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

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 8
Chapter Name : The d-and f-Block Elements

( )
Q8.1 Silver atom has completely lled d orbitals 4d 10 in its ground state. How can you say that it
is a transition element?

( )
Answer. Ag has a completely lled 4d orbital 4d 105s 1 in its ground state. Now, silver displays two
oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital.
However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital

( )
now becomes incomplete 4d 9 . Hence, it is a transition.

Page : 220 , Block Name : Intext Questions

Q8.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e.,
126kJmol − 1 Why?

Answer. The extent of metallic bonding an element undergoes decides the enthalpy of
atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy
Of atomization. In all transition metals (except Zn, electronic con guration: 3d 104s 2 ), there are
some unpaired electrons that account for their stronger metallic bonding. Due to the absence of
these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result,
it has the least enthalpy of atomization.

Page : 221 , Block Name : Intext Questions

Q8.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation
states and why?

Answer. Mn (Zn = 25) = 3d 54s 2
Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons).
Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to + 7.

Page : 223 , Block Name : Intext Questions

( )
Q8.4 The E ⊖ M 2 + / M value for copper is positive (+0.34V). What is possible reason for this?

(Hint: consider its high Δ aH ⊖ and low Δ hydH ⊖ ).

( )
Answer. The E ⊖ M 2 + / M value of a metal depends on the energy changes involved in the
following:
1. Sublimation: The energy required for converting one mole of an atom from the solid state to the
gaseous state.
M ( s ) ⟶ M ( s ) Δ sH( Sublimation energy )
2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous
state.
2+
M ( g ) ⟶ M ( g ) Δ iH( Ionization energy )
3. Hydration: The energy released when one mole Of ions are hydrated.
2+ 2+
M ( g ) ⟶ M ( aq )
Δ hydH(Hydration energy)

(
Now, copper has a high energy of atomization and low hydration energy. Hence, the E θ M 2 + / M )
value for copper is positive.

Page : 224 , Block Name : Intext Questions

Q8.5 How would you account for the irregular variation of ionisation enthalpies ( rst and second)
in the rst series of the transition elements?

Answer. Ionization enthalpies are found to increase in the given series due to a continuous lling
of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the
extra stability of con gurations such as d 0, d 5, d 10. Since these states are exceptionally stable,
their ionization enthalpies are very high.
In case of rst ionization energy, Cr has low ionization energy. This is because after losing one

( )
electron, it attains the stable con guration 3d 5 . On the other hand, Zn has exceptionally high
rst ionization energy as an electron has to be removed from stable and fully- lled orbitals

(3d 4s ).
10 2

Second ionization energies are higher than the rst since it becomes dif cult to remove an
electron when an electron has already been taken out. Also, elements like Cr and Cu have
exceptionally high second ionization energies as after losing the rst electron, they have attained

( )
the stable con guration Cr + : 3d 5 and Cu + : 3d 10 . Hence, taking out one electron more from this
stable con guration will require a lot of energy.

Page : 226 , Block Name : Intext Questions

Q8.6 Why is the highest oxidation state of a metal exhibited in its oxide or uoride only?

Answer. Both oxide and uoride 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.

Page : 227 , Block Name : Intext Questions

Q8.7 Which is a stronger reducing agent Cr 2 + or Fe 2 + and why ?

Answer. The following reactions are involved when Cr2+ and Fe2+ act as reducing agents.
Cr 2 + ⟶ Cr 3 + Fe 2 + ⟶ Fe 3 +
∘ ∘ 2 + can be easily
The E Ct 3 + / cr 2 + value is -0.41 V and E Fe 3 + / Fe 2 + is +0.77 V. This means that Cr

oxidized to Cr 3 + , but Fe 2 + does not get oxidized to Fe 3 + easily. Therefore, Cr 2 + is a better
reducing agent that Fe 3 +

Page : 227 , Block Name : Intext Questions

2+
Q8.8 Calculate the ‘spin only’ magnetic moment of M aq ion (Z = 27).

Answer.
Z = 27
⇒ [Ar]3d 74s 2

∴ M 2 + = [Ar]3d 7

i.e., 3 unpaired electrons
∴n=3
⇒ √n(n + 2) = μ

⇒ √3(3 + 2) = μ

⇒ √15 = μ
μ ≈ 4BM

Page : 229 , Block Name : Intext Questions

Q8.9 Explain why Cu + ion is not stable in aqueous solutions?

Answer. In an aqueous medium, Cu 2 + is more stable than Cu + . This is because although energy is
required to remove one electron from to Cu + to Cu 2 + high hydration energy of Cu 2 + compensates
for it. Therefore, Cu + ion in an aqueous solution is unstable. It disproportionates to give Cu 2 + and
Cu.
+ 2+
2Cu ( aq ) ⟶ Cu ( aq ) + Cu ( s )

Page : 231 , Block Name : Intext Questions

Q8.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer. In actinoids, 5f orbitals are lled. 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 than that experienced by lanthanoids. Hence, the size
contraction in actinoids is greater as compared to that in lanthanoids.

