NCERT Solutions for Class 12 Chemistry Chapter 4 The d-block and f-block Elements

Chapter 4 The d-block and f-block Elements NCERT Solutions Class 12 Chemistry- PDF Download

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.

Sc (+2) = d1

Ti (+2) = d2

V (+2) = d3

Cr (+2) = d4

Mn (+2) = d5

+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

	Electronic configuration in ground state	Stable oxidation states
1.	3d3 (Vanadium)	+2, +3, +4 and +5
2.	3d5 (Chromium)	+3, +4, +6
3.	3d5 (Manganese)	+2, +4, +6, +7
4.	3d8 (Cobalt)	+2, +3
5.	3d4	There is no3d4 configuration in ground state

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.)

4.8. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

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:

Cr2+/Cr	−0.9V
Cr3/Cr2+	−0.4 V
Mn2+/Mn	−1.2V
Mn3+/Mn2+	+1.5 V
Fe2+/Fe	−0.4V
Fe3+/Fe2+	+0.8 V

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.

Element	Atomic Number	Ionic State	Electronic configuration in ionic state
Ti	22	T13+	[Ar] 3d1
V	23	V3+	[Ar] 3d2
Cu	29	Cu+	[Ar] 3d10
Sc	21	Sc3+	[Ar]
Mn	25	Mn2+	[Ar] 3d5
Fe	26	Fe3+	[Ar] 3d5
Co	27	Co2+	[Ar] 3d7

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

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2	+3
Zn		+2

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.

4.20. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) electronic configuration

(ii) oxidation state

(iii) atomic and ionic sizes

(iv) chemical reactivity.

Solution

(i) Electronic configuration - The general electronic configuration for lanthanoids is:

[Xe]⁵⁴ 4f^0-14 5d^0-1 6s²

and that for actinoids is

[Rn]⁸⁶ 5f^1-14 6d^0-1 7s²

Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states - 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.

(iii) Atomic and lonic sizes - Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.

(iv) Chemical reactivity - In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

4.21. How would you account for the following:

(i) Of the d⁴species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.

(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

(iii) The d¹ configuration is very unstable in ions.

Solution

(i) 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 t23g 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 is highly stable.

(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3^rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.

(iii) The ions in d¹ configuration tend to lose one more electron to get into stable d⁰ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

4.22. What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

Solution

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation-reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

For example,

Cr(V) is oxidized to Cr(VI) and reduced to Cr(III).

Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

4.23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Solution

In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d¹⁰. The completely filled d-orbital makes it highly stable.

4.24. Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?

Solution

	Gaseous ions	Number of unpaired electrons
1	Mn3+,[Ar]3d4	4
2	Cr3+,[Ar]3d3	3
3	V3+,[Ar]3d2	2
4	Ti3+,[Ar]3d1	1

Cr³⁺ is the most stable in aqueous solutions owing to a t_2g³ configuration.

4.25. Give examples and suggest reasons for the following features of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Solution

(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge.

As a result, it can accept electrons and behave as an acid.

For example, Mn^IIO is basic and is Mn₂^VIIOacidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF₆ and V₂O₅, the oxidation states of Os and V are +6 and +5 respectively.

 

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in MnO4-, the oxidation state of Mn is +7.

4.26. Indicate the steps in the preparation of

(i) K₂Cr2O7 from chromite ore.

(ii) KMnO₄ from pyrolusite ore.

Solution

(i) Dichromates are generally prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr₂O₄) with sodium or potassium carbonate in free 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. Therefore, the latter is 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 are interconvertible in an aqueous solution, depending on the pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.

2CrO₄^2- + 2H⁺ ⟶ Cr₂O₇^2- + H₂O

Cr₂O₇^2- + 2OH⁻ ⟶ 2CrO₄^2- + H₂O

Hence, when the pH is increased, making the solution alkaline, dichromate ions (orange in colour) are converted into chromate ions, and the solution becomes yellow.

