The d- and f- Block Elements-Solutions

 CBSE Class–12 Subject Chemistry

NCERT Solutions
Chapter – 08
The d and f Block Elements

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In-text question

1. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?

Ans. The elements which have partially filled d or f subshells in any common oxidation state are called as the transition elements. Silver (the Atomic number is 47 and electronic configuration is [Kr] 4d105s1) has a completely filled 4d orbital in its ground state but has two oxidation states (+1, +2).
In the +1 oxidation state, an electron is removed from the s-orbital and in +2 oxidation state, one electron from d-orbital is also removed. Thus, the d-orbital now becomes partially filed (4d9). Hence, it is a transition element.

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2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126kJmol-1. Why?

Ans. The enthalpy of atomization depends on the strength of the metallic bonding. Stronger the metallic bonding, greater is the enthalpy of atomization. The metallic bonding is strong when there are more unpaired electrons in the atom. All transition metals (except Zn, electronic configuration: [Kr] 3d10 4s2), have at least one unpaired electron that is responsible for their stronger metallic bonding. Since the Zn atom does not have an unpaired electron, the metallic bonding is weak and hence the enthalpy of atomization is low.

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3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Ans. Manganese (Z = 25) exhibits the largest number of oxidation states. This is because its electronic configuration is 3d54s2. Because it has the maximum number of electrons (5d electrons and 2s electrons) to easily lose and share. Therefore, it can exhibit an oxidation state of +2 to +7.
The compounds are as follows:
Mn(0) as Mn(s), Mn(II) as MnO, Mn(II,III) , Mn etc.

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4. The Eo(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHo${\mathrm{\Delta }}_a{\mathrm{H}}^o$and low ΔhydHo${\mathrm{\Delta }}_{hyd}{\mathrm{H}}^o$)

Ans. The Eo(M2+/M) value of a metal depends on the energy changes involved in the following:
(i) Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.
M(s)→M(g)${\mathrm{M}}_{(\mathrm{s})}\to {\mathrm{M}}_{(\mathrm{g})}$; ΔgH${\mathrm{\Delta }}_g\mathrm{H}$ (Sublimation energy)
(ii) Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.
M(g)→M2+(g)${\mathrm{M}}_{(\mathrm{g})}\to {\mathrm{M}}_{(\mathrm{g})}^{2+}$; ΔiH${\mathrm{\Delta }}_i\mathrm{H}$ (Ionization energy)
(iii) Hydration: The energy released when one mole of ions are hydrated.
M(g)→M2+(aq)${\mathrm{M}}_{(\mathrm{g})}\to {\mathrm{M}}_{(\mathrm{a}\mathrm{q})}^{2+}$; ΔhydH${\mathrm{\Delta }}_{hyd}\mathrm{H}$ (Hydration energy)
Now, copper has a high energy of ionization and it is not compensated by hydration energy.
Hence, the Eo(M2+/M) value for copper is positive.

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5. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

Ans. The ionization energy increases due to the gradual filling of electrons in the d- subshells. The irregular variation of ionization energy is due to the fact that half-filled and completely filled subshells are more stable and have very high ionization energy.
In case of first ionization energy, Cr([Ar]3d54s1) attains the stable configuration (3d5) by losing one electron from s-subshell and hence, it has low ionization energy.
Whereas, Zn has high ionization energy because it has completely filled subshells and are very stable.
Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has to be removed from a stable structure, which was formed due to the removal of the first electron.

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6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Ans. 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.

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7. Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Ans. The following reactions are involved when Cr2+ and Fe2+ act as reducing agents:
Cr2+⇒ Cr3+ (E⁰Cr3+/ Cr2+ = - 0.41 V)
Fe2+⇒ Fe3+ (E⁰Fe3+/ Fe2+ = +0.77 V)
Since Cr has less potential value, Cr2+ gets oxidised easily than Fe2+. Therefore, Cr2+ is a better reducing agent that Fe3+.

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8. Calculate the 'spin only' magnetic moment of M2+(aq)$M_{(aq)}^{2+}$ ion (Z = 27).

