Exercises

4.1. Write down the electronic configuration of:

(i) Cr³⁺

(ii) Pm³⁺

(ii) Cu⁺

(iv) Ce⁴⁺

(v) Co²+

(vi) Lu²⁺

(vii) Mn²⁺

(viii) Th⁴⁺

Solution

(i) Cr³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ Or, [Ar]¹⁸ 3d³

(ii) Pm³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f⁴ Or, [Xe]⁵⁴ 3d³

(iii) Cu⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ Or, [Ar]¹⁸ 3d¹⁰

(iv) Ce⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ Or, [Xe]⁵⁴

(v) Co²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or, [Ar]¹⁸ 3d⁷

(vi) Lu²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹ Or, [Xe]⁵⁴ 2f¹⁴ 3d³

(vii) Mn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or, [Ar]¹⁸ 3d⁵

(viii) Th⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6s⁶ Or, [Rn]⁸⁶

4.2. Why are Mn²⁺compounds more stable than Fe²⁺ towards oxidation to their +3 state?

Solution

Electronic configuration of Mn²⁺ is [Ar]¹⁸ 3d⁵.

Electronic configuration of Fe²⁺ is [Ar]¹⁸ 3d⁶.

It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stabled⁵ configuration. This is the reason Mn²⁺ shows resistance to oxidation to Mn³⁺.

Also, Fe²⁺ has 3d⁶configuration and by losing one electron, its configuration changes to a more stable 3d⁵ configuration. Therefore, Fe²⁺ easily gets oxidized to Fe⁺³ oxidation state.

4.3. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Solution

The oxidation states displayed by the first half of the first row of transition metals are given in the table below.

It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of delectrons in (+2) state also increases from Ti(+2) to Mn(+ 2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn (+2) has d⁵ electrons (that is half-filled d shell, which is highly stable).

Or,

Here, after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺

4.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Solution

In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d⁵ 4s². It shows oxidation states +2 to +7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d⁵. Similarly Sc³⁺ is more stable then Sc+ and Fe³⁺ is more stable than Fe²⁺ due to half filled f-orbitals.

4.5. What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³,3d⁵, 3d⁸and 3d⁴?

Solution

4.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Solution

Cr₂O₇^2- and CrO₄^2- (Group number = Oxidation state of Cr = 6).

MnO₄^– (Group number = Oxidation state of Mn = 7).

4.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Solution

Lanthanoid Contraction: In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f-electrons However, the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is known as lanthanoid contraction.

Consequences of lanthanoid contraction:

• There is similarity in the properties of second and third transition series.
• Separation of lanthanoids is possible due to lanthanoid contraction.
• It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from La(OH)₃ to Lu(OH)₃.)

Solution

Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements.

Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.

4.9. In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Solution

Transition metals have a partially filled d−orbital. Therefore, the electronic configuration of transition elements is (n − 1)d^1-10 ns^0-2.

The non-transition elements either do not have a d−orbital or have a fully filled d−orbital. Therefore, the electronic configuration of non-transition elements is ns^1-2 or ns² np^1-6.

4.10. What are the different oxidation states exhibited by the lanthanoids?

Solution

In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

4.11. Explain giving reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.

(ii) The enthalpies of atomisation of the transition metals are high.

(iii) The transition metals generally form coloured compounds.

(iv) Transition metals and their many compounds act as good catalysts.

Solution

(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.

The paramagnetic character is measured in terms of magnetic moment and is given by:

where,

n: number of unpaired electrons.

(ii) Transition metals have high effective nuclear charge, greater number of valence electrons and some unpaired electrons. They thus have strong metal–metal bonding. Hence, transition metals have high enthalpies of atomisation.

(iii) 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.

(iv) Transition elements act as good catalysts in chemical reactions 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.

4.12. What are interstitial compounds? What are interstitial compounds? Why do these compounds have higher melting points than corresponding pure metals?

Solution

Interstitial compounds are formed when small atoms such as H, C or N are trapped inside the crystal lattices of metals. They are usually nonstoichiometric and are neither typically ionic nor covalent.
They have melting point higher than metals due to stronger metal-non-metal bonds or compared to metal-metal bonds in pure metals.

