configuration.
In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and has an extra-stability.
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Why V and Mn has E° value more than E° value of Cr?
Compare qualitatively the first and second ionisation potentials of copper and zinc. Explain the above observation.
Electronic configuration of copper is [Kr] 3d10 4s1 and electronic configuration of zinc is [Kr] 3d10 4s2 .
On the basis of Cu and Zn. The first ionization potential of zinc is higehr than Cu.
Because in the zinc electron is removed form 4s2 configuration (which is in stable state) while in copper it is removed form 4s1 configuration thus more energy is required for the removal of 4s2 electron than that of 4s1.
While in second ionization potential of copper is higher than that of zinc because Cu+ has 3d10 configuration (stable) in comparsion of 3d10 4s1of Zn+.
What does the E° value of M3+/M2+ show for Mn3+ and Co3+ (+ 1.57 and 1.97 V respectively)?
For the first row transition metals the E° values are:
E° V Cr Mn Fe Co Mc Cu
(M2+/M) –1.18 –0.91 –1.18 –0.44 –0.28 –0.25 +0.34
Explain the irregularity in the above values.
Calculate the magnetic moment of a divalent ion in aqueous solution its atomic number is 25.
Why does Mn(II) ion show maximum paramagnetic character amongst bivalent ions of first transition series?
Paramagnetic character depend on the unpaired electron. The electronic configuration of Mn(II) ion (at. no. of Mn25) is
1s2 2s2 2p6 3s2 3p6 3d5 4s0
It has five unpaired electrons in its d-orbitals which is maximum value for a transition metal ion. The paramagnetic character is due to presence of unpaired electron.
Calculate the magnetic moment of Fe3+ ion (at. no. of Fe = 26).
Electronic configuration of Fe3+ ion is 1s2 2s2 2p6 3s2 3p6 3d5 4s0
It has 5 unpaired electrons
∴ Magnetic moment,
Why Cd2+ salts are white?
Why is copper sulphate pentahydrate coloured?
In copper sulphate pentahydrate (CuSO4.5H2O) copper is divalent (Cu2+). Copper(II) have the configuration
1s2 2s2 2p6 3s2 3p6 3d9
An ion or compound is coloured when it has unpaired electron in its d-orbitals. Cu(II) has one unpaired electron in its d-orbital, hence it is coloured.
An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
The two types of alloy formed by transition elements are (i) alloy steels and (ii) stainless steel. Nickel steel is an example of smooth alloys.(i) They have high melting points, higher than those of pure metals.
(ii) They are very hard, some borides approach diamond in hardness.
A reaction in which the same species is simultaneously oxidised as well as reduced is called disproportionation reaction.
When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(Vll) and manganese(IV) in acidic solution.
3MnVI O42– + 4H+ → 2MnVIII O4– + MnIV O2 + 2H2O
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Give chemical equation of reaction when pyrolusite ore is fused with KOH in presence of air.
Why is hydrochloric acid not used to acidify a permanganate solution in volumetric estimation of Fe2+ or C2O42–?
i)When the aqueous solution of potassium dichromate is treated with calculated amount of KOH, potassium chromate is formed.
K2Cr7O7 + 2KOH → 2K2CrO4 + H2O
ii) When an alkali is added to an orange red solution containing dichromate ions, a yellow solution is obtained due to the formation of chromate ions.
K2Cr2O7 + K2CO3 → 2 K2CrO4 + CO2
Sulfur dioxide gas turns acidified potassium dichromate solution from orange to green reduced chromium +4 to +3.
K2Cr2O7 + 2H2SO4 + 3SO2 → 2Cr2(SO4)3 + K2SO4 + H2O
K2Cr2O7 + 7H2SO4 + 6FeSO4 → Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O
Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Give the general electronic configurations of lanthanoids and actinoids.
What is the composition of mischmetal alloy?
Give two uses of lanthanoid compounds.
(i) Mischmetal (an alloy) is used in gas lighters, tracer bullets and shell.
(ii) Oxides of neodymium and praseodymium are used for making colour glasses.
In permagnate ion, all the bonds formed between Mn and O are covalent. Give reasons.
What is the maximum oxidation state shown by actinoids?
The maximum oxidation state of actinoids +7.
Why Sm2+, Eu2+ and Yb2+ ions in solutions are good reducing agents but an aqueous solution of Ce4+ is a good oxidizing agent?
One among the lanthanoids, Ce(III) can be easily oxidized to Ce(IV) (at. No. of Ce = 58). Explain why?
The +3 oxidation states of lanthanum (Z = 57), gadolinium (Z = 64) and lutetium (Z = 71) are especially stable. Why?
Which trivalent ion has the largest size in lanthanoid series?
What are the different oxidation states exhibited by the lanthanoids?
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Copper metal (Cu, at. no. 29) shows +1 oxidation state i.e., it exists as Cu+ in large number of copper compounds e.g., cuprous oxide (Cu2O), cuprous sulphide (Cu2S); cuprous chloride (Cu2Cl2) etc. The electronic configuration of Cu+ is [Ar] 3d10 4s0.
This configuration is very stable as all five 3d orbitals are fully filled.
Write the electronic configurations of the elements with the atomic numbers 61, 91, 101 and 109.
Atomic number (61) = [Xe]54 4f5 6s2.
Atomic number (91) = [Rn]86 5f2 6d1 7s2
Atomic number (101) = [Rn]86 5f14 7s1
Atomic number (109) = [Rn]86 5f 14 6d7 7s2.
Write chemical equations for the reactions involved in the manufacture of perman-ganate from pyrolusite ore.
Why is the + 2 oxidation state of manganese quite stable while the same is not true for iron? |Mn = 25, Fe = 26]
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What is misch metal (misch metal)? Mention its two important uses.
Misch metal consists of lanthanoid (95%), iron, traces of S, C, Ca and Al.
Uses:
(i) It used to produce bullets.
(ii) It is used in shells and lighter flints.
What is the effect of increasing pH in K2Cr2O7, solution?
Within a periodic group of transition elements the possibility of exhibiting maximum oxidation state increases with atomic number. Why?
(i) Cr3+ = 1s2 2s2 2p6, 3s2 3p6 3d3
(ii) Cu+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10
(iii) Co2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d7
(iv) Mn2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
(v) Pm3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6 4f4
(vi) Ce4+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6.
(vii) Lu2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6 4f14, 5d1.
(viii) Th4+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 4f14, 5s2, 5p6, 5d10, 6s2, 6p6.
Exactly half-filled and completely filled subshells are relatively more stable. Species try to attain the state of exactly half-filled and completely filled subshells configuration.
Mn2+ has extra stability as it has exactly half filled configuration of 4s° 3d5 in its outermost shell while Fe2+ has 45° 3d6 configuration. Therefore, Fe2+ has a tendency to loose one electron (or tends to oxidise to Fe3+) to acquire extra stability by having 4s0 3d5configuration.
Fe2+ → Fe3+ + e–(4s03d6) (4s03d5)
Since Mn2+ has half filled orbital thus it is more stable to Fe2+.
In transition elements, there are greater horizontal similarities in the properties in contrast to the main group elements because of similar ns2 common configuration of the outermost shell.
An examination of common oxidation states reveals that excepts scandium, the most common oxidation state of first row transition elements is +2 which arises from the loss of two 4s electrons. This means that after scandium, d-orbitals become more stable than the s-orbital.
Further, +2 state becomes more and more stable in the first half of first row transition elements with increasing atomic number because 3d orbitals acquire only one electron in each of five 3d orbitals (i.e. remains half filled) and electronic repulsion is the least and nuclear charge increases. In 2nd half of first row transition elements, electrons starts pairing up in 3d orbitals.
(Ti2+ to Mn2+ electronic, configuration changes from 3d2 to 3d5 but in 2nd half i.e. Fe2+ to Zn2+ it changes from d6 to d10).
One of the main characteristic of a transition element is that it can show large variety of oxidation states in its compounds. It is due to its characteristic electronic configuration i. e., (n – 1)d and ns electrons take part in bond formation either by loosing or by sharing of electrons with other combining atoms.
The stability of oxidation state depends mainly on electronic configuration and also on the nature of other combining atom.
The elements which show largest number of oxidation states occur in or near the middle of series (i.e., 4s23d3 to 4s23d7 configuration). For example, Mn exhibits all oxidation states from +2 to +7 as it has 4s23d5 configuration. The lesser number of oxidation states at extreme ends arise from either too few electrons to loose or share (e.g. Sc and Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher elements at upper end of first transition series (i.e., Cu and Zn). Thus electronic configuration, to large extent, the existence and stability of oxidation states.
The other factors which determine stability of oxidation state are:
(i) Enthalpy of atomisation (ii) Ionisation energy (iii) Enthalpy of solvation (iv) E.N. of other element.
The stable oxidation state of transition element with the d electron configuration in ground state of atoms are as follows:
|
S. No. d-electron |
configuration |
Symbol of element |
Stable oxidation states |
|
1. 2. 3. 4. |
3d3 3d4 3d5 3d8 |
Vanadium[V](4s23d3) Chromium Cr(4s23d4) Manganes Mn(4s23d5) Nickel Ni(4s23d8) |
+2, +3, +5 +2, +3, +6 +2,+7 +2,+4 |
It should be noted that lower stable oxidation state generally leads to ionic bond and higher oxidation state corresponds to covalent bond.
|
Name of oxometal anion |
Name of metal with oxidation state |
Group no. to which metal belong |
|
1. CrO42– 2. MnO4– |
Cr in +6 state of oxidation |
6th group of periodic table |
A group of fourteen elements following lanthanum i.e. from 58Ce to 71Lu placed in 6th period of long form of periodic table is known as lanthanoids (or lanthanide series). These fourteen elements are represented by common general symbol ‘Ln’. In these elements, the last electron enters the 4f-subshells (pre pen ultimate shell). It may be noted that atoms of these elements have electronic configuration with 6s2 common but with variable occupancy of 4f level. However, the electronic configuration of all the tripositive ions (the most stable oxidation state of all lanthanoids) are of the form 4f n(n = 1 to 14 with increasing atomic number). These elements constitute one of the two series of inner transition elements or f-block.
Lanthanoid contraction: In the lanthanoide series with the increase in atomic number, atomic radii and ionic radii decrease from one element to the other, but this decrease is very small. The regular small decrease in atomic radii and ionic radii of lanthanides with increasing atomic number along the series is called lanthanoid contraction.
Cause of lanthanoid contraction: When one moves from 58Ce to 71Lu along the lanthanide series nuclear charge goes on increasing by one unit every time. Simultaneously an electron is also added which enters to the inner f subshell. The shielding effect of f-orbitals in very poor due to their diffused shape. It results in the stronger force of nuclear attraction of the 4f electrons and the outer electrons causing decrease in size.
Consequences of lanthanoid contraction:
(i) Similarly in the properties of elements of second and third transition series e.g. Sr and Hf; Nb and Ta; Mo and W. This resemblance is due to the similarity in size due to the presence of lanthanoids in berween.
(ii) Similarity among lanthanoids: Due to the very small change in sizes, all the lanthanoids resemble one another in chemical properties.
(iii) Decrease in basicity: With the decrease in ionic radii, covalent character of their hydroxides goes on increasing from Ce(OH)3 to Lu(OH)3 and so base strength goes on decreasing.
The important characteristics of transition metals are:
(i) All transition elements are metallic in nature, e.g., all are metals.
(ii) These metals exhibit variable oxidation states.
(iii) Transition metal atoms or ions generally form the complexes with neutral, negative and positive ligands.
(iv) Compounds of transition metals are usually coloured.
(v) The compounds of these metals are usually paramagnetic in nature.
(vi) Transition metals and their compounds act as good catalysts, i.e., they show catalytic activities.
(vii) These metals form various alloys with other metals of the series.
(viii) These metals form interstitial compounds with C, N, B and H.
The presence of partially filled d-orbitaIs in the electronic configuration of atomic and ionic species of these elements is responsible for the characteristic properties of transition elements. They are called transition elements because of their position in the periodic table. These elements lie in the middle of periodic table between s and p-blocks (i.e., between group 2 and group 13). A transition element may be defined as a element whose atom or at least one of its simple ions contain partially filled d-orbitals, e.g., iron, copper, chromium, nickel etc.
The general characteristic electronic configuration may be written as (n – 1)d1–10ns1–2.
The elements of group 12 i.e., Zinc, Cadmium, and Mercury are generally not regarded as transition elements as their atoms and all ions formed have completely filled d-orbitals i.e., these do not have partially filled d-orbitals in atomic state or common oxidation state (Zn2+, Cd2+, Hg2+).
Zn (30) = [Ar] 4s2 3d10 Zn2+ = [Ar] 3d104s°
Cd (48) = [Kr] 5s2 4d10 Cd2+ = [Kr] 4d105s°
Hg (80) = [Xe] 6s2 5d10 Hg2+ = [Xe] 5d106s°
The melting points depend upon the inter atomic bonding, more the number of electrons given by a metal higher in its interatomic bonding (metallic bonding) and thus higher will be its melting point.
In d-block elements due to the involvement of both ns and (n – 1)d electrons, the number of electrons involved are higher and therefore stronger interatomic bonding and higher melting point.
The metallic bond is formed as a result of the interaction of electrons in the outermost shell. Greater the number of valence electrons, stronger is the metallic bond.
The most important reason transition metals are good catalysts is that they can lend electrons or withdraw electrons from the reagent, depending on the nature of the reaction. The ability of transition metals to be in a variety of oxidation states, the ability to interchange between the oxidation states and the ability to form complexes with the reagents and be a good source for electrons make transition metals good catalysts.
What are interstitial compounds? Why are such compounds well known for transition metals?
Interstitial compounds are those which are formed when small atoms like H, C, N, B etc. are trapped inside the crystal lattices of metals. They are generally non-stoichiometric and neither typically ionic nor covalent.
Most of transition metals form interstitial compounds with small non-metal atoms such as hydrogen, boron, carbon and nitrogen. These small atoms enter into the void sites between the packed atoms of crystalline transition metals and form chemical bonds with transition metals. For Example, steel and cast iron become hard by forming interstitial compound with carbon.
The existence of vacant (n – 1) d orbitals in transition elements and their ability to make bonds with trapped small atoms in the main cause of interstitial compound formation. Other examples are : VH0.56, TiH1.7 Some main characteristics of these compounds are:
(i) They have high melting and boiling points, higher than those of pure metals.
(ii) They are very hard. Some borides of transition elements approach diamond in hardness.
(iii) They are chemically inert but retain metallic conductivity.
The variability in oxidation states is a fundamental characteristic of transition elements and it arises due to incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity. For example, vanadium, V show the oxidation states of + 2, + 3, + 4 and + 5. Similarly, Cr shows oxidation states of +2, +3, +4, +5 and +6; Mn shows all oxidation states from +2 to +7.
This is contrasted with variability of oxidation states of non-transition elements where oxidation states generally differ by units of two. For example, S shows oxidation states of -2, +2, + 4, +6 while P shows +3 and +5 oxidation states. Halogenes like Cl, Br and I show oxidation states of –1, +1, +3, +5 and +7 states. In non-transition elements variability of oxidation states is caused due to unpairing of electrons in ns or np orbitals and their promotion to np or nd vacant orbitals.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
The following steps are involved in preparation of K2Cr2O7 from iron chromite (FeCr2O4) ore:
(i) Preparation of sodium chromate: The chromite ore (FeO.Cr2O3) is finely powdered and mixed with sodium carbonate and quick lime and then heated to redness in a reverberatory furnace with free supply of air.