Ans. Given, Z = atomic number = 27 = [Ar]3d74s2
⇒ M2+ = [Ar]3d7
Hence, 3 unpaired electrons are present.
The spin only magnetic moment μ= n(n+2)−−−−−−−√$\sqrt{n(n+2)}$
where n is the number of unpaired electrons.
Hence, μ = 3(3+2)−−−−−−−√$\sqrt{3(3+2)}$ = 15−−√$\sqrt{15}$ = 3.87 BM 

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9. Explain why Cu+ ion is not stable in aqueous solutions?

Ans. The stability in aqueous condition depends on the hydration energy of the ions when they bond to the water molecules. And, the hydration energy is the amount of heat released as an ionic substance is dissolved and its constituent ions are hydrated or surrounded by water molecules.

Now, in Cu2+ and Cu+ ion, Cu2+ has a greater charge density than the Cu+ ion and so forms much stronger bonds releasing more energy. Therefore, in an aqueous medium, Cu2+ ion is more stable than Cu+ ion. This is because the energy required to remove one electron from Cu+ to Cu2+, is compensated by the high hydration energy of Cu2+.

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10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans. In actinoids, 5f orbitals 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. As the effective nuclear charge experienced is high, the electrons are attracted with much force, hence the size of the atom decreases. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Chapter End Question

1. Write down the electronic configuration of:

(i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+

Ans. (i) The atomic number of Cr is 24 and the electronic configuration is [Ar] 3d54s1
When 3 electrons are removed, it becomes Cr3+.
The electronic configuration of Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or [Ar]3d3

(ii) The atomic number of Pm is 61 and the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f5 5s2 5p6 6s2 or [Xe] 4f56s2.
When 3 electrons are removed, it becomes Pm3+, having the electronic configuration,
Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4 Or, [Xe]544f4

(iii) The atomic number of Cu is 29 and has the electronic configuration of [Ar] 3d104s1.
On removing one electron, the Cu+is obtained with the electronic configuration,
1s2 2s2 2p6 3s2 3p6 3d10 Or [Ar] 3d10

(iv)The atomic number of Ce is 58 and has the electronic configuration of [Xe] 4f15d16s2.
When the valence electrons are removed, the Ce4+ ion with configuration, 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or [Xe] is obtained.

(v) The atomic number of Co is 27 and has the electronic configuration of [Ar] 3d74s2.
When 2 electrons are removed from s-orbital, it becomes Co2+with electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar] 3d7.

(vi)The atomic number of Lu is 71 and has the electronic configuration of [Xe] 4f145d16s2.
When 2 electrons are removed, Lu2+with the configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d 1 Or, [Xe] 4f145d1

(vii) The atomic number of Mn is 25 and has the electronic configuration of [Ar] 3d54s2 .
When 2 electrons are removed, Mn2+ ion is obtained and has an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 .

(viii) The atomic number of Th is 90 and has the electronic configuration of [Rn] 6d2 7s2 .
When 4 electrons are removed, the electronic configuration becomesTh4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d10 6s2 6p6 Or, [Rn]

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

Ans. Electronic configuration of Mn2+ is [Ar]183d5
and Electronic configuration of Fe2+ is [Ar]183d6

It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe3+ oxidation state.

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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?

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

	Sc	Ti	V	Cr	Mn
Oxidation state		+2	+2	+2	+2
	+3	+3	+3	+3	+3
		+4	+4	+4	+4
			+5	+5	+6
				+6	+7

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 d electrons 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).

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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.

Ans. The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s23d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar]3d5.

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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: 3d3,3d5,3d8 and 3d4?

Ans. For answering this question, we can compare the electronic configuration of standard elements and then write their corresponding oxidation states.

	Electronic Configuration in Ground State	Stable oxidation states
(i)	3d3(Vanadium)	+2, +3, +4 and +5
(ii)	3d5(Chromium)	+3, +4 and +6
(iii)	3d5(Manganese)	+2, +4 , +6 , +7
(iv)	3d8(Cobalt)	+2, +3
(v)	3d4	There is no 3d4 configuration in ground state.

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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.

Ans. (i) Vanadate, VO−3${\mathrm{V}\mathrm{O}}_3^{-}$

Oxidation state of V is +5.