Page : 239 , Block Name : Intext Questions

Q8.1 Write down the electronic con guration of:
(i) Cr 3 +
(ii) Pm 3 +
(iii) Cu +
(iv) Ce 4 +
(v) Co 2 +
(vi) Lu 2 +
(vii) Mn 2 +
(viii) Th 4 +

Answer.
(i) Cr 3 + : 1s 22s 22p 63s 23p 63d 3
Or, [Ar] 183d 3

(ii) Pm 3 + : 1s 22s 22p 63s 23p 63d 104s 24p 64d 105s 25p 64f 4
Or, [Xe] 543d 3

(iii) Cu + : 1s 22s 22p 63s 23p 63d 10
Or, [Ar] 183d 10

(iv) Ce 4 + : 1s 22s 22p 63s 23p 63d 104s 24p 64d 105s 25p 6
Or, [Xe] 54

(v) Co 2 + : 1s 22s 22p 63s 23p 63d 7
Or, [Ar] 183d 7

(vi) Lu 2 + : 1s 22s 22p 63s 23p 63d 104s 24p 64d 105p 64f 145d 1
Or, [Xe] 542f 143d 3

(vii) Mn 2 + : 1s 22s 22p 63s 23p 63d 5
Or, [Ar] 183d 5

Th 4 + : 1s 22s 22p 63s 23p 63d 104s 24p 64d 104f 104s 25p 65d 106s 26s 6
Or, [Rn] 86

Page : 241 , Block Name : Exercise

Q8.2 Why are Mn 2 + compounds more stable than Fe 2 + towards oxidation to their +3 state?

Answer. Electronic con guration of Mn 2 + is [Ar] 183d 5
Electronic con guration of Fe 2 + is [Ar] 183d 6
It is known that half- lled and fully- lled orbitals are more stable. Therefore, Mn in (+2) state has
a stable d 5 con guration. This is the reason Mn2+ shows resistance to oxidation to
Mn 3 + . Also, Fe 2 + has 3d 6 con guration and by losing one electron, its con guration changes to a
more stable 3d 5 con guration. Therefore, Fe 2 + easily gets oxidized to Fe + 3 oxidation state.

Page : 241 , Block Name : Exercise

Q8.3 Explain brie y how +2 state becomes more and more stable in the rst half of the rst row
transition elements with increasing atomic number?

Answer. The oxidation states displayed by the rst half of the rst 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) = d 1
Ti( + 2) = d 2
V( + 2) = d 3
Cr( + 2) = d 4
Mn( + 2) = d 5

+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- lled). Mn (+2) has d 5 electrons (that is half- lled d
shell, which is highly stable).

Page : 241 , Block Name : Exercise

Q8.4 To what extent do the electronic con gurations decide the stability of oxidation states in the
rst series of the transition elements? Illustrate your answer with examples.

Answer. The elements in the rst-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 lled in
the d-orbital. However, Sc does not show +2 oxidation state. Its electronic con guration is 4s 23d 1.
It loses all the three electrons to form Sc 3 + . +3 oxidation state of Sc is very stable as by losing all
three electrons, it attains stable noble gas con guration, [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- lled, [Ar]3d 5.

Page : 241 , Block Name : Exercise

Q8.5 What may be the stable oxidation state of the transition element with the following d
electron con gurations in the ground state of their atoms : 3d 3, 3d 5, 3d 8 and 3d 4?

Answer.

Page : 241 , Block Name : Exercise

Q8.6 Name the oxometal anions of the rst series of the transition metals in which the metal
exhibits the oxidation state equal to its group number.

Answer.
−
(i) Vanadate, VO 3
Oxidation state of V is + 5 .

2−
(ii) Chromate, CrO 4
Oxidation state of Cr is + 6 .

−
(iii) Permanganate, MnO 4
Oxidation state of Mn is + 7 .

Page : 241 , Block Name : Exercise

Q8.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Answer. 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.

Page : 241 , Block Name : Exercise

Q8.8 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?

Answer. Transition metals have a partially lled d—orbital. Therefore, the electronic con guration
of transition elements is (n − 1)d 1 − 10ns 0 − 2.
The non-transition elements either do not have a d—orbital or have a fully lled d-orbital.
Therefore, the electronic con guration of non-transition elements is ns 1 − 2 or ns 2np 1 − 6.

Page : 241 , Block Name : Exercise

Q8.9 In what way is the electronic con guration of the transition elements different from that of
the non transition elements?