The mass is then extracted with water, when sodium chromate is completely dissolved while Fe2O3 is left behind.
(ii) Conversion of sodium chromate into sodium dichromate (NaCr2O7) : The sodium chromate extracted with water in previous step is acidified.
3Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4), + H2O
On cooling Na2SO4 separates out as Na2SO4. 10H2O and Na2Cr2O7, is remains in solution.
(iii) Conversion of Na2Cr2O7 into K2Cr2O7: The solution containing Na2Cr2O7 is treated with KCl
Na2Cr2O7 + KCl → K2Cr2O7 + 2NaCl
Sodium chloride (NaCl) being less soluble separates out on cooling. On crystallising the remaining solution, orange coloured crystals of K2Cr2O7 separate out.
Effect of Change of pH: When pH of solution of K2Cr2O7 is increased slowly the medium changes from acidic to basic. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of solution.
At low pH (acidic medium), K2Cr2O7 solution is oranged coloured while at higher pH (alkaline medium) it changes to yellow due to formation of chromate ions.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(a) iodide (b) iron (II) solution and (c) H2S.
Potassium dichromate, K2Cr2O7 is a strong oxidising agent and is used as a primary standard in volumetric analysis involving oxidation of iodides, ferrous ion and S2– ions etc. In acidic solution, its oxidising action can be represented as follows:
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O;
(E+ = 1.33 V)
(a) It oxidises potassium iodide to iodine.
Cr2O72– + 14H+ 61– → 2Cr3 + 7H2O + 3I2
(b) It oxidises iron(II) salt to iron (III) salt
Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O
(c) It oxidises H2S to S
Cr2O72– + 8H+ + 3H2S → 2Cr3+ + 7H2O + 3S
Preparation of KMnO4 from pyrolusite ore (MnO2) involves the following steps:
(i) Fusion of ore with alkali in presence of air : Pyrolusite ore is fused with alkali in the presence of air when potassium manganate is obtained as green mass.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(green mass)
The green mass is dissolved in water to obtain aqueous solution of potassium manganate. The insoluble impurities of sand and other metal oxides are removed by filtration.
(ii) Oxidation of manganate into permanganate : The aqueous solution of K2MnO4 is oxidised electrolytically or by using ozone or Cl2 to obtain potassium permanganate. The process is carried out till green colour disappear and solution acquires distinct pink colour.
Potassium permanganate is crystallised out from the solution.
Oxidising Properties: It acts as a powerful oxidising agent in different media differently. In acidic medium, it oxidises iron(II) salts to iron(III) salts,
SO2 to H2SO4 and oxalic acid to CO2 and H2O.
(a) It oxidises iron(II) salt to iron(III) salts.
2MnO42–+ 16H++ 10Fe2+ → 2Mn2+ + 8H2O + 10Fe3+
(b) It oxidised sulphur dioxide to sulphuric acid.
2MnO4– + 5SO2 + 2H2O → 5SO42– + 2Mn2+ 4H+
(c) It oxidises oxalic acid to CO2 and H2O
2MnO4– + 16H+ + 5C2O42 → 2Mn2+ + 8H2O + 10CO2
For M2+/M and M3+/M2+ systems the E° values for some metals are as follows:
Cr2+/Cr – 0.9 V Cr3+/Cr2+ – 0.4 V
Mn2+/Mn – 1.2 V Mn3+/Mn2+ + 1.5 V
Fe2+/Fe – 0.4 V Fe3+/Fe2+ + 0.8 V
Use this data to comment upon
(a) The stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and
(b) In case with which iron can be oxidised as compared to the similar process for either chromium or manganese metal
∴ Mn is oxidised most readily to Mn2+ and Fe is oxidised least readily among given metals Mn, Cr and Fe.
A transition metal ion in coloured if it has one or more unpaired electron in (n-\)d orbitals i.e. 3d orbitals in case of first transition series. When such species are exposed to visible radiation, d-d transition can take place.
In order to predict colour, let us examine (n – 1)d configuration of each specie.
|
Ti3+ |
3d1 configuration |
Species is coloured (purple) |
|
V3+ |
3d2 configuration |
Species is coloured (green) |
|
Cu+ |
3d10 4s°configuration |
Species is colourless. |
|
Sc3+ |
3d° 4s0 |
It does not contain unpaired e– in 3d orbitals. |
|
Mn2+ |
3d5 |
Colourles. It does not contain unpaired electron in 3d-orbitals coloured (pink). All 5 electrons unpaired in five 3d orbitals. |
|
Fe3+ |
3d5 |
Coloured (yellow). All 5 electrons unpaired in 3d orbitals. |
|
Co2+
|
3d7
|
Coloured (pink). Three e– unpaired in 3d-orbitals.
|
In the beginning of 3d transition series, Sc2+ is virtually not known or in other words it is not stable in comparison to Sc3+, Ti2+, V2+, Cr2+ are known but less stable in comparison to their most common oxidation state of +3.
In the middle Mn2+, Fe2+, Co2+ are known and quite common. In fact Mn2+ and Mn7+ are most stable in Mn. Fe2+ is less stable in comparison to Fe3+ but is due to fact that Fe3+tends to loose one electron to aquire d5 structure which has extra stability. Co2+ is less stable as compared to Co3+. Ni2+ is most common and most stable among its +2, +3, +4 states. Cu2+ is more stable and is most common species as compared to Cu1+. At end Zn forms only Zn2+ which is highly stable as it has 3d10 structure.
Compare the chemistry of actinides with that of the lanthanoids with special reference to:
(i) electronic configuration, (ii) atomic and ionic sizes, (iii) oxidation state (iv) chemical reactivity.
|
Characteristics |
Lanthanoids |
Actinides |
|
(a) Electronic configuration |
It may be represented by [Xe]4fx 5dy 6s2, where x varies from 0 to 14 and y = 0 or 1. |
It may be represented by [Rn]5fx 6dy 7s2,where x varies from 0 to 14 and y = 0 or 1. |
|
(b) Oxidation state
|
Show +3 oxidation state only except in few cases where it is +2 or +4. They never show more than +4 state. |
Show higher oxidation states such as +4, +5, +6, +7 also in addition to +3.
|
|
(c) atomic and ionic sizes
|
The ionic radii of M3+ ions in lanthanoids series show a regular decrease in size of ions with increase in atomic number. This decrease is known as lanthanoid contraction. |
There is a greater and gradual decrease in the size of atoms or M3+ ions across the series. This greater decrease is known as actinoid contraction.
|
|
(d) Chemical reactivity |
These are less reactive metals and form oxides, sulphides, nitrides, hydroxides and halides etc. These also form H2 with acids. They show a lesser tendency for complex formation |
These are highly reactive metals especially when are in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temperature. They show a greater tendency for complex formation. |
How would you account for the following:
(a) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.
(b) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidized.
(c) The d1 configuration is very unstable in ions.
(a) Of d4 species, Cr2+ has 3d4 configuration and tends to loose one electron to acquire d3 configuration as it is highly stable and best metallic specie available for complex formation. Cr3+can accommodate six lone pair of electrons from ligands due to sp3d2 hybridisation e.g. [Cr(NH3)6]3+ Mn3+ although have d4 configuration but tends to become Mn2+ stable specie by acquiring one electron to attain d5 configuration. It becomes exactly half filled on one hand and more energy is released in gain of electron due to higher nuclear charge.
(b) Co2+ is stable in aqueous solution because it get surrounded and weakly bonded to water molecules. In presence of strong ligands and air it gets oxidised to Co(III) as strong ligands get coordinated more strongly with Co(III). The electronic configuration of Co(II) and Co(III) are:
Co(II) = [Ar]18 4s03d7 and
Co(III) = [Ar]184s03d6
In Co(III) specie, 6 lone pairs of electrons from ligands are accommodated by sp3d2hybridisation which is not possible in Co(II).
(c) Some species with d1 configuration are reducing and tends to loose one electron to acquire d4 stable configuration. Some other species with d1 configuration like Cr(V) and Mn(VI) undergo disproportionation.
Electronic configuration of Mn is [Ar] 4s2 3d5 and Configuration of iron [Ar] 4s2 3d5.
Mn3+ + 3e– → Mn2+
(more spontaneous due to higher stability of Fe3+) Fe3+ + e– → Fe2+
(less spontaneous due to higher stability of Fe3+)
Due to stability of half filled d-orbitals, Mn2+ is more stable than Mn3+ and thus its reduction is more spontaneous. Similarly Fe3+ is more stable than Fe2+ and thus its reduction is less spontaneous.
Calculate the number of unpaired electrons in following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+, which one of these is the most stable in aqueous solution?
The number of unpaired electrons can be determined from their electronic configurations and are tabulated below:
|
Specie |
Electronic configuration |
No. of unpaired electron |
|
1. Mn3+ |
[Ar] 3d44s0 |
4 |
|
2. Cr3+ |
[Ar] 3d34s0 |
3 |
|
3. V3+ |
[Ar] 3d24s0 |
2 |
|
4. Ti3+ |
[Ar] 3d14s0 |
1 |
Out of these, Cr3+ is most stable in aqueous solution as its hydration energy is highest
.
Give example and suggest reasons for the following features of the transition metal chemistry:
The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
Give example and suggest reasons for the following features of the transition metal chemistry:
A transition metal exhibits highest oxidation state in oxides and fluroides.
The oxidation state of an element is related to the number of electrons that an atom loses, gains, or appears to use when joining with another atom in compounds. It also determines the ability of an atom to oxidize (to lose electrons) or to reduce (to gain electrons) other atoms or species. Oxidation results in an increase in the oxidation state. Reduction results in a decrease in the oxidation state. If an atom is reduced, it has a higher number of valence shell electrons, and therefore a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants.Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
Give example and suggest reasons for the following features of the transition metal chemistry:
The highest oxidation state is exhibited in oxoanions of a metal.
Indicate the steps in the preparation of K2Cr2O7 from chromite ore.
The preparation of potassium dichromate from chromite involves the following main steps:
(i) The chromate ore is finely ground and heated strongly with molten alkali in the presence of air.
(ii) The solution of sodium chromate is filtered and acidified with dilute sulphuric acid so that sodium dichromate is obtained.
(iii) A calculated quantity of potassium chloride is added to a hot concentrated solution of sodium dichromate. Potassium dichromate is less soluble therefore it crystallizes out first.
Indicate the steps in the preparation of KMnO4 from pyrolusite ore.
Pottassium Permanganate (KMnO4) is prepared from Pyrolusite ore (MnO2). The finely powdered Pyrolusite ore (MnO2) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidizing agent like KNO3 to give the dark green potassium Manganate (K2MnO4). Potassium manganate disproportionate in a neutral or acidic solution to give potassium permanganate.
2 MnO2 + 4 KOH + O2 ----------> 2K2MnO4 + 2H2O
3 MnO42- + 4H+ ------------> 2MnO4- + MnO2 + 2H2O
Commercially potassium permanganate is prepared by the alkaline oxidative fusion ofPyrolusite ore (MnO2) followed by the electrolytic oxidation of manganate (4) ion.
2 MnO2 + 4KOH + O2 -----------> 2K2MnO4 + 2H2O
MnO42- ------(electrolytic oxidation)----> MnO4- + e-
An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.
An important alloy which contains some of the lanthanoid is mischmetall. Mischmetall consists of a lanthanoid metal (~95%) and iron (~5%) and traces of S, C, Ca and Al.
Uses:
(i) Mischmetall is used in Mg based alloy to produce bullets, shell and lighter flint.
Some individual Ln oxides are used asjphosphorus in television screens and similar fluorescing surfaces.
Inner transition metals are those elements in which the last electron enters the f-orbital also transition elements are the elements which have partly filled f-orbitals. These are also called f-block elements. There are two series (each of 14 elements) of inner transition elements..
(i) Lanthanoids (also called 4f series). These are from atomic numbers 58-71.
(ii) Actinoids (also called 5f series ). These are from atomic numbers 90-103
Among the given atomic numbers, only 59, 95 and 102 are the atomic numbers of inner transition elements. The other three atomic numbers represent the transition elements.
Among the actinoids, there is a greater range of oxidation states as compared to lanthanoids. This is in part due to the fact that 5f, 6d and 7s levels are of very much comparable energies and the frequent electronic transition among these three levels is possible. This 6d-5f transition and larger number of oxidation states among actinoids make their chemistry more complicated particularly among the 3rd to 7th elements. Following examples of oxidation states of actinoids. Justify the complex nature of their chemistry.
(i) Uranium exhibits oxidation states of+3, +4, +5, +6 in its compounds. However, the dominant oxidation state in actinoides is +3.
(ii) Nobelium, No is stable in +2 state because of completely filled f14 orbitals in this state.
(iii) Berkelium, BK in +4 oxidation state is more stable due to f7 (exactly half filled) configuration.
(iv) Higher oxidation states are exhibited in oxo ions are UO22+, PuO22+, NpO+ etc.
Lawrencium (Lr) is the last element in the series of antinoids.
Its electronic configuration is [Rn] 5f14 6d1 7s2. The possible oxidation state of Lawrencium is +3.
Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin only' formula.
The electronic configuration of Ce and Ce3+ ion is :
The electronic configuration of Ce : 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2
The electronic configuration of Ce3+ : 1s22s22p63s23p63d104s24p64d105s25p64f1
OR
Ce(Z = 58) = 54[Xe] 4f15d16s2
Ce3+ = 54[Xe] 4f1
The no. of unpaired electron = 1 ‘Spin only’ formula for magnetic moment of a specie,
Cerium (Ce) and Terbium (Tb) show +4 oxidation state. Their electronic configurations are given below:
Ce = [Xn] 4f1 5d1 6s2
Tb = [Xn] 4f0 6s2
It is clear from the configuration of Ce that Ce+4 is favoured by its noble gas configuration i.e., [Xn] 4f0 5d0 5s0, but can be easily converted into Ce3+ ([Xn] 4f1 5d0 6s0). Due to this reason Ce+4 is an oxidising agent.
Tb4+ ion is stabilized due to half filled f-subshell i.e., [Xn] 4f7. It also acts as an oxidant.
Europium (63) and ytterbium (70) show +2 oxidation state, this acts as reducing agents because they can be converted into common oxidation state +3. The electronic configuration of Eu and Y are as follows:
Eu = [Xn] 4f 6s2 Y = [Xn] 4f14 6s2 Formation of Eu2+ ion leaves 4f7 configuration and Y2+ ion leaves 4f14 configuration. These configurations can stable due to half filled and full filled f-subshell. Samarium, Sm (62) 4f6 6s2 also shows both +2 and +3 oxidation states like europium.
|
Characteristics |
Lanthanoids |
Actinides |
|
(a) Electronic configuration |
It may be represented by [Xe]4fx 5dy 6s2, where x varies from 0 to 14 and y = 0 or 1. |
It may be represented by [Rn]5fx 6dy 7s2,where x varies from 0 to 14 and y = 0 or 1. |
|
(b) Oxidation state
|
Show +3 oxidation state only except in few cases where it is +2 or +4. They never show more than +4 state. |
Show higher oxidation states such as +4, +5, +6, +7 also in addition to +3.
|
|
(c) atomic and ionic sizes
|
The ionic radii of M3+ ions in lanthanoids series show a regular decrease in size of ions with increase in atomic number. This decrease is known as lanthanoid contraction. |
There is a greater and gradual decrease in the size of atoms or M3+ ions across the series. This greater decrease is known as actinoid contraction.
|
|
(d) Chemical reactivity |
These are less reactive metals and form oxides, sulphides, nitrides, hydroxides and halides etc. These also form H2 with acids. They show a lesser tendency for complex formation |
These are highly reactive metals especially when are in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temperature. They show a greater tendency for complex formation. |
Explain:
CrO42– is a strong oxidizing agent while MnO42– is not.