(ii) Chromate, CrO2−4${\mathrm{C}\mathrm{r}\mathrm{O}}_4^{2-}$

Oxidation state of Cr is +6.

(iii) Permanganate, MnO−4${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}$

Oxidation state of Mn is +7.

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

Ans. 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 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.

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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?

Ans. 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 in the ground state and most stable oxidation state.

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

Ans. Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d1-10ns1-2
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 ns1-2 or ns2np1-6.

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

Ans. 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.

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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 a good catalyst.

Ans. (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) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d-orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small 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) The catalytic activity of the transition elements can be explained by two basic facts:
(a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.
(b) Transition metals also provide a suitable surface for the reactions to occur.

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12. What are interstitial compounds? Why are such compounds well known for transition metals?

Ans. Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

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

Ans. 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.

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

Ans. Potassium dichromate is prepared from chromite ore (FeCe2O4) in the following steps:
Step (1): Preparation of sodium chromate
4FeCr2O4+16NaOH+7O2→8Na2CrO4+2Fe2O3+8H2O$4{\mathrm{F}\mathrm{e}\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_4+16\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+7{\mathrm{O}}_2\to 8{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4+2{\mathrm{F}\mathrm{e}}_2{\mathrm{O}}_3+8{\mathrm{H}}_2\mathrm{O}$
Step (2): Conversion of sodium chromate into sodium dichromate
2Na2CrO4+conc⋅H2SO4→Na2Cr2O7+Na2SO4++H2O$2{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\cdot {\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4\to {\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+{\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4++{\mathrm{H}}_2\mathrm{O}$
Step (3): Conversion of sodium dichromate to potassium dichromate
Na2Cr2O7+2KCl→K2Cr2O7+2NaCl${\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{K}\mathrm{C}\mathrm{l}\to {\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}$
Potassium dichromate being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.
The dichromate ion (Cr2O2−7)$\left({\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}\right)$ exists in equilibrium with chromate(CrO2−4)$\left(\mathrm{C}\mathrm{r}{\mathrm{O}}_4^{2-}\right)$ ion at pH 4. However, by changing the pH, they can be interconverted.
2CrO2 - 4Chromate(Yellow)⟷AcidAlkali2HCrO - 4Hydrogen Chromate⟷AcidAlkaliCrO2 - 7Dichromate(Orange)$\underset{\text{(Yellow)}}{\underset{\text{Chromate}}{{\text{2CrO}}_{\text{4}}^{\text{2 - }}}}\underset{\text{Alkali}}{\stackrel{\text{Acid}}{⟷}}\underset{\text{Hydrogen Chromate}}{{\text{2HCrO}}_{\text{4}}^{\text{ - }}}\underset{\text{Alkali}}{\stackrel{\text{Acid}}{⟷}}\underset{(\text{Orange)}}{\underset{\text{Dichromate}}{{\text{CrO}}_7^{\text{2 - }}}}$

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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

Ans.  K2Cr2O7 acts as a very strong oxidising agent in the acidic medium.
K2Cr2O7 takes up electrons to get reduced and acts as an oxidising agent.
The reaction of K2Cr2O7 with 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.

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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.

Ans. Potassium permanganate can be prepared from (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.
2MnO2 + 4KOH + O2 −→−heat2K2MnO4 (Green) $\stackrel{heat}{\to }\begin{array}{l}2{\mathrm{K}}_2{\mathrm{M}\mathrm{n}\mathrm{O}}_4\\ \text{ (Green) }\end{array}$+ 2H2O