4.13. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Solution

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe²⁺ and Fe³⁺; Cu⁺ and Cu²⁺). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.

4.14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution

Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr₂O₃) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂→ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na₂Cr₂O₇.2H₂O can be crystallised.

2Na₂CrO₄ + 2H⁺→ Na₂Cr₂O₇ + 2Na⁺ + H₂O

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.

Na₂Cr₂O7 + 2KCl → K₂Cr₂O₇ + 2NaCl

Orange crystals of potassium dichromate crystallise out. The chromates and dichromates depending upon p^H of the solution. If p^H of potassium dichromate is increased it is converted to yellow potassium chromate

2CrO₄^2- + 2H⁺→ Cr₂O₇^2- + H₂O

2CrO₇^2- + 2OH^-→ Cr₂O₄^2- + H₂O

4.15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

(ii) iron(II) solution and

(iii) H₂S

Solution

K₂Cr₂O₇ acts as a very strong oxidising agent in the acidic medium.

K₂Cr₂O₇ + 4H₂SO₄→ K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]

K₂Cr₂O₇ takes up electrons to get reduced and acts as an oxidising agent. The reaction of K₂Cr₂O₇with other iodide, iron (II) solution, and H₂S are given below.

(i) K₂Cr₂O₇ oxidizes iodide to iodine

(ii)K₂Cr₂O₇ oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.

(iii) K₂Cr₂O₇ oxidizes H₂S to sulphur.

4.16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO₂ and (iii) oxalic acid? Write the ionic equations for the reactions.

Solution

Potassium permanganate can be prepared from pyrolusite (MnO₂). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO₃ or KClO₄, to give K₂MnO₄.

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.

Electrolytic oxidation

K₂MnO₄↔ 2K⁺ + MnO₄^2-

H₂O ↔ H⁺ + OH^-

At anode, manganate ions are oxidized to permanganate ions.

Oxidation by chlorine

2K₂MnO₄ + Cl₂→ 2KMnO₄ + 2KCl

2MnO₄²- + Cl₂→ 2MnO₄^- + 2Cl^-

Oxidation by ozone

2K₂MnO₄ + O₃ + H₂O → 2KMnO₄ + 2KOH + O₂

2MnO₄^2- + O₃ + H₂O → 2MnO₄^2- + 2OH^- + O₂

(i) Acidified KMnO₄ solution oxidizes Fe(II) ions to Fe(III) ions i.e., ferrous ions to ferric ions.

(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

4.17. For M²⁺/M and M³⁺/M²⁺ systems, the Eθ values for some metals are as follows:

Use this data to comment upon:

(i) The stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and

(ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution

(i) The E^θ value for Fe³⁺/Fe^{2+ ­­} is higher than that for Cr³⁺/Cr²⁺ and lower than that for Mn³⁺/Mn²⁺. So, the reduction of Fe³⁺ to Fe²⁺ is easier than the reduction of Mn³⁺ to Mn²⁺, but not as easy as the reduction of Cr³⁺ to Cr²⁺. Hence, Fe³⁺ is more stable than Mn³⁺, but less stable than Cr³⁺.

These metal ions can be arranged in the increasing order of their stability as:

Mn³⁺ < Fe³⁺ < Cr³⁺

(ii) The reduction potentials for the given pairs increase in the following order.

Mn²⁺/Mn < Cr²⁺/Cr < Fe²⁺/Fe

So, the oxidation of Fe to Fe²⁺ is not as easy as the oxidation of Cr to Cr²⁺ and the oxidation of Mn to Mn²⁺.

Thus, these metals can be arranged in the increasing order of their ability to get oxidised as:

Fe < Cr < Mn

4.18. Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.

Solution

Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured. The ions in which d-orbitals are empty or completely filled will be colourless as no d-d transition is possible in those configurations.

From the above table, it can be easily observed that only Sc³⁺ has an empty d-orbital and Cu⁺ has completely filled d-orbitals. All other ions, except Sc³⁺ and Cu⁺, will be coloured in aqueous solution because of d−d transitions.

4.19. Compare the stability of +2 oxidation state for the elements of the first transition series.

Solution

From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.