Explain:
Zr and Hf have identical sizes.
Explain:
The lowest oxidation state of manganese is basic while the highest is acidic.
Explain:
Mn(II) shows maximum paramagnetic character amongst the divalent ions of the first transition series.
(i) Electronic configuration:
(ii) Oxidation states:
|
Sc Ti |
V |
Cr |
Mn |
Fe |
Co |
Ni |
Cu |
Zn |
|
|
+2 (+2) +3 +3 +4 |
+2 +3 +4 +5 |
+2 +3 (+4) (+5) +6 |
+2 (+3) +4 (+5) (+6) +7 |
(+1) +2 (+4) (+5) (+6) |
(+1) +2 (+3) (+4) |
+ 1 +2 (+3) (+4) |
+ 1 +2 (+3) |
(+1) +2 |
(iii) Ionization enthalpies
In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4felectrons in the third transition series.
Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2ndtransition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1sttransition series.
|
Element |
Sc |
Ti |
V |
Cr |
Mn |
Fe |
Co |
Ni |
Cu |
Zn |
|
IE, (kJ mol–1) |
631 |
656 |
650 |
652 |
717 |
762 |
758 |
736 |
745 |
906 |
(iv) Atomic Sizes
Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
|
Element |
Sc |
Tc |
V |
Cr |
Mn |
Fe |
Co |
Ni |
Cu |
Zn |
|
Atomic sizes (pm) |
144 |
132 |
122 |
117 |
117 |
117 |
116 |
115 |
117 |
125 |
|
Ions |
Number of 3d electron |
Filling of d- orbital |
|
Ti2+ |
|
|
|
V2+ |
3d3 |
|
|
Cr3+ |
3d5 |
|
|
Mn2+ |
d5 |
|
|
Fe2+ |
3d6 |
|
|
Fe3+ |
3d5 |
|
|
Co2+ |
3d7 |
|
|
Ni2+ |
3d8 |
|
|
Cu2+ |
3d9 |
|
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2ndand 3rdtransition series).
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.
(iii) The enthalpies of atomization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.
(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding (M-M bonding).
(v) The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.
Mention the direct consequence of the following factors on the chemical behaviour of the transition elements.
(i) They have in completely filled d-orbitals in the ground state or in one of the oxidised states of their atoms.
(ii) They contribute more valence electrons per atom in the formation of metallic bonds?
(i) The element which is completely filled d- orbital in the ground state or in the oxidised state can show these properties:
(a) They can form complex compounds.
(b) They show variable oxidation state.
(c) They are used as catalyst.
(ii)
(a) They are hard.
(b) They have high melting and boiling point.
(c) They have high enthalpy of atomization.
(d) They are malleable and ductile.
The transition element exhibit color and magnetic property due to unpaired electron.Co2+ has electronic configuration [Ar] 4s03d7 has seven unpaired electron, Cr3+ has [Ar] 4s03d3 has three unpaired electrons, therefore, they will form coloured solution whereas Sc3+ [Ar] 4s03d0 has no unpaired electron, it will form colourless solution.
Co2+ and Cr3+ will be attracted by magnetic field due to presence of unpaired electrons whereas Sc3+ will be repelled by magnetic field because it does not have unpaired electrons.
What is the general trend of the ionisation enthalpies in the lanthanoids?
Explain the following facts:
Transition metals acts as catalyst.
Transition metals acts as catalyst due to the following reasons:
(i) Their partially empty d-orbitals provide surface area for reactant molecules.
(ii) They combine with reactant molecules to form transition states and lowers their activation energy.
(iii) They show multiple oxidation states and by giving electrons to reactants they form complexes and lower their energies.
Explain the following facts:
Chromium group elements have highest melting points in their respectively series.
Explain the following facts:
Transition metals form coloured complexes.
Give reasons:
Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal.
Give reasons:
Ce4+ is used as an oxidising agent in volumetric analysis.
Ce4+ = [Xe], 4f0, 5d0, 6s0.
Ce4+ has the tendency to accept one electron to get the + 3 oxidtion state, hence Ce4+ is a good oxidising agent.
Give reasons:
Zn2+ salts are white while Cu2+ salts are blue.
Transition element exhibit colour due to d-d transition. d-d tansition is possible only when d- subshell have unpaired electron. In Cu2+ salts (3d9) d–d transition is possible. Therefore Cu2+ salts are coloured.
In Zn2+ salts (3d10) no. d–d transition is possible due to completely filled d-orbitals. Hence Zn2+ salts are white.
Complete the following reaction equations:
(i) Cr2O7 2 + Sn2+ + H+ →
(ii) MnO4– + Fe2+ + H+ →
(i) Cr2O72– + 3Sn2++ 14H+ → 2Cr3+ + 3Sn4+ + 7H2O
(ii) MnO4– + 5Fe2– + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Actinide cations are coloured? Why?
|
Cation of actinide |
U3+ |
Np3+ |
Pu3+ |
Am3+ |
|
Unpaired electrons |
3 |
4 |
5 |
6 |
|
Colour observed |
Red |
Violet |
Black violet |
Pink |
The colour is due to electronic transition within the 5f levels. The electronic transitions of actinides are about ten times more intense than those of lanthanides. The difference is due to difference in 4f and 5f electrons.
Why Am and Cm have exceptional configuration in actinoids?
|
Lanthanides |
Actinides |
|
1. Lanthanides show +3 oxidation state only except in few cases where it is +2 or +4. Oxidation states exhibited by lanthanides is never more than +4. |
1. Actinides show higher oxidation states such as +4, +5, +6, +7 also in addition to +3. |
|
2. Paramagnetic properties of lanthanides can be easily explained.
|
2. Paramagnetic properties are difficult to interpret.
|
|
3. Lanthanides do not form complexes.
|
3. They have a greater tendency to complex formation. Even -complexes are formed by actinides.
|
|
4. Lanthanide compounds are less basic.
|
4. Compounds of these are sufficiently basic.
|
|
5. Lanthanides do not form oxo ions.
|
5. These in higher oxidation states form oxo ions. |
|
6. Except promethium, these are ion-radio-active. |
6. Such as UO22+, NbO+, PuO,2+etc. |
The decrease in the atomic as well as ionic sizes of actinoids along the series is called actinoid contraction. The actinoid in +3 oxidation state shows regular decrease in their atomic size.
It is different from the lanthanoid contraction.
As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
A compound has more thermodynamic stability if the ionization energy of the metal is low. Since the ionization energy of Ni2+ is less than that for Pt2+, therefore Ni2+, compounds are thermodynamically more stable than Pt2+ compounds.
Ni → Ni2+ + 2e– ΔH = 2490 kJ mol–1
(low energy more stable)
Pt → Pt2+ + 2e– ΔH = 2660 kJ mol–1
(high energy less stable)
Pt4+ compounds are stable than Ni4+ compounds because the energy needed to remove 4 electrons in Pt is less than that in Ni.
Pt → pt4+ + 4e– ΔH = 6770 kJ mol–1
(low energy more stable)
Ni → Ni4+ + 4e– ΔH = 8800 kJ mol–1
(high energy, less stable)
The magnetic behaviour of actinoids is more complex and cannot be explained, where for lanthanoids it can be explained and calculated by spin only formula.
Though the magnetic behaviour variation is similar to that of lanthanoids but still lanthanoids have greater magnetic moments.
Compare the chemical reactivity of actinoids with lanthanoids.
Actinoids are far more reactive than lanthanoids especially in the powdered form. They react with non-metals at moderate temperatures whereas lanthanoids needs high temperature.
They are not affected by nitric acid as protective oxide layer prevents their further reaction. They do not react with alkalies. With water they form both oxides and hydrides whereas lanthanoids form hydroxide and H2 gas.
Account for the following facts:
The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
The entrophy is higher if the metal is in liquid state than when it is in solid state.In solid state, the entropy or randomness will be less due to high intermolecular force of attraction while in liquid state, the entropy will be more comparatively. Therefore value of entropy change (ΔS) of the reduction process is more +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ΔG° becomes more on negative side and the reduction becomes easier.
Account for the following facts:
The reduction of Cr2O3 with Al is the thermo-dynamically feasible, yet it does not occur at room temperature.
Account for the following facts:
Pine oil is used in froth floatation process.
What is the trend of ionisation enthalpies of actinoids?
What are the inner transition elements? Decide which of the following atomic numbers at the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104?
Inner transition metal are those element in which the last electron enters the f- orbital.
The element in which the 4f and 5f orbital are progressively filled are called f- block element.
The f-block elements are called inner transition elements. They consist of two series lanthanoids series having elements with atomic number form 57 to 71 and actinoid series having elements with atomic number from 89 to 103.
Hence, the elements with atomic number 59, 95 and 102 are inner transition elements and rest of them are not.
A green chromium compound (A) on fusion with alkali gives a yellow compound (B) which on acidification gives an orange coloured compound (C). ‘C’ on treatment with NH4Cl gives an orange coloured product (D), which on heating decomposes to give back (A). Identify A, B, C and D. Write equations for reactions.
Reaction with mineral acids : Lanthanoids when treated with mineral acids liberates H2 gas as they all have reduction potential of –2.0 to –2.4 V.
2Ln + 6HCl → 2LnCl3 + 3H2
Reaction with water: When treated with water they form consequently hydroxides and liberates H2 gas:
Ln + H2O → Ln(OH)3 + H2
Reaction with sulphur: Heating lanthanoids with sulphur form corresponding sulphides.
2Ln + 3S → Ln2S3
Give Possible reason for the fact that the radii of Mn2+ ions of the first row transition metals. Ti2+ (Z = 22) to Cu2+ (Z = 29) decrease with increasing atomic number.
The radii M2+ ions decrease from Ti2+ Cu2+. This is due to the following reasons:
(i) The nuclear charge increases in going from Ti to Cu.
(ii) The added electrons enter the same inner 3d orbitals. The net effect of these factors is a greater inward pull on the electron shell and therefore reduction in size.
Explain the catalytic action of Fe(III) between the reaction of I– and persulphate ion S2O82–.
The transition metals have great tendency to form complex compounds and they form large number of such compounds due to:
(i) Small sizes of the metal ions.
(ii) High ionic charge in them.
(iii) Availability of partially filled d-orbitals for bond promotion.
The most important reason transition metals are good catalysts is that they can lend electrons or withdraw electrons from the reagent, depending on the nature of the reaction. The ability of transition metals to be in a variety of oxidation states, the ability to interchange between the oxidation states and the ability to form complexes with the reagents and be a good source for electrons make transition metals good catalysts.
What can be inferred from the magnetic moment values of the following complex species?
|
Example |
Magnetic Moment (BM) |
|
K4[Mn(CN)6] |
2.2 |
|
[Fe(H2O)6]2+ |
5.3 |
|
K3 [Mn Cl4] |
5.9 |
(i) K4[Mn(CN)6]
For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,
We can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
Hence, we can say that CN - is a strong field ligand that causes the pairing of electrons.
(ii) [Fe(H2O)6]2+
We can see from the above calculation that the given value is closest to n = 4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.
Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.
(iii) K2[MnCl4]
We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
Hence, we can say that Cl - is a weak field ligand and does not cause the pairing of electrons.
In the crystal lattices of transition metals-small atoms like C, H, B, N etc. are trapped non-stoichiometrically. Such compounds are called interstitial compounds, e.g., TiC, Mn4N, Fe3H, TiH2 etc. Their main characteristics are:
(i) They have high melting points than their metals.
(ii) They become much harder (e.g., borides).
(iii) Metallic conductivity is retained.
(iv) They become chemically inert.
(v) They become less malleable and ductile.
Explain the structures of dichromate and chromate ion.
Why are transition metal fluorides ionic in nature, whereas bromides and chlorides are covalent in nature.
Why are the ionization enthalpies of 5d elements greater than those of 3d and 4d elements?
A blackish brown coloured solid ‘A’, when fused with alkali hydroxide in presence of air produces a dark green coloured compound ‘B’. When electrolytic oxidation in alkaline medium gives a dark purple coloured compound. Identify A, B and C and write the reaction involved.
A balckish brown coloured solid is when it is fused with alkali hydroxide in presence of air it produce Potassium manganate,(B) is
At cathode:
2H+ + 2e– → H2
The overall reaction is
2K2MnO4 + H2O + O → 2KMnO4 + 2KOH
What happens when an acidic solution of the green compound ‘B’ is allowed to stand for sometime? Give the equation involved. What is the type of reaction called?
When an acidic solution of the green compound B(K2MnO4) is allowed to stand for some time potassium manganate is oxidised to potassium permangnate.
The type of reaction is oxidation reaction
3K2MnO4 + 2CO2 → 2K2CO3 + 2MnO2 ↓ + 2KMnO4
Chemical equation :
6Fe2+ + Cr2O72– + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
Chemical equation:
2MnO4– + 3S2O32– + H2O → 2MnO2 + 3SO42– + 3S + 2OH–.
It can be explained on the basis of E° values of Cr3+ | Cr2+ (–0.4 A)
Mn3+ | Mn2+(+ 1.5 V).
On the basis of these values, it is clear that Cr2+ has a greater tendency to oxidise into Cr3+, thus it acts as reducing agent.
Cr2+(aq) → Cr3+(aq) + e–
While Mn3+ has a greater tendency to reduce into Mn2+, thus it acts as oxidising agent.
Mn3+(aq) + e– → Mn2+(aq)
Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
Give reasons:
(i) The d-orbital of Cu is completely filled (3d104s1) still it is considered as transition metal but Zn (3d104s2) is not.
(ii) Zinc salts are colourless at room temperature but nickel salts are coloured.
(iii) The atomic radii of europium and ytterbium do not fit into the concept of lanthanide contraction but ionic radii fit in.
(iv) The atomic radius of Cu is greater than that of Cr but ionic radius of Cr2+ is greater than that of Cu2+.
(v) Cu2+ is colourless while Cu2+ is blue in aqueous solution.
(i) The quantum mechanical definition of a transition element demands that there should be incomplete d-orbital either in the atom or ion of that element. In Cu atom (3d104s1) the 3d10 1s completely filled but in its Cu2+ ion (3d94s0) one of the 3d-orbitals is only half filled. Therefore, copper is a transition elements. On the other hand, in Zn (3d104s2) or Zn2+ (3d104s0) the 3d-orbital is completely filled. Therefore, zinc is not considered as a true transition element, but its electronic configuration does not fit in the quantum mechanical definition.
(ii) Zinc salts are colourless because in Zn2+ ion (3d04s0) there is no unpaired electron in its d-orbital rather 3d orbital is completely filled. Therefore, the visible light is not absorbed but it is transmitted through zinc salts. On the other hand, nickel salts are coloured because in Ni2+ ion (3d84s0) there are 2 unpaired electrons in 3d orbital
Therefore, some light is absorbed for d-d transition and the complementary part is transmitted. The colour of nickel salt is because of the partly transmitted radiation.
(iii) Ca. In europium (Eu = 4f75d0 6s2) the f-orbital is half filled. There are only 2 electrons in 6s-orbitals available for the formation of metallic bond in the crystal lattice of Europium. So, the crystal lattice is less compact. The atomic separation in Eu is large as compared to other metals in which 3 or more electrons can take part in the formation of metallic bonds.
In Ytterbium (Yb = 4f145d06s2) the f-orbitals is completely filled. Therefore, the crystal lattice is less compact and the atomic separation is large. Therefore, the metallic radius of Yb is larger than other members of lanthanide series.