 

(ii) Potassium permanganate KMnO₄ 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 the 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 ⟶ 2 MnO₄^2- + 2OH^- + O₂

4.27. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Solution

An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements.

An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of S, C, Si, Ca, and Al.

Uses:

• Mischmetal is used in cigarettes and gas lighters.
• It is used in flame throwing tanks.
• It is used in tracer bullets and shells.

4.28. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

Solution

Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.

4.29. 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

Solution

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.

4.30. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Solution

The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is [Rn] 5f146d17s2. The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f¹⁴ configuration.

4.31. Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Solution

Ce: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹5d¹6s²

Magnetic moment can be calculated as:

where,

n = number of unpaired electrons

The electronic configuration of Ce³⁺: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹

In Ce³⁺, n = 1

4.32. Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements.

Solution

The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.

+2	+4
Nd (60)	Ce (58)
Sm (62)	Pr (59)
Eu (63)	Nd (60)
Tm (69)	Tb (65)
Yb (70)	Dy (66)

Ce after forming Ce⁴⁺ attains a stable electronic configuration of [Xe].

Tb after forming Tb⁴⁺ attains a stable electronic configuration of [Xe] 4f⁷.

Eu after forming Eu²⁺ attains a stable electronic configuration of [Xe] 4f⁷.

Yb after forming Yb²⁺ attains a stable electronic configuration of [Xe] 4f¹⁴.

4.33. Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) chemical reactivity.

(ii) electronic configuration

(iii) oxidation states

Solution

(i) Chemical reactivity: In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

(ii) Electronic configuration: The general electronic configuration for lanthanoids is:

[Xe]⁵⁴ 4f^0-14 5d^0-1 6s²

and that for actinoids is:

[Rn]⁸⁶ 5f^1-14 6d^0-1 7s².

Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(iii) Oxidation states: 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.

4.34. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Solution

Atomic number	Electronic configuration
61	[Xe]544f55d06s2
91	[Rn]865f26d17s2
101	[Rn]865f135d7s2
109	[Rn]865f146d7s2

4.35. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(i) electronic configurations,

(ii) oxidation states,

(iii) ionisation enthalpies, and

(iv) atomic sizes.

Solution

(i) In the 1^st, 2^nd and 3^rd transition series, the 3d, 4d and 5d orbitals are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations:

Cr(24) = 3d⁵4s¹

Cu(29) = 3d¹⁰4s¹

Similarly, there are exceptions in the second transition series. These are:

Mo(42) = 4d⁵5s¹

Tc(43) = 4d⁶5s¹

Ru(44) = 4d⁷5s¹

Rh(45) = 4d⁸5s¹

Pd(46) = 4d¹⁰5s⁰

Ag(47) = 4d¹⁰5s¹

There are some exceptions in the third transition series as well. These are:

W(74) = 5d⁴6s²

Pt(78) = 5d⁹6s¹

Au(79) = 5d¹⁰6s¹

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.

For example,

 are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example: WCl₆, ReF₇, RuO₄, etc.

(iii) 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 4f electrons 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 2^nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1^st transition series.

(iv) 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.

4.36. Write down the number of 3d electrons in each of the following ions:

Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, CO²⁺, Ni²⁺ and Cu²⁺.

Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Solution

Metal ion	Number of d-electrons	Filling of d-orbitals
Ti2+	2	t2g2
V2+	3	t2g3
Cr3+	3	t2g3
Mn2+	5	t2g3 eg2
Fe2+	6	t2g4 eg2
Fe3+	5	t2g3 eg2
CO2+	7	t2g5 eg2
Ni2+	8	t2g6 eg2
Cu2+	9	t2g6 eg3

4.37. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Solution

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 2^nd and 3^rd 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.

(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 atomisation 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.

4.38. 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
K2[MnCl4]	5.9

Solution

Magnetic moment (μ) is given as: 

For value n = 1,

For value n = 2,

For value n = 3,

For value n = 4,

For value n = 5,

(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.