Answer. Transition metals have a partially lled d—orbital. Therefore, the electronic con guration
of transition elements is (n − 1)d 1 − 10ns 0 − 2.
The non-transition elements either do not have a d—orbital or have a fully lled d-orbital.
Therefore, the electronic con guration of non-transition elements is ns 1 − 2 or ns 2np 1 − 6.

Page : 241 , Block Name : Exercise

Q8.10 What are the different oxidation states exhibited by the lanthanoids?

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

Page : 241 , Block Name : Exercise

Q8.11 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 catalyst.

Answer. (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 rst 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,k 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 re ected,
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, E a , for the reaction.
(b) Transition metals also provide a suitable surface for the reactions to occur.

Page : 241 , Block Name : Exercise

Q8.12 What are interstitial compounds? Why are such compounds well known for transition
metals?

Answer. 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.

Page : 241 , Block Name : Exercise

Q8.13 How is the variability in oxidation states of transition metals different from that of the non
transition metals? Illustrate with examples.

Answer. 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

(Fe and Fe ; Cu and Cu ). In non-transition elements, the oxidation states differ by 2, for
2+ 3+ + 2+

example, +2 and +4 or +3 and +5, etc.

Page : 241 , Block Name : Exercise

Q8.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the
effect of increasing pH on a solution of potassium dichromate?

( )
Answer. Potassium dichromate is prepared from chromite ore FeCr 2O 4 in the following steps.
Step (1): Preparation of sodium chromate
4FeCr 2O 4 + 16NaOH + 7O 2 ⟶ 8Na 2CrO 4 + 2Fe 2O 3 + 8H 2O
Step (2) : Conversion of sodium chromate into sodium dichromate
2Na 2CrO 4 + conc. H 2SO 4 ⟶ Na 2Cr 2O 7 + Na 2SO 4 + H 2O
Step(3): Conversion of sodium dichromate to potassium dichromate
Na 2Cr 2O 7 + 2KCl ⟶ K 2Cr 2O 7 + 2NaCl

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange
coloured crystals and can be removed by ltration.

(
The dichromate ion Cr 2O 7
2−
) exists in equilibrium with chromate (CrO ) ion at pH 4.
2−
4

However, by changing the pH, they can be interconverted.

Page : 241 , Block Name : Exercise

Q8.15 Describe the oxidising action of potassium dichromate and write the ionic equations for its
reaction with:
(i) iodide
(ii) iron(II) solution and
(iii) H 2S

Answer.
K 2Cr 2O 7 acts as a very strong oxidising agent in the acidic medium.

K 2Cr 2O 7 + 4H 2SO 4 ⟶ K 2SO 4 + Cr 2 SO 4 ( )3 + 4H2O + 3[O]
K 2Cr 2O 7 takes up electrons to get reduced and acts as an oxidising agent. The reaction of K 2Cr 2O 7
with other iodide, iron (II) solution, and are given below.

(i) K 2Cr 2O 7 oxidizes iodide to iodine.

(ii) K 2Cr 2O 7 oxidizes iron (II) solution to iron (Ill) solution i.e., ferrous ions to ferric ions.

(iii) K 2Cr 2O 7 K oxidizes H 2S to sulphur.

Page : 241 , Block Name : Exercise

Q8.16 Describe the preparation of potassium permanganate. How does the acidi ed permanganate

solution react with (i) iron(II) ions (ii) SO 2 and (iii) oxalic acid?

( )
Answer. Potassium permanganate can be prepared from pyrolusite MnO 2 . The ore is fused with
KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO 3 or
KClO 4, to give K 2MnO 4
heat
2MnO 2 + 4KOH + O 2 ⟶ 2K 2MnO 4 + 2H 2O

The green mass can be extracted with water and then oxidized either electrolytically or
by passing chlorine/ozone into the solution.
Electrolytic oxidation
2−
K 2MnO 4 ⟷ 2K + + MnO 4

H 2O ⟷ H + + OH −
At anode, manganate ions are oxidized to permanganate ions.
2− −
MnO 4 ⟷ MnO 4 + e −
Oxidation by chlorine
−
2K 2MnO 4 + Cl 2 ⟶ 2MnO 4 + 2KCl
2− −
2MnO 4 + Cl 2 ⟶ 2MnO 4 + 2C −
Oxidation by ozone
2−
2K 2MnO 4 + O 3 + H 2O ⟶ 2MnO 4 + 2OH − + O 2
2− 2−
2MnO 4 + O 3 + H 2O ⟶ 2MnO 4 + 2OH − + O 2

(i) Acidi ed KMn04 solution oxidizes Fe (II) ions to Fe (Ill) ions i.e., ferrous Ions to ferric ions.

(ii) Acidi ed potassium permanganate oxidizes SO 2 to sulphuric acid.