The lanthanide contraction in ionic radii r(M3+) of tri-positive cations is quite smooth because three electrons are removed from the valence orbitals.
(iv) The atomic radius of Cu is greater than that of Cr because in Cu atom (3d104s1) the d-d electron repulsion is large due to pairing of electrons
But in Cr atom (3d34s1) the d-d electron repulsion is minimum as the 3d orbitals are singly occupied
The ionic radius of Cu2+ is less that of Cr+. Because in Cu2+ ion (3d9) the d-d electron repulsion is reduced due to unpairing of electron spins and the other electrons are attracted by 29 protons of the nucleus.
Therefore, the attraction of nucleus on the outer electron has become strong and d-electrons are pulled inward.
In Cr2+ the 3d4 electrons are still unpaired and there are attracted by only 24 protons of nucleus.
(v) The ion which has unpaired electron is coloured in aqueous solution. Thus Cu+(ion) [s22s22p63s23p63d104s0] does not have any unpaired electron, hence its aqueous solution is colourless.
Cu2+ (ion) [1s22s22p63s23p63d94s0] has a unpaired electron in one of the 3d orbitals, therefore its aqueous solution is coloured (blue).
Give reasons:
(i) Scandium (At. No. 21) is a transition element but calcium (At. No. 20) is not.
(ii) The transition metal ions such as Cu+ Ag+ and Sc3+ are colourless.
(iii) The melting points of transition metals of 3d series are much higher than the melting point of zinc of the same period.
(iv) The radius of Fe2+ is less than that of Mn2+.
(v) Actinides cations are coloured.
(i) An element is called a transition element if its atom or ion has incomplete i.e., partly filled d-orbital. In scandium (1s22s22p63s23p63d14s2) the penultimate 3d orbital is partly filled and it is a transition element. In calcium (1s22s22p63s2 3p63d04s2) the 3d orbital does not contain any electron, therefore, it is not a transition element.
(ii) An ion appears colourless if it does not contain any unpaired electron in its d-orbitals.
(a) Cu+(3d104s0) is colourless because it does not contain any unpaired electron and 3d10orbital is completely filled.
(b) Ag+(14d105s10) is colourless because it does not contain any unpaired electron in its 4d-orbital and 4d10 orbital is completely filled.
(c) Sc3+ (3d04s0) is colourless because it does not contain any unpaired electron in its 3d orbital and 3d-orbital is empty.
(iii) The high values of melting points indicate that the atoms in transition metals are held together by strong metallic bonds. This is attributed to the availability of vacant d-orbitals in transition metals. This facilitates the formation of strong metallic bonds by delocalization of electron clouds in the metallic crystal lattice. In the 3d-series, Zinc has exceptionally low melting point and low boiling point, because it has completely filled d-orbitals. Therefore, the atoms in the crystal lattice of zinc are held less strongly as compared to the other members in their respective crystal lattices.
(iv) The radius of Fe2+ (1s22s22p63s23p63d64s0) is less than that of Mn2+(1s22s22p63s23p63d54s0) because in Fe2+ the electrostatic attraction between its nucleus and the outer electron clouds is stronger than that in case of Mn2+. Strong attraction causes a decrease in the distance between the nucleus and outer electron clouds.
(v) The cations of actinides are coloured because of the presence of unpaired electrons. The colour observed depends on the number of unpaired electrons. The colour is due to electronic transition within the 5f-levels.
Paramagnetic: A material is said to be paramagnetic substance when it is attracted by a magnet when kept in a magnetic field. Paramagnetism occurs due to the presence of unpaired electrons.
Ferromagnetic: A material is said to be ferromagnetic substance when it exhibits permanent magnetism even when the magnetic field is removed.
The number of unpaired electrons present in bivalent ions of transition metal series from
Z = 22 to Z = 29 is given below:
From this it is clear that the paramagnetism character of bivalent ion of transition series varies as Mn2+ > Cr2+ = Fe2+ > V2+ = Co2+ > Ti2+ = Ni2+ > Cu2+
Inner transition elements are called actinides. This includes elements Actinium to Lawrencium. Their general electronic configuration may be represented as:
(n – 2) f1–14 (n – 1) d0 – 1ns2
In what way are the observed oxidation states of the lanthanides related to their electronic configurations?
The lanthanum (At. No. 57) has the outer electronic configuration 5d1 6s2 and has the oxidation number 3.
Similarly next element cerium (At. No. 58) has the outer electronic configuration 4f15d16s2 and has the oxidation number 4. Thus we see that the oxidation number of lanthanides are well related to electronic configurations of lanthanides.
Describe the general characteristics of the transition elements with special reference to their tendency to:
(i) Exhibit paramagnetism.
(ii) Form complex compounds.
(iii) Their catalytic behaviour.
(i)
Transition elements show paramagnetism. Paramagnetism is due to the presence of unpaired electrons in the d-orbitals of transition metal atoms, ions or molecules. The greater the number of unpaired electrons, the greater is the paramagnetism.
(ii)
Transition elements have strong tendency to form complex ion. This is because transition elements form small, highly charged ions, which have vacant (n – 1) d-orbitals of approximately the appropriate energy to accept lone pairs of electrons donated by other groups or molecules, such as cyanide ion, water and ammonia molecules.
(iii)
Most of the transition elements act as catalyst e.g., finely divided nickel is used in the hydrogenation of vegetable oils, iron in the manufacture of ammonia by Haber’s Process. This is because transition elements form transient intermediate complexes utilising empty d-orbitals. Consequently, low energy path-ways for slow reactions are provided which increase the rate of the reaction.
Why are transition metals able to form alloys?
An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics
of transition metals, alloys are readily formed by these metals.
How is the variability in the oxidation states of transition metals different from that of the nontransition metals? Illustrate with examples.
Absorption bonds of lanthanoids narrow because of the shielding of the 4f orbitals by the filled 5s2 and 5p6 sub-shells.
Assign reasons for the following:
The enthalpies of atomisation of transition metals are high.
Transition metal has high heat of atomisation due to presence of strong metallic bond which arises due to presence of unpaired electron in the (n - 1) d subshell. This is because the atoms in these elements are closely packed and held together by strong metallic bonds. The metallic bond is formed as a result of the interaction of electrons in the outermost shell. Greater the number of valence electrons, stronger is the metallic bond.
Assign reasons for the following:
Transition metals and their many compounds act as good catalyst.
Transition metals acts as catalyst due to the following reasons:
(i) Their partially empty d-orbitals provide surface area for reactant molecules.
(ii) They combine with reactant molecules to form transition states and lowers their activation energy.
(iii) They show multiple oxidation states and by giving electrons to reactants they form complexes and lower their energies.0
V2O5 (in contact process for manufacture of H2SO4), finely divided iron (in Haeber’s process for NH3 manufacture) and Ni (in catalytic hydrogenation) are examples of their good catalytic activities.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
In actinoids, 5f orbitals are filled. These 5forbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contaction in actinoids is greater as compared to that in lanthanoids.
Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as a transition element.
Electronic configuration of Scandium is 1s2 2s2 2p6 3s2 3p6 3d1 4s2.
On the basis of incompletely filled 3d orbital in case of scandium atom in its ground state (3d1), it is regarded as a transition element.
What may be the possible oxidation states of the transition metals with the following d electronic configurations in the ground state of their atoms: 3d34s2,3d54s2 and 3d64s2. Indicative relative stability of oxidation states in each case.
|
(a) Configuration 3d34s2 3d5 4s2 3d6 4s2 |
(a) Stable oxidation state +3, +3 +2, +7 +2, +8 |
The stability of any oxidation state of any element generally depends upon the electronic configuration of that particular ion. If any ion has stable configuration (d5, d10 etc.), the particular oxidation state will be more stable.
Colour exhibited by the transition element cations in the _________ of the absorbed wavelength.
The general configuration of the last two energy levels of d-block elements is __________.
A.
All the transition elements have unfilled d-orbitals.B.
Iron belongs to 3d series of transition elements.C.
Ferric iron has a half-filled 3d subshell.D.
Steel is an interstitial compound.E.
Zn, Cd, Hg are not considered as true transition metals as their d subshell is full.A.
The general electronic configuration of d-block elements is (n – 1)1– 8 ns1–2.B.
Pd is called a transition element.C.
Zn, Cd and Hg are harder and non-volatile metals.D.
The E° value of Zn2+/Zn and Mn2+/Mn are quite high.E.
The common symbol used for lanthanoids is Lu.Cr has electronic configuration
B.
3s23p63d54s1All actinides
B.
Exhibit variable valencyWith the increase in atomic number along a transition series the nuclear charge increase which tends to decrease the size of atom.
But te addition of electron in the d- subshell increase the screening effect which counter balances the increased nuclear charge.
Hence along a transition series the atomic radius does not change very much.
Why is radius of Fe2+ less than that of Mn2+?
Fe2+ has lesser radius than Mn2+ because the effective nuclear charge is more in Fe3+ as compared to mn2+.
Write outer electronic configuration of lanthanoids.
Outer electronic configuration of lanthanoids is 4f0 5d1 6s2
Write the general electronic configuration of the transition elements of d-block elements.
The electronic configuration of the transition element (n-1)d1–10ns1–2
What happens when chromates are kept in acidic solution and dichromate in alkaline solution?
What type of magnetic behaviour is shown by the molecules containing unpaired electrons?
Paramagnetism
As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction. Consequences of lanthanoid contraction.
(i) There is similarity in the properties of second and third transition series.
(ii) Separation of lanthanoids not is possible due to lanthanide contraction.
(iii) It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)
The overall reaction is
6I- + Cr2O7 2- + 14H+ ----> 3 I2 + 2Cr3+ + 7H2O
Which out of V(IV) and V(V) is paramagnetic and why?
Vanadium has its electronic configuration as [Ar] 3d3 4s2.
In case of V (v) has no unpaired electron.
In case of V(iv) has one unpaired electron 3d1, Hence V(iv) is paramagnetic.
Why does a transition series contain only ten elements?
The transition elements are also known as the d-block elements, because while the outermost level contains at most two electrons, their next to outermost main levels have incompletely filled d sub-orbitals, which are filled-up progressively on going across the periodic table from 8to 18 electrons.
General electronic configuration of d - block element is (n – 1) d1–10 ns1-2.
Why are Cd2+ salts white?
Transition element exhibit colour to d-d transition.
d-d transition will be possible only when d- subshell is not fully filled.
The electronic configuration of Cd2+ is 5d10 .
It means it has full filled d- subshell. thus there will be not d-d transition take place, so Cd2+ is white.
Which one of the two Co or Zn would be repelled by a magnetic field?
Electronic configuration of Co is [Ar] 4s2 3d7 And the electronic configuration of Zn [Ar] 4s2 3d10. Since zinc has no unpaired electron electron so it will be repelled by magnetic field.
Why is electronic configuration 1s22s22p63s23p64s23d4 not correct for ground state of Cr (24)?
The electronic configuration of a Chromium is 1s22s22p63s23p64s23d4 not correct for ground state.
Because its unstable and every metal tends to be stable thus the most stable configuration is 1s22s22p63s23p64s1 3d5
How is the magnetic moment of a species related to number of unpaired electrons?
Any ion molecule or atom that has unpaired electron exhibit the stronger magnetic property known as paramagnetism.
Because the magnetic effect of all the electronic with spin'up' are not canceled by the opposed effect of an equal number of electron with spin down the species is attracted to magnetic field.
Which property of the transition metals makes them good catalyst?
Transition metals acts as catalyst due to the following reasons:
(i) Their partially empty d-orbitals provide surface area for reactant molecules.
(ii) They combine with reactant molecules to form transition states and lowers their activation energy.
(iii) They show multiple oxidation states and by giving electrons to reactants they form complexes and lower their energies.
What is the effect of increasing pH on K2Cr2O7 solution?
The potassium dichromate (K2Cr2O7), solution contain dichromate (Cr2O72- ion . The ion is less basic and so increase in PH increase in the basicity of the compound.
In simple words, by decreasing hydrogen ion concentration the dichromate ion is converted to chromate ion or vice versa.
Cr2O72- +2OH- à 2CrO42- +H2O
2CrO42- +2H+ à Cr2O72- +H2O
Which is more basic, La(OH)3 or Lu(OH)3? Why?
Why is Ce4+ in aqueous solution is good oxidising agent.
Why do Zr and Hf exhibit similar properties?
Zr and Hf exhibit similar properties due to lanthanoid contraction. Electrons present in f subshell didn't do good shielding due to which with the increasing atomic number or increasing effective nuclear charge size gets constricted and size of Halfenium and Zirconium becomes almost equal.
Write any two uses of pyrophric alloys.
1)Pyrophoric alloys emit sparks when struck. Hence they are used in making flints for lighter.
2) It used as sparking mechansim in lighter and vorious toys.
Why do transition metals show catalytic properties?
The most important reason transition metals are good catalysts is that they can lend electrons or withdraw electrons from the reagent, depending on the nature of the reaction. The ability of transition metals to be in a variety of oxidation states, the ability to interchange between the oxidation states and the ability to form complexes with the reagents and be a good source for electrons make transition metals good catalysts
Transition metals and their compounds are extensively used as catalysts due to the following reasons:
(i) Their partially empty d-orbitals provide surface area for the reactant molecules.
(ii) They combine with reactant molecules to form transition states and lowers their activation energy.
(iii) They show multiple oxidation states and by giving electrons to reactants they form complexes and lowers their energies.
Why do transition metals display variable oxidation series?
Transition metals have electrons of similar energy in both the 3d and 4s levels. This means that one particular element can form ions of roughly the same stability by losing different numbers of electrons. Thus, all transition metals from titanium to copper can exhibit two or more oxidation states in their compounds.
Oxidation states of some Transition Metals:
Titanium- +2, +3, +4
Vanadium- +2, +3, +4, +5
Chromium- +2, +3, +6
Manganese- +2, +3, +4, +5, +6, +7
Iron- +2, +3
Cobalt- +2, +3
Nickel- +2, +3, +4
Copper- +1, +2
When Transition Metals form positive ions they loose their electrons from the 4s sub-shell first, then the 3d sub-shell
What are transition elements? Discuss their magnetic properties.
The transition elements are also known as the d-block elements, because while the outermost level contains at most two electrons, their next to outermost main levels have incompletely filled d sub-orbitals, which are filled-up progressively on going across the periodic table from 8to 18 electrons.
General electronic configuration of d - block element is (n – 1) d1–10 ns1-2.
Magnetic property depend on the number of unpaired electron present in it. In transition element magnetic property increase upto Mn and decrease, because first number of electron increase upto Mn and after that pairing occur and thus number of unpaired electron decrease and hence magnetic property decrease.
Describe the oxidation states and variable valencies of elements of 3d series.
The ability of the transition metals to exhibit variable valency is generally attributed to the availability of more electrons in the (n-1)d orbitals which are closer to the outermost ns orbital in energy levels.
Transition metals have electrons of similar energy in both the 3d and 4s levels. This means that one particular element can form ions of roughly the same stability by losing different numbers of electrons. Thus, all transition metals from titanium to copper can exhibit two or more oxidation states in their compounds.
Oxidation states of some Transition Metals:
Titanium- +2, +3, +4
Vanadium- +2, +3, +4, +5
Chromium- +2, +3, +6
Manganese- +2, +3, +4, +5, +6, +7
Iron- +2, +3
Cobalt- +2, +3
Nickel- +2, +3, +4
Copper- +1, +2
When Transition Metals form positive ions they loose their electrons from the 4s sub-shell first, then the 3d sub-shell
Assigns reasons for the following:
(i) Transition metals exhibit enthalpy of atomisation.