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++MnO2−4${\mathrm{K}}_2{\mathrm{M}\mathrm{n}\mathrm{O}}_4\leftrightarrow 2{\mathrm{K}}^{+}+{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{2-}$
H2O↔H++OH−${\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{H}}^{+}+{\mathrm{O}\mathrm{H}}^{-}$
At anode, manganate ions are oxidized to permanganate ions.
MnO2−4(Green)↔MnO−4(Purple)+e−$\underset{\text{(Green)}}{{\text{MnO}}_4^{2-}}\leftrightarrow \underset{(\text{Purple)}}{{\text{MnO}}_4^{-}}+{\text{e}}^{-}$
Oxidation by chlorine:
2K2MnO4+Cl2⟶2KMnO4+2KCl$2{\mathrm{K}}_2{\mathrm{M}\mathrm{n}\mathrm{O}}_4+{\mathrm{C}\mathrm{l}}_2⟶2{\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4+2\mathrm{K}\mathrm{C}\mathrm{l}$
2MnO2−4+Cl2⟶2MnO−4+2Cl−$2\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{2-}+{\mathrm{C}\mathrm{l}}_2⟶2{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}+2{\mathrm{C}\mathrm{l}}^{-}$
Oxidation by ozone:
2K2MnO4+O3+H2O⟶2KMnO4+2KOH+O2$2{\mathrm{K}}_2{\mathrm{M}\mathrm{n}\mathrm{O}}_4+{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{K}\mathrm{M}\mathrm{n}\mathrm{O}}_4+2\mathrm{K}\mathrm{O}\mathrm{H}+{\mathrm{O}}_2$
2MnO2−4+O3+H2O⟶2MnO2−4+2OH−+O2$2\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{2-}+{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{M}\mathrm{n}\mathrm{O}}_4^{2-}+2{\mathrm{O}\mathrm{H}}^{-}+{\mathrm{O}}_2$

(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.

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17. For M2+/M and M3+/M2+ systems, the Eo 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:
(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.

Ans. (i) The Eo value for Fe3+/Fe2+ is higher than that for Cr3+/Cr2+ and lower than that of Mn3+/Mn2+. So converstion of  Mn3+ to Mn2+ is easier than Fe3+ to Fe2+ and since Cr3+ is stable so its converstion to Cr2+ is difficult. 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

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

Ans. 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	Ti3+	[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.

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

Ans.

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.

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20. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity

Ans. (i) Electronic configuration
The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2 and that for actinoids is [Rn]865f1-146d0-17s2. 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).

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Chapter End Question

21. How would you account for the following:
(i) Of the d4 species, 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.

Ans. (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 t32g$t_{2g}^3$ 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 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) and hydration enegy of Co(III) 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.

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22. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Ans. The reactions in which same substance gets oxidized as well as reduced due to unstable oxidation state.
Example:
(i) 2MnO42- + 4H+→ 2MnO4- + MnO2 + 2H2O
Mn (VI) is oxidised to Mn(VII) and also reduced to Mn(IV).
(ii) 2CrO43- + 2H+→ CrO42- + Cr3+ + 4H2O
Cr(V) is oxidised to Cr(VI) and also reduced to Cr(III).

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

Ans. 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.

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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?

Ans.

	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 t32g$t_{2\mathrm{g}}^3$ configuration.

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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.

Ans. (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 and lower oxide are also ionic in nature.
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 higher metal oxide are covalent and acidic in nature.
For example, MnO is basic and Mn2O7 is acidic.

(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 MnO−4${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}$, the oxidation state of Mn is +7.

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26. Indicate the steps in the preparation of:
(i) K2Cr2O7 from chromite ore.
(ii) KMnO4 from pyrolusite ore.

Ans. (i)  The chromite ore (FeCr2O4) is fused with sodium hydroxide (NaOH)
4FeCr2O4+16NaOH+7O2→8Na2CrO4+2Fe2O3+8H2O$4{\mathrm{F}\mathrm{e}\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_4+16\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+7{\mathrm{O}}_2\to 8{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4+2{\mathrm{F}\mathrm{e}}_2{\mathrm{O}}_3+8{\mathrm{H}}_2\mathrm{O}$
The yellow solution of sodium chromate is then filtered and acidified with sulphuric acid giving its dichromate.
2Na2CrO4+concH2SO4→Na2Cr2O7+Na2SO4+H2O$2{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}{\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4\to {\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+{\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O}$
On cooling, sodium sulphate crystallizes out as Na2SO4.10H2O and is removed.
Na2Cr2O7+2KCl→K2Cr2O7+2NaCl${\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{K}\mathrm{C}\mathrm{l}\to {\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}$

(ii) The pyrolusite ore(MnO2) is oxidised in the presence of potassium hydroxide by heating.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
The green potassium manganate (K2MnO4) is then treated with a current of chlorine or ozone to oxidise potassium manganate to potassium permanganate.
2K2MnO4 + Cl2 → 2KCl + 2KMnO4
K2MnO4 + O3 + H2O → 2KMnO4 + 2KOH + O2

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

Ans. 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:
(i) Mischmetal is used in cigarettes and gas lighters.
(ii) It is used in flame throwing tanks.
(iii) It is used in tracer bullets and shells.