(ii) Transition metals form interstitial compounds.
(i) Transition metal has high heat of atomisation due to presence of strong metallic bond which arises due to presence of unpaired electron in the (n - 1) d subshell. This is because the atoms in these elements are closely packed and held together by strong metallic bonds. The metallic bond is formed as a result of the interaction of electrons in the outermost shell. Greater the number of valence electrons, stronger is the metallic bond.
ii)Transition metals often have close packed (ccp, hcp) lattices that have N oct interstices (holes) for N close packed metal atoms. Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent, for example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc.
Explain why:
(i) Transition elements are metals?
(ii) Lanthanides are uniformly trivalent?
i) As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat. The interesting thing about transition metals is that their valence electrons, or the electrons they use to combine with other elements, are present in more than one shell.
Thus transition element are metals.
ii) The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.
Write the preparation of :
(i) Potassium dichromate from chromite.
(ii) KMnO4 from pyrolusite.
What is lanthanide contraction? How does it occur? What are its consequences?
A group of fourteen elements following lanthanum i.e. from 58Ce to 71Lu placed in 6th period of long form of periodic table is known as lanthanoids (or lanthanide series). These fourteen elements are represented by common general symbol ‘Ln’. In these elements, the last electron enters the 4f-subshells (pre pen ultimate shell). It may be noted that atoms of these elements have electronic configuration with 6s2 common but with variable occupancy of 4f level. However, the electronic configuration of all the tri positive ions (the most stable oxidation state of all lanthanoids) are of the form 4f n (n = 1 to 14 with increasing atomic number). These elements constitute one of the two series of inner transition elements or f-block.
Lanthanoid contraction: In the lanthanoide series with the increase in atomic number, atomic radii and ionic radii decrease from one element to the other, but this decrease is very small. The regular small decrease in atomic radii and ionic radii of lanthanides with increasing atomic number along the series is called lanthanoid contraction.
Cause of lanthanoid contraction: When one moves from 58Ce to 71Lu along the lanthanide series nuclear charge goes on increasing by one unit every time. Simultaneously an electron is also added which enters to the inner f subshell. The shielding effect of f-orbitals in very poor due to their diffused shape. It results in the stronger force of nuclear attraction of the 4f electrons and the outer electrons causing decrease in size.
Consequences of lanthanoid contraction:
(i) Similarly in the properties of elements of second and third transition series e.g. Sr and Hf; Nb and Ta; Mo and W. This resemblance is due to the similarity in size due to the presence of lanthanoids in berween.
(ii) Similarity among lanthanoids: Due to the very small change in sizes, all the lanthanoids resemble one another in chemical properties.
(iii) Decrease in basicity: With the decrease in ionic radii, covalent character of their hydroxides goes on increasing from Ce(OH)3 to Lu(OH)3 and so base strength goes on decreasing.
How do you explain the paramagnetism and colour exhibited by compounds of transition metals?
Transition element exhibit colour due to d-d transition, d-d transition is possible only when d- subshell have unpaired electron. Thus transition metal compounds are shows coloure because of the presence of one or more unpaired electrons in the d-orbital of the metal ion. A compound of transition metal appears coloured when one frequency of definite energy is absorbed (E = hv) for d-d transition of the unpaired electron and other frequencies of visible region are transmitted. The colour of the compound corresponds to the frequency of the transmitted light. Thus, the observed colour (frequency) is complementary to the absorbed frequency.
Write the electronic configuration of 25Mn2+.
The electronic configuration of Mn is 4s2 5d10.
Mn2+ electronic configuration is 4s0 5d10.
Vanadium ion has a magnetic moment of 1.73 BM. What is the symbol of the ion? (Atomic number of V = 23)
Magnetic moment can be calculated by the using formula
As, we have given magnetic moment 1.73B.M so
1.73 =
taking both side saqure
(1.73)2 =( )2
2.99 =n(n+2)
2.99 =n2 +2n
n2 +2n-2.99 =0
solving above equation, we get
n=1
Thus, Vanadium ion must have one unpaired electron .
electronic configuration is [Ar] 4s2 3d1.
What is the highest oxidation state shown by transition metals?
Why are the compounds of transition elements usually coloured, both in the solid state and in aqueous solution?
The compounds of transition element are usually coloured both in solid state and in aqueous solution. The colour of these complex is due to adsorption of some radiation form visible light which is used in promoting an electron form one of the d- shells to another.
Why do transition elements form complexes?
Transition- metal ions form coordinationcomplexes
because they have empty valence-shell orbitals that can accept pairs of electrons from a Lewis base. Ligands must therefore be Lewis bases: They must contain at least one pair of nonbonding electrons that can be donated to a metal ion.
Many transition metals are paramagnetic. Why?
Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum
also always transition metals have unpaired d-electron.
Distinguish between paramagnetism and ferro-magnetism. How does ferro magnetism arise?
Paramagnetism: Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired electrons which are attracted by the magnetic field. O2, Cu2+, Fe3+, Cr3+ are some examples of such substances.
Ferromagnetism: A few substances like iron, cobalt, nickel, gadolinium and CrO2 are attracted very strongly by a magnetic field. Such substances are called ferromagnetic substances.
Besides strong attractions, these substances can be permanently magnetised. In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called
domains. Thus, each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance the domains
are randomly oriented and their magnetic moments get cancelled. When the substance is placed in a magnetic field all the domains
get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.
What are actinoids? Mention two characteristics of actinoids.
The actinoid metals are all silvery in appearance but display a variety of structures. Actinides consist of a family of 14 elements that range in atomic numbers from 89 to 103. It may be represented by [Rn]5fx 6dy 7s2,where x varies from 0 to 14 and y = 0 or 1.
Two characteristic of actinoids:
1) These are highly reactive metals especially when are in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temperature. They show a greater tendency for complex formation.
2) It Show higher oxidation states such as +4, +5, +6, +7 also in addition to +3.
(i) Breifly explain the physical properties of lanthanoids.
(ii) The chemical reactivity of lanthanoids resembles to which other elements of the periodic table.
All the lanthanoids are silvery white soft metals and tarnish rapidly in air. The hardness increases with increasing atomic number, samarium being
steel hard. Their melting points range between 1000 to 1200 K but samarium melts at 1623 K. They have typical metallic structure and are
good conductors of heat and electricity. Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm.
The chemical reactivity of the starting lanthanoids resmbles calcium(due to similar 1st and 2nd ionisation enthlpies) but latter lanthanoids resmbles Al (due to ability of showing and similarity in
How does the acidified KMnO4 react with the following:
(a) Iron(II) solution, (b)SO32–, (c) Iodide (I–) ion, (d) oxalic acid, (e) NO2– (Nitrite ion).
a) Fe2+ ion (green) is converted to Fe3+ (yellow):
5Fe2+ + MnO4– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+
b) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:
5SO32– + 2MnO4– + 6H+ ——> 2Mn2+ + 3H2O + 5SO42–
c) Iodine is liberated from potassium iodide :
10I– + 2MnO4– + 16H+ ——> 2Mn2+ + 8H2O + 5I2
d) Oxalate ion or oxalic acid is oxidised at 333 K:
5C2O42– + 2MnO4– + 16H+ ——> 2Mn2+ + 8H2O + 10CO2
e) Nitrite is oxidised to nitrate:
5NO2– + 2MnO4– + 6H+ ——> 2Mn2+ + 5NO3– + 3H2O
Explain the oxidising property of KMnO4 in neutral medium with example.
The oxidising property of KMnO4 in neutral medium
(a) A notable reaction is the oxidation of iodide to iodate:
2MnO4– + H2O + I– ——> 2MnO2 + 2OH– + IO3–
(b) Thiosulphate is oxidised almost quantitatively to sulphate:
8MnO4– + 3S2O32– + H2O ——> 8MnO2 + 6SO42– + 2OH–
(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:
2MnO4– + 3Mn2+ + 2H2O ——> 5MnO2 + 4H+
Explain any one of following statement:
(i) The transition metals are well known for the formation of interstitial compounds.
(ii) The largest number of oxidation states are exhibited by mangnese in the first series of transition elements.
i) The transition metals appear to be unique in their ability to react with small atoms of non-metal to give interstitial compound. These metals form various interstitial compound in which small non- metal like H,C,B,N and He occupy the empty space in their lattice and also form bonds with them.
These new compounds are known as hydrides, carbides, borides, nitrides and halids respectively.
These interstial compound are hard and rigid. for example steel and cast iron which are interstitial compound of Fe and C are hard.
ii) Mn (Z = 25) = 3d5 4s2
Mn+2 is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn(s), Mn(II) as MnO, Mn(II,III) as Mn3O4, Mn(IV) as MnO2, or manganese dioxide, Mn(VII) in the permanganate ion MnO4-, and so on. Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
Answer the following questions:
(i) Which element in the first series of transition elements does not exhibit variable oxidation states and why?
(ii) Why do actinoids in general exhibit a greater range of oxidation states than the Lanthanoids?
i) The ability of the transition metals to exhibit variable valency is generally attributed to the availability of more electrons in the (n-1)d orbitals which are closer to the outermost ns orbital in energy levels.
Scandium and zinc does not show variable oxidation states.
ii) Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
i)Describe how potassium permanganate is made from pyrolusite. Write the chemical equations for the involved reactions.
ii) Describe with an example each of the oxidising actions of permanganate ion in alkaline and acidic media. What acid and alkali are usually used?
Pottassium Permanganate (KMnO4) is prepared from Pyrolusite ore (MnO2). The finely powdered Pyrolusite ore (MnO2) is fused with an alkali metal hydroxide like KOH in the presensce of air or an oxidizing agent like KNO3 to give the dark green potassium Manganate (K2MnO4). Potassium manganate disproportionate in a neutral or acidic solution to give potassium permanganate.
2 MnO2 + 4 KOH + O2 ----> 2K2MnO4 + 2H2O
3 MnO42- + 4H+ -----> 2MnO4- + MnO2 + 2H2O
Commercially potassium permanganate is prepared by the alkaline oxidative fusion ofPyrolusite ore (MnO2) followed by the electrolytic oxidation of manganate (4) ion.
2MnO2 + 4KOH + O2 ---> 2K2MnO4 + 2H2O
MnO42- ------(electrolytic oxidation)----> MnO4- + e-
ii) Equation Reduction of KMnO4 in acidic medium:
8H(+) + MnO4(2-) + 5e(-) ----------> Mn(2+) + 4H2O
In basic medium:
MnO4(-) + e(-) ------------> MnO4(2-)
Thus, you can see that oxidizing effect of KMnO4 is maximum in acidic medium and least in basic medium as in acidic medium the reduction in oxidation state of Mn is max while it is the least in basic medium.
H2SO4 and NaOH is used as acid and alkali.
Describe how potassium dichromate is made from chromate ore and give the equations for the chemical reactions involved.
Write balanced ionic equations for reacting ions to represent the action of acidified potassium dichromate solution on:
(a) Potassium iodide solution.
(b) Acidified ferrous sulphate solution. Write two uses of potassium dichromate.
1)Potassium dichromate (K2Cr2O7) is prepared from chromite ore FeCr2O4. The chromite ore is fused with sodium or potassium carbonate in free access of air.
4FeCr2O4 + 8Na2CO3 + 7O2 -------> 8Na2CrO4 + 2FeO3 + 8CO2
The yellow solution of sodium chromate is filtered and acidified with sulfuric acid to give a solution from which orange sodium dichromate , Na2Cr2O7 .2H2O can be crystallized.
2Na2CrO4 + 2H+ ---------> Na2Cr2O7 + 2Na + H2O
Sodium dichromate is more soluble than potassium dichromate. Hence sodium dichromate when fused with KCl forms orange crystals of potassium dichromate.
Na2Cr2O7 + 2KCl --------> K2Cr2O7 + 2NaCl
a) Iodine is liberated from potassium iodide :
10I– + 2MnO4– + 16H+ ——> 2Mn2+ + 8H2O + 5I2
b) Fe2+ ion (green) is converted to Fe3+ (yellow):
5Fe2+ + MnO4– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+
uses of potassium dichromate
an oxidant, potassium permanganate can act as an antiseptic.
(a) Comment on the following tendencies of transition elements of first series:
(i) They exhibit variable oxidation states.
(ii) They easily form alloys.
(iii) They often act as catalyst.
(b) State the different characteristic of actinoids and lanthanoids which places them in f-block elements.
(c) How many unpaired electrons are there in Ni2+(aq)?
i) The ability of the transition metals to exhibit variable valency is generally attributed to the availability of more electrons in the (n-1)d orbitals which are closer to the outermost ns orbital in energy levels.
ii) An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
iii) Transition metals acts as catalyst due to the following reasons:
(a) Their partially empty d-orbitals provide surface area for reactant molecules.
(b) They combine with reactant molecules to form transition states and lowers their activation energy.
(c) They show multiple oxidation states and by giving electrons to reactants they form complexes and lower their energies.
b)
|
Characteristics |
Lanthanoids |
Actinides |
|
(a) Electronic configuration |
It may be represented by [Xe]4fx 5dy 6s2, where x varies from 0 to 14 and y = 0 or 1. |
It may be represented by [Rn]5fx 6dy 7s2,where x varies from 0 to 14 and y = 0 or 1. |
|
(b) Oxidation state
|
Show +3 oxidation state only except in few cases where it is +2 or +4. They never show more than +4 state. |
Show higher oxidation states such as +4, +5, +6, +7 also in addition to +3.
|
|
(c) atomic and ionic sizes
|
The ionic radii of M3+ ions in lanthanoids series show a regular decrease in size of ions with increase in atomic number. This decrease is known as lanthanoid contraction. |
There is a greater and gradual decrease in the size of atoms or M3+ ions across the series. This greater decrease is known as actinoid contraction.
|
|
(d) Chemical reactivity |
These are less reactive metals and form oxides, sulphides, nitrides, hydroxides and halides etc. These also form H2 with acids. They show a lesser tendency for complex formation |
These are highly reactive metals especially when are in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temperature. They show a greater tendency for complex formation. |
Account for the following:
(a) Transition elements have high boiling point and high enthalpy of atomisation.
(b) Zn, Cd, Hg are normally not regarded as transition elements.
(c) Transition metals generally form alloys with other transition metals.
(d) Transition metals form a number of interstitial compounds.
(e) Ni2+ compounds are thermodynamically more stable than Pt2+ compounds.
a) The melting points of transition elements are high due to the presence of strong intermetallic bonds (formed by valence electrons) and covalent bonds (formed due to d-d overlapping of unpaired d-electrons). Transition metal has high heat of atomisation due to presence of strong metallic bond which arises due to presence of unpaired electron in the (n - 1) d subshell. This is because the atoms in these elements are closely packed and held together by strong metallic bonds. The metallic bond is formed as a result of the interaction of electrons in the outermost shell. Greater the number of valence electrons, stronger is the metallic bond.
b) The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.
c) An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.
d) The transition metals appear to be unique in their ability to react with small atoms of non- metal to give interstitial compound. These metals form various interstitial compound in which small non- metal like H,C,B,N and He occupy the empty space in their lattice and also form bond with them.
e) Ni2+ is smaller in size and thus has smaller the ionization enthalpy of a metal greater thermodyanmic stability of its compound.
(i) What are inner transition elements? Write their general electronic configuration.