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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.

Ans. 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.

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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.

Ans. 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.

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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.

Ans. 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.

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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.

Ans. The atomic number of Cerium (Ce) is Z = 58.
The electronic configuration of 58Ce= [Xe]54 4f1 5d1 6s2
And, the electronic configuration of Ce3+= [Xe]54 4f1 , i.e., there is only one unpaired electron, i.e., n = 1.
Now, the magnetic moment on the basis of spin only formula is given as:
μ=n(n+2)−−−−−−−√$\mu =\sqrt{n(\mathrm{n}+2)}$
where, n = number of unpaired electrons
In Ce3+, n = 1
Therefore,
μ=1(1+2)−−−−−−−√$\mu =\sqrt{1(1+2)}$ 
=3–√$=\sqrt{3}$ = 1.73 BM

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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.

Ans. 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.

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33. Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic configuration
(ii) oxidation states and
(iii) chemical reactivity

Ans. (i) Electronic Configuration

The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2 and that for actinoids is [Rn]865f1-146d0-17s2. 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 7slevels 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) Chemical reactivity

In the lanthanide series, the earlier members of the series are more reactive similar to calcium. 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 the case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

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

Ans.

Atomic number	Electronic Configuration
61	[Xe]544f55d06s2
91	[Rn]865f26d17s2
101	[Rn]865f146d07s2
109	[Rn]865f146d77s2

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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
(iv) atomic sizes

Ans. (i) In the 1st, 2nd and 3rd 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, [Fe2(Cn)6]4-, [Co3(NH3)6]3-, [Ti(H2O)6]3+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 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st 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.

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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).

Ans.

Metal ion	Number of d-electrons	Filling of d-orbitals
Ti2+	2	t22g$t_{2\mathrm{g}}^2$
V2+	3	t32g$t_{2\mathrm{g}}^3$
Cr3+	3	t32g$t_{2\mathrm{g}}^3$
Mn2+	5	t32ge2g$t_{2g}^3{e}_g^2$
Fe2+	6	t42ge2g$t_{2g}^4{e}_g^2$
Fe3+	5	t32ge2g$t_{2g}^3{e}_g^2$
Co2+	7	t52ge2g$t_{2g}^5{e}_g^2$
Ni2+	8	t62ge2g$t_{2g}^6{e}_g^2$
Cu2+	9	t62ge3g$t_{2g}^6{e}_g^3$

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

Ans. 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 the 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.

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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

Ans. Magnetic moment (μ)$(\mu )$ is given as μ=n(n+2)−−−−−−−√$\mu =\sqrt{n(n+2)}$.
For value n = 1, μ=1(1+2)−−−−−−−√=3–√$\mu =\sqrt{1(1+2)}=\sqrt{3}$ = 1.732
For value n = 2, μ=2(2+2)−−−−−−−√=8–√$\mu =\sqrt{2(2+2)}=\sqrt{8}$ = 2.83
For value n = 3, μ=3(3+2)−−−−−−−√=15−−√$\mu =\sqrt{3(3+2)}=\sqrt{15}$ = 3.87
For value n = 4, μ=4(4+2)−−−−−−−√=24−−√$\mu =\sqrt{4(4+2)}=\sqrt{24}$ = 4.889
For value n = 5, μ=5(5+2)−−−−−−−√=35−−√$\mu =\sqrt{5(5+2)}=\sqrt{35}$ = 5.92

(i) K4[Mn(CN)6]
The given magnetic moment is 2.2.
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+
The given magnetic moment is 5.3.
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]
The given magnetic moment is 5.9.
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.