(ii) The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements?
i) The last electron enters into (n - 2)f - orbital which is inner to the penultimate shell (For d-block elements), they are also called inner transition elements. Hence, the general electronic configuration for these elements is (n-2)f 1-14, (n-1)d0-1, ns2.
ii) Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
Describe the general characteristics of the transition elements with special reference to their tendency to
(i) exhibit paramagnetism, (ii) form complex compounds, (iii) their catalytic behaviour.
i) Paramagnetism is due to presence of one or more unpaired electrons which are attracted by the magnetic. The electronic configuration of transition element is (n-1)d1–10ns1–2. Thus in transition element number of unpaired electron exist. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum
also always transition metals have unpaired d-electron.
ii) The transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d orbitals for bond formation. A few examples are: [Fe(CN)6]3–, [Fe(CN)6]4–,[Cu(NH3)4]2+ and [PtCl4]2–.
iii) The most important reason transition metals are good catalysts is that they can lend electrons or withdraw electrons from the reagent, depending on the nature of the reaction. The ability of transition metals to be in a variety of oxidation states, the ability to interchange between the oxidation states and the ability to form complexes with the reagents and be a good source for electrons make transition metals good catalysts.
What are the transition elements? Write two characteristics of the transition elements ?
Elements that possess incompletely filled d-orbitals either in their ground state or in any of their oxidation states are known as transition elements. The name transition element given to the elements of d-block is only because of their position between s-block and p-block elements.
Characteristics of transition elements are as follows:
1. They show paramagnetic behaviour.
2. They show variable oxidation states.
3. They exhibit catalytic properties.
4. They generally form complex compounds.
5. They generally form coloured compounds.
How would you account for the following?
Actinoid contraction is greater than lanthanoid contraction.
In actinoids, 5 f-orbitals are filled. These 5 f-orbitals have a poorer shielding effect than 4 f-orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more than that experienced by electrons in valence shells in case of lanthanoids. Hence, the size contraction in actinoids is greater than that in lanthanoids.
How would you account for the following?
Transition metals form coloured compounds.
In the presence of ligands, the d-orbitals of transition metal ions split up into two sets of orbitals having different energies. Thus, the transition of electrons takes place from one set to another. The energy required for these transitions is quite less and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.
Which metal in the first transition series (3d series) exhibits + 1 oxidation state most frequently and why?
Cu is the only metal in the first transition series (3d series) which shows +1 oxidation state most frequently. This is because the electronic configuration of Cu is 3d10 4s1 and after losing one electron it acquires a stable 3d10 configuration.
Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+ (Atomic number Sc = 21, V = 23, Ti = 22, Mn = 25)
The colour of cations is dependent on the number of unpaired electrons present in d-orbital. The electronic configuration of the following cations is as follows:
Sc (Atomic number 21) = 3d1 4s2 and Sc3+ = 3d0 4s0. As d-orbital is empty, it is colourless.
V (Atomic number 23) = 3d3 4s2 and V3+ = 3d2 4s0. As d-orbital is having 2 unpaired electrons, it undergoes d-d transition and shows green colour
Ti = (Atomic number 22) = 3d2 4s2 and Ti4+ = 3d0 4s0. As d-orbital is empty, it is colourless
Mn = (Atomic number 25) = 3d5 4s2 and Mn2+ = 3d5 4s0. As d-orbital is having 5 unpaired electrons, it shows pink color.
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst(i) The variable oxidation states of transition elements are due to the participation of ns and (n-1) d-electrons in bonding. Lower oxidation state is exhibited when ns-electrons take part in bonding. Higher oxidation states are exhibited when (n-1) d-electrons take part in bonding.
(ii) This is because the atomic radii of 4d and 5d transition elements are nearly same. This similarity in size is a consequence of lanthanide contraction. Because of this lanthanide contraction, the radii of Hf becomes nearly equal to that of Zr.
(iii) Transition elements act as a good catalyst in chemical reaction because they can lend electrons or withdraw electrons from the reagent, depending on the nature of the reaction. The ability of transition metals to be in a variety of oxidation states, the ability to interchange between the oxidation states and the ability to form complexes with the reagents and be a good source for electrons make transition metals good catalysts.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved.
The phenomenon of catalysis can be divided into two groups.
Homogeneous catalyst: In the homogeneous catalyst is present in the same phase as the reactant, it is called a homogeneous catalyst and this type of catalyst is called is called homogeneous catalysis.
For example: 2SO2(g) + O2(g) 
2SO3(g)
Heterogeneous catalyst: In the heterogeneous catalyst present in a different phase as a reactant, it is called a heterogeneous and this type of catalyst is called heterogeneous catalysis.
For example: N2(g) + 3H2(g) 
2NH3(g)
Explain the following giving an appropriate reason in each case.
O2 and F2 both stabilise higher oxidation states of metals but O2 exceeds F2 in doing so.
O2 and F2 both stabilise high oxidation states with metal but the tendency is greater in oxygen than fluorine. It is because O2 bears -2 charges for each oxygen atom while F2 bears only -1 for each atom thus the force of attraction between the metal atom and O2-ion is greater than the force between the same metal atom and F- ion. Thus, oxygen has the ability to form multiple bonds with transition element whereas fluorine does not have the ability to form multiple bonds with transition elements. Hence, O2 gets the higher state of metals.
Explain the following giving an appropriate reason in each case.
Structures of Xenon fluorides cannot be explained by Valence Bond approach.
According to the valence bond approach, covalent bonds are formed by the overlapping of the half-filled atomic orbital. But xenon has a fully filled electronic configuration. Hence the structure of xenon fluorides cannot be explained by VBT.
For example: In case of XeF2
Hybridization is:
Total, valence electron =8 for Xe
Monovalent atom F=2
So, total hybrid orbitals =1/2 [valence electron + monovalent atom + cation + anion] as no cation and anion so term is zero
So hybrid orbital are ½[8+2+0+0] = 5
So hybridization = sp3d
According to sp3d hybridization, the structure should be Trigonal bipyramidal, but the actual structure is linear. So, VBT fails to answer this however VSPER theory explains the liner shape.
How would you account for the following?
i) Many of the transition elements are known to form interstitial compounds.
ii)The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group member of the second (4d) series.
iii) Lanthanoids from primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even + 6 being typical.
(i) Formation of the interstitial compounds: Transition elements exist in CCP and HCP structure which are known to possess vacant position (Holes),such that transition elements form a few interstitial compounds with elements possessing small atomic radii, like hydrogen, carbon boron, and nitrogen. Small atoms of these type of elements get entrapped in between the void spaces (called as interstices) of the metal lattice. Some of the characteristics of the interstitial compound. These compounds show basically the same chemical properties as the parent metals but vary in the physical properties such as density and hardness. For example, the best known is tungsten carbide WC that is extremely hard and used in steel cutting tools, armour and jewellery.
Explanation: Interstitial compounds are hard and dense in nature. This is because; the smaller atoms of the lighter elements occupy the interstices in the lattice, leading to the much closely packed structure. Because of greater electronic interactions, the strength of metallic bonds also increases.
(ii) Metallic radii of third (5d) series of transition metals are virtually same as those of (4d) series because of the lanthanoid contraction.
Lanthanoid contraction: it is the phenomenon occurring in the 3rd series of the transition element. As they have electron filling up the 4f shell which is filled with the 5d series. The 4f shell has very weak shielding effect and is highly attracted towards the Nucleus due to high nuclear charge so the filling of 4f orbitals before 5d orbital resulting in their atomic radius which is very similar to the elements of the 2nd series i.e. Zr=160 atomic radius and Hf=159 atomic radius.
(iii) The wide range of oxidation states of actinoids is attributed to the fact that the 5f, 6d and 7senergy levels are comparable energies. Therefore all these three sub shells can participate. But the most common oxidation state of actinoids is +3.
Assign reasons for the following:
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibit greater range of oxidation states than Lanthanoids.
Stability in the aqueous medium depends on the hydration energy of ions when they attract to water to water molecules. In aqueous solution, Cu+ ion undergoes oxidises and reduces simultaneously
in aqueous to give Cu and Cu2+ ion.
The relative stability of different oxidation states can be seen from their
Electrode potentials 
Due to more reduction electrode potential value of Cu+, it undergoes oxidation reaction quite feasible. Hence, the copper (I) ion is not known in aqueous solution.
(ii) The actinoids show a larger number of oxidation states because of the very small energy gap between the 5f, 6d and 7s subshells. The energies are decided on the basis of the (n+1) rule.
The (n+1) values of the three orbitals are under:
5f=5+3=8
6d=5+2=8
7s=7+ 1=8
All of the value comes to be same. Hence they have the same energy.
|
Element/Ion |
Electronic Configuration |
|
Zn |
3d10 4s2 |
|
Zn2+ |
3d10 |
|
Cu |
3d10 4s1 |
|
Cu2+ |
3d9 |
The colours of salts of transition metals are due to the d-d transition that depends on the presence of incompletely filled d-orbitals. Zn2+ has completely filled d-orbitals (3d10), while Cu2+ has incompletely filled d-orbitals (3d9); therefore, d-d transition is possible in Cu2+, which imparts colour to copper (II) salts.
(a) Account for the following:
(i) Zr and Hf have almost similar atomic radii.
(ii) Transition metals show variable oxidation states.
(iii) Cu+ ion is unstable in aqueous solution.
(b) Complete the following equations:
(i) 2 MnO2 + 4 KOH + O2--->
(ii) 2 Na2CrO4 + 2 H + -->
(i) Due to lanthanide contraction, Zr and Hf have almost similar atomic radii. It can be explained on the basis of shielding effect. The electrons present in inner shells, shield the outer electrons from the nuclear charge, making them experience a low effective nuclear charge. The shielding effect exerted by the electrons decreases in the orders > p > d > f. The f subshell poorly shields the outer electrons from nuclear attraction, which results in the most attractive pull of nucleus on the outer electron. In the case of post lanthanide elements like Hf, 4f subshell is filled and it is not very effective at shielding the outer shell electrons. Therefore, Zr and Hf have almost similar atomic radii.
(ii) The transition metals have their valence electrons in (n-1)d and ns orbitals. Since there is very little energy difference between these orbitals, both energy levels can be used for bond formation. Thus, transition elements exhibit variable oxidation states.
(iii) Cu2+ is more stable than Cu+ in an aqueous medium. Cu2+ has high hydration energy that compensates for the energy required to remove one electron from Cu+to form Cu2+. So in an aqueous medium, Cu+ gives Cu2+ and Cu.
Chemical reaction: 2 Cu+(aq) à Cu2+(aq) + Cu(s)
(b)
(i) 2 MnO2 + 4 KOH + O2 ---> 2 K2MnO4 + 2 H2O
(ii) 2 Na2CrO4 + 2 H+ ----> Na2Cr2O7 + 2 Na+ + H2O
|
|
Cr |
Mn |
Fe |
Co |
Ni |
Cu |
|
-0.91 |
-1.18 |
-0.44 |
-0.28 |
-0.25 |
-0.34 |
From the given data of E0 values answer the following Question:
(i) Which is strongest why is value exceptionally positive?
(ii) Why is value highly negative as compared to other elements?
(iii) Reducing agent Cr2+ or Fe2+? Give reason
(b) Why do actinoids show wide range f oxidation states? Write on the similarity between the chemistry of lanthanoids and actinoids.
The E0 (M2+/M) value of a metal depends on the energy changes involved in the formation of the M2+ ion:
1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state
M(s) ---> M (g) sH (Sublimation energy)
2. Ionisation: The energy required to take out electrons from one mole of atoms in the gaseous state to form the corresponding cation in the gaseous state
M (g) --->M2+ (g) iH (Ionization energy)
3. Hydration: The energy released when one mole of ions are hydrated
M (g)---> M2+ (aq) hydH (Hydration energy)
Now, copper has high energy of atomization and low hydration energy. Hence, theE0(Cu2+/Cu) is exceptionally positive.
(ii) The 
value is highly negative as compared to other elements because of the extra stability of Mn2+ due half -filled d orbitals.
(iii) The following reactions are involved when Cr2+ and Fe+ act as reducing agent:
Cr2+ ---> Cr3+ + e-
Fe2+ ----> Fe3+ + e-
The 
value is -0.41 V and 
is + 0.77V. This means that Cr2+ can be easily oxidised to Cr3+ but Fe2+ does not get oxidised to Fe3+ easily, therefore, Cr2+ is a stronger reducing agent than Fe3+.
(b) In actinoids, the 5f, 6d, 7s shells are present. These three shells are of comparable energies; therefore electrons can remove from these shells. This gives rise f variable oxidation states in actinoids.
Following are the similarities between actinoids and lanthanoids:
(i) The size of atom: It decreases across the series in both actinoids (due to actinoid contraction) and lanthanoids(due to lanthanoids contraction).
(ii) Oxidation states: Lanthanoids and actinoids generally show +3 oxidation states. However, some element in the actinoids series is capable of exhibiting oxidation states higher than +3.
How are interhalogen compounds formed? What general compositions can be assigned to them?
An interhalogen compound is a molecule which contains two or more different halogen atoms. A most interhalogen compound known are binary. Their formula is generally XYn Where n=1, 3, 5, or 7 and X is the less electronegative of two halogens. On hydrolysis, they ionise to give rise to poly -halogen ion. For example
Br2 (l) + F2 (g) --> 2BrF (g)
(a) Give reasons for the following:
(i) Mn3+ is a good oxidising agent.
(ii) 
Values are not regular for first row transition metals (3d series).
(iii) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn2O7.
(b) Complete the following equations:
(a) Outer electronic configuration f Mn is 3d5 4s2.
Outer electronic configuration of Mn3+ is 3d4 4s0.
Now Mn3+ is a strong oxidising agent. A good oxidizing agent reduces itself . i.e. gains electrons from other. Its tends to gain one more electron to acquire stable electronic configuration. If it gains one electron, its configuration will be 3d5, which is stable .this is the reason, it acts as a good reducing agent.
(ii) Values are not regular which can be explained by the irregular variation of ionisation enthalpies i.e. iH1 + iH2 and also the sublimation enthalpies which are relatively much less for manganese and vanadium.
(iii) The ability of oxygen to stabilise the higher oxidation state exceeds that of fluorine. Also, the ability of oxygen to form multiple bonds with metals favours Mn2O7. Therefore, the highest Mn fluoride is MnF4 whereas highest oxide is Mn2O7. In Mn2O7, each Mn is tetrahedrally surrounded by O's including a Mn—O—Mn bridge.
(a) Why do transition elements show variable oxidation states?
(i) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
(ii) Name the element which shows only +3 oxidation state.
(b) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals.
(a) The ability of the transition metal to exhibit variable valency is generally attributed to the availability of more electrons in the (n-1)d orbitals which are closer to the outermost ns orbital in energy levels.
Thus in the case of iron, we get the divalent Fe(II) state when only the 2 electrons in the 4s orbital are removed. And we get the trivalent Fe(III) state when one more 3d electron is removed, in addition to the two 4s electrons from the neutral Fe atom.
(i) Mn shows a maximum number of oxidation states among the first series of transition metals from Sc to Zn. Mn exhibits all the oxidation states from +2 to +7.
(ii) Scandium shows only +3 oxidation state.
(b) The regular decrease in the size of the atoms and ions with increasing atomic number is known as lanthanide contraction. It arises because as we move along the lanthanide series, the nuclear charge increases by one unit at each successive element, the new electron is added into the same subshell (viz., 4f). As a result, the attraction on the electrons by the nucleus increases and this tends to decrease the size. Further, as the new electron is added into the f-subshell, there is imperfect shielding of one electron by another in this subshell due to the shapes of these f-orbitals. This imperfect shielding is unable to counterbalance the effect of the increased nuclear charge. Hence, the net result is a contraction in the size though the decrease is very small.
Explain each of the following observations:
(i) With the same d-orbital configuration (d4), Cr2+ is a reducing agent while Mn3+ is an oxidising agent.
(ii) Actinoids exhibit a much larger number of oxidation states than the lanthanoids.
(iii) There is hardly any increase in atomic size with increasing atomic number in a series of transition metals.
(i) Outer electronic configuration f Mn is 3d5 4s2.
Outer electronic configuration of Mn3+ is 3d4 4s0.
Now Mn3+ is a strong oxidizing agent. A good oxidizing agent reduces itself. I.e. gains electrons from other. Its tends to gain one more electron to acquire stable electronic configuration. If it gains one electron, its configuration will be 3d5, which is stable .this is the reason, it acts as a good reducing agent.
Cr2+ is strongly reducing agent in nature. It has d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ such as d3. This d3 configuration can be written as t32g configuration, which is more stable configuration.
(ii) In actinoids, the 5f, 6d, 7s shells are present. These three shells are of comparable energies; therefore electrons can remove from these shells. This gives rise f variable oxidation states in actinoids.
Following are the similarities between actinoids and lanthanoids:
(a) The size of atom: It decreases across the series in both actinoids (due to actinoid contraction) and lanthanoids (due to lanthanoids contraction).
(b) Oxidation states: Lanthanoids and actinoids generally show +3 oxidation states. However, some element in the actinoids series is capable of exhibiting oxidation states higher than +3
(iii) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series).
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
Give reasons:
(i)Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(ii)Transition metals show variable oxidation states.
(iii)Actinoids show irregularities in their electronic configurations.
(i) Mn shows the highest oxidation state of +7 with oxygen because it can form p-pi−d-pi multiple bonds using 2p orbital of oxygen and 3d orbital of Mn. On the other hand, Mn shows the highest oxidation state of +4 with fluorine because it can form a single bond only.
ii) Transition metals show variable oxidation states due to the participation of ns and (n-1)d- electrons in bonding. Lower oxidation state is exhibited when the ns- electrons take part in bonding and higher oxidation states are exhibited when the (n-1) d-electrons take part in bonding.
(iii)Actinoids show irregularities in their electronic configurations because the energy differences between 5f, 6d and 7s subshells are very small. hence, an electron can be occupied in any of the subshells. They have the electronic configuration of 7s2 with a variation of occupancy in 5f and 6d orbitals.
What is meant by ‘lanthanoid contraction’?
Lanthanoid contraction, also called lanthanide contraction, the steady decrease in the size of the atoms and ions of the rare earth elements with increasing atomic number from lanthanum (atomic number 57) through lutetium (atomic number 71).
Explain giving reasons:
(i) Transition metals and their compounds generally exhibit a paramagnetic behaviour.
(ii) The chemistry of actinoids is not as smooth as that of lanthanoids.(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.
(ii)
(a)The general electronic configuration for lanthanoids is [Xe] 54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn] 86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent. In actinoids, the 5forbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids.
(b) Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 states are the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less .hence actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3. Hence, the chemistry of actinoids is not as smooth as that of lanthanoids.Complete the following chemical equations:
Or
State reasons for the following:
(i) Cu (I) ion is not stable in an aqueous solution.
(ii) Unlike Cr3+, Mn2+, Fe3+ and the subsequent other M2+ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species.
(i) In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion. The relative stability of different oxidation states can be seen from their electrode potentials.
Cu+(aq) + e- --> Cu(s) , E0red = 0.52V
Cu2+(aq) + 2e- ---> Cu(s), E0red = 0.34V
Thus overall reaction is :
2Cu+ (aq) ---> Cu2+ (aq) + Cu(s)
Due to more reduction electrode potential value of Cu+, it undergoes oxidation reaction quite feasibly. Hence, copper (I) ion is not stable in aqueous solution.
(ii) The valence shell electronic configurations of Cr3+, Mn2+, Fe3+ are d3, d5 and d3respectively. Owing to the symmetrical distribution of the d electrons, these ions can form stable cationic complexes. The atomic radii of the 4d and 5d transition elements are more than that of the 3d series elements. Hence, generally, they do not form stable cationic species.
(a) Complete the following equations :
(i) Cr2O72- + 2OH- --->
(ii) MnO4- + 4H+ + 3e- --->
(b) Account for the following :
(i) Zn is not considered as a transition element.
(ii) Transition metals form a large number of complexes.
(iii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple.
OR
(i) With reference to structural variability and chemical reactivity, write the differences between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation :
MnO4- + 8H+ + 5e---->
iv) Mn3+ is more paramagnetic than Cr3+.
(a) The balanced chemical equations are the following:
(i) Cr2O2-7 + 2OH- ---> 2CrO2-4 + H2O
(ii) MnO-4 + 4H+ + 3e- ---> MnO2 + 2H2O
(b)
(i) The electronic configuration of zinc is 1s22s22p63s23p63d104s2. Zinc has the stable filled valence shell d10 electronic configuration in its ground state as well as in its most common oxidation state of +2. Hence, it is not considered as a transition element.
(ii) Ions formed by transition metals have small sizes and high ionic charges. Also, they possess vacant d-orbitals to accommodate lone pairs of electrons for bond formation. As a result transition, metals form a large number of complexes.
(iii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for a Cr3+/Cr2+ couple. This is because Mn2+ ion is particularly stable due to the extra stability of its half filled valence electronic configuration (d5). Thus Mn3+ ion has a very high tendency to gain an electron and form the much more stable Mn2+ ion.
OR
(i) The differences between lanthanoids and actinoids, with reference to structural variability and chemical reactivity, are as follows:
a) Actinoids has far greater irregularities in metallic radii than lanthanoids, as a result actinoids display a variety of structures.
b) The ionisation enthalpies of the early actinoids are lower than those of the early lanthanoids.
c) Actinoids are more reactive and have more complex magnetic properties than lanthanoids.
(ii) Cerium is the lanthanoid element which is well known to exhibit +4 oxidation state.
(iii) The complete equation is as follows:
MnO4- + 8H+ + 5e----> Mn2+ + 4H2O
(iv) Mn3+ has four unpaired electrons (d4) in its valence shell whereas Cr3+ (d3) has three unpaired electrons. Thus, Mn3+ is more paramagnetic than Cr3+.
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.
The formula of an oxo anion of Manganese in which it shows the oxidation state equal to its group member is Permanganate MnO−4 /KMnO4
The oxidation state of Mn is + 7.
What happens when
(NH4)2Cr2O7 is heated?
Ammonium dichromate, (NH4)2Cr2O7, decomposes when heated to produce N2, H2O, and Cr2O3
(NH4)2Cr2O7 + heat →N2 + 4H2O + Cr2O3
Account for the following :
Transition metals form large number of complex compounds.
Transition metals form a large number of complex compounds because of:
(i) small in size
(ii)availability of empty d-orbital.
Account for the following :
The lowest oxide of transition metal is basic whereas the highest oxide is
amphoteric or acidic.
The lowest oxide of transition metal is basic because of some valence electron are not involved in bonding thus act as a base due to the availability of free electrons. The highest oxide of transition metal electrons of metal is involved in the bonding. Therefore, these electrons are not available for donation. Hence, they are acidic in nature.
Account for the following :
E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compared to Cr3+/Cr2+.
Mn2+ exists in 3d5 configuration which is a half-filled configuration which provides extra stability to the Metal Ion. While Mn3+ will exist in 3d4 configuration which is less stable than 3d5. Hence the conversion from 3+ to 2+ is very feasible. Hence E° value is more. while in
Cr+3 exist in 3d3 half-filled d orbital (3 e; electrons in t2g ) extra stability is attained by Cr+3 than Cr+2. Hence Cr+3 →Cr+2 is less feasible. And hence it has less reduction potential as compare to Mn+3/Mn+2.
Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.
Similarities between lanthanides and actinides
Differences between lanthanides and actinides
Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why?
Cu+2 is more stable in aqueous solution because of more hydration energy which compensates to the ionization energy of Cu+2 →Cu+1.
Cu+1 in aqueous solution undergoes disproportionation reaction.
2Cu+→ Cu + Cu2+
The Orange colour of Cr2O72– ion changes to yellow when treated with an
alkali. Why ?
The Orange colour of Cr2O72– ion changes to yellow when treated with an
alkali because of the formation of chromate ion CrO42- ion, which is yellow in colour.
Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons.
Lanthanides
Actinides
Give reason:
E° value for a Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.
Mn3+/ Mn2+ has large + value because Mn3+ reduced to Mn+2 which has a half-filled d-orbital configuration ([Ar]4s03d5).
On the other hands, Fe3+/ Fe2+ has small +ve value because Fe3+ is much more stable than Fe2+. Electronic configuration of Fe3+ is [Ar]4s2 3d5 and Fe2+ [Ar]4s2 3d6.
Give reason:
Iron has a higher enthalpy of atomization than that of copper.
Transition metal has the high heat of atomisation due to the presence of strong metallic bond which arises due to the presence of an unpaired electron in the (n - 1) d subshell. This is because the atoms in these elements are closely packed and held together by strong metallic bonds. The metallic bond is formed as a result of the interaction of electrons in the outermost shell. Greater the number of valence electrons, stronger is the metallic bond.
The pair in which phosphorous atoms have a formal oxidation state of +3 is:
Pyrophosphorous and hypophosphoric acids
Orthophosphorous and hypophosphoric acids
Pyrophosphorous and pyrophosphoric acids
Orthophosphorous and pyrophosphorous acids
D.
Orthophosphorous and pyrophosphorous acids
Orthophosphorous H3PO4
H3PxO3 = 3 +x + 3(-2) = 0 x = +3
pyrophosphorous acids H4P2O6
H4P2xO5 = 4 + 2x + 5 (-2) = 0
4 + 2x -10 = 0
x = +3
Which among the following is the most reactive?
Cl2
Br2
I2
ICI
D.
ICI
Interhalogen compounds are more reactive then homonuclear diatomic.
Match the catalysis to the correct processes.
| Catalyst | Process |
| (A)TiCl3 | (i) Wacker Process |
| (B) PdCl2 | (ii) Ziegler -Natta |
| (C) CuCl2 | (iii) Contact process |
| (D) V2O5 | (iv)Deacon's process |
(A)- (iii), (B)-(ii), (C)-iv, (D)- (i)
(A) - (ii), (B) - (i), (C) -(iv), (D) - (iii)
(A)- (ii), (B)- (iii), (C)- (iv), (D) - (i)
(A) - (iii) , (B) - (i), (C) - (ii), (D)- (iv)
B.
(A) - (ii), (B) - (i), (C) -(iv), (D) - (iii)
(i) TiCl3 is used as Ziegler- Natta catalyst for the polymerisation of ethene.
(ii) PdCl2 is used in Wacker process, in which alkene changed into aldehyde via catalytic cyclic process initiated by PdCl2.
(iii) CuCl2 is used in Deacon's process (for Cl2)
(iv) V2O5 is used in contact process manufacturing sulphuric acid.
Which of the following arrangements does not represent the correct order of the property stated against it?
V2+ < Cr2+<Mn2+<Fe2+: paramagnetic behaviour
Ni2+ < Co2+ < Fe2+ < Mn2+ ; Ionic size
Co3+ < Fe3+< Cr3+ < Sc3+ : stability in aqueous solution
Sc< Ti < Cr< Mn : Number of oxidation states
A.
V2+ < Cr2+<Mn2+<Fe2+: paramagnetic behaviour
a) V2+ = 3 unpaired electrons;
Cr2+ = 4 unpaired electrons
Mn2+ = 5 unpaired electrons;
Fe2+ =4 unpaired electrons
Iron exhibits + 2 and +3 oxidation states. Which of the following statements about iron is incorrect?
Ferrous oxide is more basic in nature than the ferric oxide.
Ferrous compounds are relatively more ionic than the corresponding ferric compounds.
Ferrous compounds are less volatile than the corresponding ferric compounds.
Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds.
D.
Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds.
FeO → More basic, more ionic, less volatile
The correct order of E°M2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is
Cr > Mn > Fe > Co
Mn > Cr > Fe > Co
Cr > Fe > Mn > Co
Fe > Mn > Cr > Co
B.
Mn > Cr > Fe > Co
Across the first transition series, the negative values for standard electrode potential decrease except for Mn due to the stable d5 configuration.
Mn > Cr > Fe > Co
Which of the following species is not paramagnetic
NO
CO
O2
B2
B.
CO
NO ⇒ One unpaired electron is present in π*molecular orbital.
CO ⇒ No unpaired electron is present
O2 ⇒ Two unpaired electrons are present in π* molecular orbitals.
B2 ⇒ Two unpaired electrons are present in π bonding molecular orbitals.
Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements in incorrect
The ionic sizes of Ln (III) decrease in general with increasing atomic number.
Ln (III) compounds are generally colourless.
Because of the large size of the Ln (III) ions, the bonding in its compounds is predominantly ionic in character
Ln (III) hydroxides are mainly basic in character.
B.
Ln (III) compounds are generally colourless.
Most of the Ln(III) compounds are coloured due to f – f transition.
In context with the transition elements, which of the following statements is incorrect?
In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.
In the highest oxidation states, the transition metal shows basic character and form cationic complexes.
In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding.
Once the 5 d configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.
B.
In the highest oxidation states, the transition metal shows basic character and form cationic complexes.
In highest oxidation states transition metals form anionic complexes rather than that of cationic complexes.
Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being
4f orbitals more diffused than the 5f orbitals
lesser energy difference between 5f and 6d than between 4f and 5d orbitals
more energy difference between 5f and 6d than between 4f and 5d orbitals
more reactive nature of the actinoids than the lanthanoids
B.
lesser energy difference between 5f and 6d than between 4f and 5d orbitals
Being lesser energy difference between 5f and 6d than 4f and 5d orbitals.
Identify the incorrect statement among the following –
d-Block elements show irregular and erratic chemical properties among themselves
La and Lu have partially filled d orbitals and no other partially filled orbitals
The chemistry of various lanthanoids is very similar
4f and 5f orbitals are equally shielded
D.
4f and 5f orbitals are equally shielded
4f and 5f belong to different energy levels, hence the shielding effect is on them is not the same.Shielding of 4f is more than 5f
The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because
the 5f orbitals are more buried than the 4f orbitals
there is a similarity between 4f and 5f orbitals in their angular part of the wave function
the actinoids are more reactive than the lanthanoids
the 5f orbitals extend further from the nucleus than the 4f orbitals
D.
the 5f orbitals extend further from the nucleus than the 4f orbitals
The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because the 5f orbitals extend further from the nucleus than the 4f orbitals.
The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence
GeX2 << SiX2 << SnX2<< PbX2
SiX2 << GeX2 << PbX2<< SnX2
SiX2 << GeX2 << SnX2<< PbX2
PbX2<< SnX2 << GeX2 << SiX2
C.
SiX2 << GeX2 << SnX2<< PbX2
Due to inert pair effect, the stability of +2 oxidation state increases as we move down this group.
A metal, M forms chlorides in it's +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?
MCl2 is more volatile than MCl4
MCl2 is more soluble in anhydrous ethanol than MCl4
MCl2 is more ionic than MCl4
C.
MCl2 is more ionic than MCl4
Lanthanoid contraction is caused due to
the appreciable shielding on outer electrons by 4f electrons from the nuclear charge
the appreciable shielding on outer electrons by 5d electrons from the nuclear charge
the same effective nuclear charge from Ce to Lu
the imperfect shielding on outer electrons by 4f electrons from the nuclear charge
D.
the imperfect shielding on outer electrons by 4f electrons from the nuclear charge
The lanthanide contraction is responsible for the fact that
Zr and Y have about the same radius
Zr and Nb have similar oxidation state
Zr and Hf have about the same radius
Zr and Zn have the same oxidation
C.
Zr and Hf have about the same radius
Due to Lanthanide contraction.
Which of the following factors may be regarded as the main cause of lanthanide contraction?
Poor shielding of one of 4f electron by another in the subshell
Effective shielding of one of 4f electrons by another in the subshell
Poorer shielding of 5d electrons by 4f electrons
Greater shielding of 5d electrons by 4f electrons
A.
Poor shielding of one of 4f electron by another in the subshell
Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect?
The common oxidation states of cerium are +3 and +4
Cerium (IV) acts as an oxidizing agent
The +4 oxidation state of cerium is not known in solutions
The +3 oxidation state of cerium is more stable than the +4 oxidation state
C.
The +4 oxidation state of cerium is not known in solutions
+3 and +4 states are shown by Ce in aqueous solution.
When copper is heated with conc. HNO3 it produces
Cu(NO3)2 and NO
Cu(NO3)2, NO and NO2
Cu(NO3)2 and N2O
Cu(NO3)2 and NO2
D.
Cu(NO3)2 and NO2
Nitric acid acts as an oxidising agent while reacting with copper.
i) when copper reacts with dilute nitric acid it forms,
3Cu + 4HNO3(dilute) --> 3Cu(NO3)2 +2NO +4H2O
ii) When copper reacts with concentrated nitric acid it forms,
Cu +4HNO3(conc.) ---> Cu(NO3)2 +NO2 +2H2O
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
The solution is decolourized
SO2 is reduced
Green Cr2(SO4)3 is formed
The solution turns blue
C.
Green Cr2(SO4)3 is formed
When SO2 is passed through the acidified K2Cr2O7 solution, the orange colour of potassium dichromate solution turns to clear green due to the formation of chromium sulphate. In this reaction, the oxidation state of Cr changes from +6 to +3.
The appearance of green colour is due to the reduction of chromium metal.
Because of lanthanoid contraction, of elements which of the following elements have nearly same atomic radii ? (Number in the parenthesis are atomic number).
Ti (22) and Zr(40)
Zr(40) and Nb (41)
Zr (40) and Hf (72)
Zr (40) and Ta(73)
C.
Zr (40) and Hf (72)
Because of the lanthanoid contraction Zr (atomic radii 160 pm) and Hf (atomic radii 158 pm) have nearly same atomic radii.
Lanthanoids include the elements from lanthanum La (Z=57) to lutetium Lu(Z =71).Zirconium Zr (40) belong to the second transition series (4d) and Hf (72) belongs to third transition series (5d). Lanthanoid contraction is associated with the intervention of the 4f orbitals which are filled before the 5d- series of elements starts. The filling of 4f if orbitals before 5d orbitals result in a regular decrease in atomic radii which compensates the expected increases in atomic size with increasing atomic number. As a result of this lanthanoid contraction, the elements of second and third transition series have almost similar atomic radii.
In acidic medium,H2O2 changes Cr2O72- to CrO5 which has two (-O-O-) bonds. Oxidation state of Cr in CrO3 is
+5
+3
+6
-10
C.
+6
When H2O2 is added to an acidified solution of a dichromate, Cr2O72-, aO deep blue coloured complex, chromic peroxide, CrO5 [ or CrO(O2)2] is formed.
Cr2O72- +2H++4H2O2 ---> 2CrO(O2)2 +5H2O
This deep blue coloured complex.
Oxidation state of Cr is +6 due to the presence of two peroxide linkages which can be calculated as
Cr peroxide normal
x+(-1)4+(-2) =0
x-6 =0
x=+6
Reason of lanthanoid contraction is
negligible Screening effect of 'f' orbitals
increasing the nuclear charge
decreasing the nuclear charge
decreasing screening effect
A.
negligible Screening effect of 'f' orbitals
Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides. This is due to the imperfect shielding [or poor screening effect] of f-orbitals due to their diffused shape, which unable to counterbalance the effect of the increased nuclear charge. Hence, the net result is a contraction in size of lanthanoids.
Which of the following does not give oxygen on heating?
KClO3
Zn(ClO3)2
K2Cr2O7
(NH4)2Cr2O7
D.
(NH4)2Cr2O7
Oxygen-rich compounds like chlorate perchlorate K2Cr2O7 etc. When heated gives oxygen but ammonium dichromate gives nitrogen gas when heated.
Which of the following lanthanoid ions is diamagnetic?
(At. number; Ce =58, Sm =62, Eu =63, Yb = 70)
Ce2+
Sm2+
Eu2+
Yb2+
D.
Yb2+
Lanthanoid ion with no unpaired electron is diamagnetic in nature.
Ce58 = [Xe]4f2 5d0 6s2
Ce2+ = [Xe]4f2 (two unpaired electrons)
Sm62 = [Xe]4f6 5d0 6s2
Sm2+ = [Xe] 4f6 (six unpaired electron)
Eu63 = [Xe]4f7 5d0 6s2
Eu2+ = [Xe] 4f7 (seven unpaired electron)
Yb70 = [Xe]4f14 5d0 6s2
Yb2+ = [Xe] 4f14 (No unpaired electron)
Because of the absence of unpaired electrons, Yb2+ is diamagnetic.
KMnO4 can be prepared from K2MnO4 as per reaction
The reaction can go to completion by removing OH- ions by adding
HCl
KOH
CO2
SO2
D.
SO2
Since OH- are generated from weak acid (H2O), a weak acid (like CO2) should be used to remove it because of a strong acid (HCl) reverse the reaction, KOH increases the concentration of OH-, thus again shifts the reaction in the backwards side.
CO2 combines with OH- to give carbonate which is easily removed.
SO2 reacts with water to give strong acid, so it cannot be used.
Which one of the following does not correctly represent the correct order of the property indicated aginst it?
Ti < V<Cr< Mn: increasing number of oxidation states
Ti< V<Cr3+<Mn3+ : increasing magnetic moment
Ti < V < Cr < Mn : Increasing melting points
Ti < V < Mn < Cr : increasing second ionisation enthalpy
C.
Ti < V < Cr < Mn : Increasing melting points
|
Element |
Oxidation states |
|
Ti |
+2,+3,+4 |
|
V |
+2,+3,+4,+5 |
|
Cr |
+1,+2,+3,+4,+5,+6 |
|
Mn |
+1,+2,+3,+4,+5,+6,+7 |
Therefore, order of oxidation states is
Ti < V < Cr < Mn
b)
|
Ion |
Outermost electronic arrangement |
No. of unpaired electrons |
|
Ti3+ |
3d1 |
1 |
|
V3+ |
3d2 |
2 |
|
Cr3+ |
3d3 |
3 |
|
Mn3+ |
3d4 |
4 |
Therefore, order of magnetic moment is
Ti3+ < V3+ < Cr3+ < Mn3+
In 3d- series melting point increases as we move from left to right. But Mn shows low melting point due to its complex formatting nature. It is unable to form metallic and covalent bonds.
|
Element |
Second ionisation enthalpy |
|
Ti |
1320 |
|
V |
1376 |
|
Mn |
1513 |
|
Cr |
1635 |
Therefore, order of second ionisation enthalpy is
Ti< V< Mn< Cr
Four successive members of the first series of the transition metals are listed below. For which one of them, the standard potential value 
has a positive sign?
Co (Z=27)
Ni (Z=28)
Cu (Z=29)
Fe (Z=26)
C.
Cu (Z=29)
In electrochemical series metals with positive standard potential are placed below hydrogen. Out of the given transition metals, only Cu is placed below hydrogen in the electrochemical series. Therefore, it must have a positive sign for standard potential.
The catalytic activity of transition metals and their compounds is ascribed mainly to
their magnetic behaviour
their unfilled d- orbitals
their ability to adopt variable oxidation states
their chemical reactivity
C.
their ability to adopt variable oxidation states
Transition metals show variable oxidation states, due to which they can perform as a catalyst.
Which of the following exhibits only +3 oxidation states?
U
Th
Ac
Pa
C.
Ac
89Ac = [Rn] 6d1, 7s2For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?
(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Fe > Mn > Co> Cr
Co > Mn > Fe > Cr
Cr > Mn > Co > Fe
Mn > Fe > Cr > Co
D.
Mn > Fe > Cr > Co
This can be understood on the basis of Eo values for M2+/M.
| Eo/V | Cr | Mn | Fe | Co |
| M2+/M | -0.90 | -1.18 | -0.44 | -0.28 |
liberates H2 from dilute acids.Acidified K2Cr2O7 solution turns green when Na2SO3 is added to it. This is due to the formation of
CrO42-
Cr2(SO3)3
CrSO4
Cr2(SO4)3
D.
Cr2(SO4)3
K2Cr2O7 +3Na2SO3 + 4H2SO4 → 3Na2SO4 +K2SO4 + Cr(SO4)3 +4H2O
Chromium(III) sulfate gives green colour.
Gadolinium belongs to 4f series. Its atomic number is 64. which of the following is the correct electronic configuration of gadolinium?
[Xe]4f8 6d2
[Xe]4f95s1
[Xe]4f7 5d16s2
[Xe]4f65d26s2
C.
[Xe]4f7 5d16s2
Atomic number of Gadolinium = 64
electronic configuration of Gd,
[Xe]4f7 5d16s2
Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation?
FeSO4
FeSO3
FeC2O4
Fe(NO2)2
A.
FeSO4
FeSO4 will require the least amount of acidified KMnO4 for complete oxidation.
Which of the statements given below is incorrect?
Cl2O7 is an anhydride of perchloric acid
O3molecule is bent
ONF is isoelectronic with NO2
OF2 is an oxide of fluorine
D.
OF2 is an oxide of fluorine
Cl2O7 is an anhydride of perchloric acid
The Shape of O3 molecule is bent.
c) Number of electrons in ONF = 24
Number of electrons in NO2 = 24
therefore, ONF and NO2 both are isoelectronic.
d) OF2 is a fluoride of oxygen because electronegativity of fluorine is more than that of oxygen
OF2 = oxygen difluoride
Which one of the following compounds is a peroxide?
KO2
BaO2
MnO2
NO2
B.
BaO2
In peroxides, the oxidation state of O is -1 and they give H2O2, with dilute acids, and have peroxide linkage.
In KO2,
+1 + (x X 2) = 0
x = -1/2 (thus, it is a superoxide, not a peroxide.)
In BaO2, + 2 + (x X 2) = 0
x = -1
Thus, it is a peroxide. Only it gives H2O2 when reacts with dilute acids and has peroxide linkage as
Ba2+ [ O- O]2-
Peroxide linkage
In MnO2 and NO2, Mn and N exhibit variable oxidation states, thus, the oxidation state of O in these is -2. Hence, these are not peroxides. Thus, it is clear that among the given molecules only BaO2 is a peroxide.
Which of the following ions will exhibit colour in aqueous solutions?
La3+ (Z = 57)
Ti3+ (Z = 22)
Lu3+
Sc3+
B.
Ti3+ (Z = 22)
Colour is obtained as a consequence of d-d (or f-f) transition, the presence of unpaired electrons is necessary conditions.
La3+ (Z = 57) = [Xe] 4f0 5d0 6s0 [ no unpaired electron]
Ti3+ (Z = 22) = [Ar] 3d1 4s0 [one unpaired electron]
Lu3+ (Z = 71) = [Xe] 4f14 5d0 6s0 [no unpaired electron]
Sc3+(Z = 21) = [Ar] 3d0 4s0 [no unpaired electron]
Hence, only Ti3+ will exhibit colour in aqueous solution.
Which of the following oxidation states is the most common among the lanthanoids
4
2
5
+3
D.
+3
The most common oxidation state exhibited by lanthanoids is +3.
The Number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is:
3/5
4/5
2/5
1
C.
2/5
Identify the incorrect statement among the following:
There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
Lanthanoid contraction is the accumulation of successive shrinkages
As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements
Shielding power of 4f electrons is quite weak
C.
As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements
The regular decrease in the radii of lanthanide ions from La3+ to Lu3+ is known as lanthanides contraction.
It is due to the greater effect of the increased nuclear charge than that of screening effect (shield) effect.
As a result of lanthanide contraction, the atomic radii of the element of 4d and 5d one close just above them in their respective group, so the properties of 4d and 5d transition element shows the similarities.
Which one of the following ions is the most stable in aqueous solution?
Cr3+
V3+
Ti3+
Mn3+
C.
Ti3+
Stability to transition metal ion is directly proportional to the unpaired electron. The exactly half filled and completely filled d - orbitals are extra stable.
Cr3+ (21) = 3d3 4s0 -3 unpaired electrons
V3+ (20) = 3d2 4so -2 unpaired electrons
Ti3+ (19) = 3d1 4s0 -1 unpaired electron
Mn3+ (22) = 3d4 4so - 4 unpaired electrons
So, Mn3+ ion is most stable in aqueous solution.
More the number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is:
more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
the lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
the greater metallic character of the lanthanoids than that of the corresponding actinoids
more active nature of the actinoids
B.
the lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
A number of oxidation states are exhibited by the actinoids than by the lanthanoids due to the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.
It is because of inability of ns2 electrons of the valence shell to participate in bonding that
Sn2+ is reducing while Pb4+ is oxidising
Sn2+ is oxidising while Pb4+ is reducing
Sn2+ and Pb2+ are both oxidising and reducing
Sn4+ is reducing while Pb4+ is oxidising
A.
Sn2+ is reducing while Pb4+ is oxidising
Inability of ns2 electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect
As a result, Pb(II) is more stable than Pb(IV)
Sn(IV) is more stable than Sn(II)
∴ Pb(IV) is easily reduced to Pb(II)
∴ Pb(IV) is oxidising agent
Sn(II) is easily oxidised to Sn(IV)
∴ Sn(II) is reducing agent
The reason for greater range of oxidation states in actinoids is attributed to
The radioactive nature of actinoids
Actinoid contraction
5f, 6d and 7s levels having comparable energies
4f and 5d levels being close in energies
C.
5f, 6d and 7s levels having comparable energies
Name the gas that can readily decolourises acidified KMnO4 solution
CO2
SO2
NO2
P2O5
B.
SO2
SO2 is readily decolourises acidified KMnO4.
The correct increasing order of ionic radii of the following Ce3+, La3+, Pm3+ and Yb3+ is
Yb3+ < Pm3+< Ce3+<La3+
Ce3+< Yb3+<Pm3+<La3+
Yb3+< Pm3+< La3+<Ce3+
Pm3+ < La3+< Ce3+ < Yb3+
A.
Yb3+ < Pm3+< Ce3+<La3+
Due to the lanthanide contraction size of +3 ions decreases from La3+ to Lu3+
The shape of gaseous SnCl2 is
Tetrahedral
Linear
Angular
T-shape
C.
Angular
Total no. of valence electron for Sn = 4
Total no. of valence electron for Cl = 7
Thus total number of valence electron for SnCl2 = 4 + 7 x 2 = 18
Bond pair = 2
Number of electrons used to complete the octets and duet = 8 x 2 = 16
No. of lone pair = 1
SnCl2 has sp2 hybridisation with one lone pair of electrons. Thus has angular shape.
Sponsor Area
Sponsor Area
[Mg(EDTA)]2+ + 6H2O