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

Page : 242 , Block Name : Exercise

Q8.17
For M 2 + / M and M 3 + / M 2 + systems, the E ⊖ values for some metals are as follows:
Cr 2 + / Cr − 0.9V
Cr 3 / Cr 2 + − 0.4V
Mn 2 + / Mn − 1.2V
Mn 3 + / Mn 2 + + 1.5V
Fe 2 + / Fe − 0.4V
Fe 3 + / Fe 2 + + 0.8V
Use this data to comment upon:
(i) the stability of Fe 3 + in acid solution as compared to that of Cr 3 + or Mn 3 + and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium
or manganese metal.

Answer. (i) The E ⊖ value for is higher than that for and lower than that for
Cr 3 + / Cr 2 + . So, the reduction of Fe 3 + to Fe 2 + is easier than the reduction of Mn 3 + to Mn 2 +
but not as easy as the reduction of Cr 3 + to Cr 2 + . Hence, Fe 3 + is more stable than Mn 3 +
but less stable than Cr 3 + . These metal ions can be arranged in the increasing order of their
stability as:Mn 3 + < Fe 3 + < Cr 3 +
(ii) The reduction potentials for the given pairs increase in the following order.
Mn 2 + / Mn < Cr 2 + / Cr < Fe 2 + / Fe
So, the oxidation of Fe to Fe 2 + is not as easy as the oxidation of Cr 2 + and the oxidation of Mn to
Mn 2 + . Thus, these metals can be arranged in the increasing order of their ability to get oxidised
as: Fe < Cr < Mn

Page : 242 , Block Name : Exercise

Q8.18 Predict which of the following will be coloured in aqueous solution? Ti 3 + , V 3 +
Cu + , Sc 3 + , Mn 2 + , Fe 3 + and Co 2 + . Give reasons for each.

Answer. Only the ions that have electrons in d-orbital will be coloured. The ions in which d-orbital
is empty will be colourless.

From the above table, it can be easily observed that only Sc 3 + has an empty d-orbital. All other
ions, except Sc 3 + , will be coloured in aqueous solution because of d-d transitions.

Page : 242 , Block Name : Exercise

Q8.19 Compare the stability Of +2 oxidation State for the elements Of the rst transition series.

Answer.

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 dif cult to remove the third electron from the d-orbital.

Page : 242 , Block Name : Exercise

Q8.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic con guration
(ii) atomic and ionic sizes
(iii) oxidation state
(iv) chemical reactivity.

Answer. (i) Electronic con guration
The general electronic con guration for lanthanoids is [Xe] 544f 0 − 145d 0 − 16s 2 and that for actinoids
is [Rn] 365f − 146d 0 − 17s 2. 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 full•r lled and half- lled

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 Ionic 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
nely 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).

Page : 242 , Block Name : Exercise

Q8.21 How would you account for the following:
(i) Of the d 4 species, Cr 2 + 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 d 1 con guration is very unstable in ions.

Answer. (i) Cr 2 + is strongly reducing in nature. It has a d 4 con guration. While acting as a
reducing agent, it gets oxidized to Cr 3 + (electronic con guration, d 3). This d 3 con guration can be
3
( )
written as t 2g con guration, which is a more stable con guration. In the case of Mn 3 + d 4 it acts

( )
as an oxidizing agent and gets reduced to Mn 2 + d 5 . This has an exactly half- lled d-orbital and is
highly stable.

(ii) Co(ll) is stable in aqueous solutions. However, in the presence of strong eld complexing
reagents, it is oxidized to Co (Ill). Although the 3rd ionization energy for Co is high, but the higher
amount Of crystal eld stabilization energy (CFSE) released in the Dresence Of strona eld liaands
overcomes this ionization enerav.

(iii) The ions in d 1 con guration tend to lose one more electron to get into stable d 0 con guration.
Also, the hydration or lattice energy is more than suf cient to remove the only electron present in
the d-orbital of these ions. Therefore, they act as reducing agents.

Page : 242 , Block Name : Exercise

Q8.22 What is meant by 'disproportionation'? Give two examples of disproportionation reaction in

aqueous solution.

Answer. 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,
3− 2−
3CrO 4 + 8H + ⟶ 2CrO 4 + Cr 3 + + 4H 2O
(i)
Cr(V) Cr(VI) Cr(III)
Cr(V) is oxidized to Cr(Vl) and reduced to Cr(lll).

2− −
3MnO 4 + 4H + ⟶ 2MnO 4 + MnO 2 + 2H 2O
(ii)
Mn(VI) Mn(vII) Mn(IV)
Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

Page : 242 , Block Name : Exercise

Q8.23 Which metal in the rst series of transition metals exhibits +1 oxidation state most
frequently and why?

Answer. In the rst transition series, Cu exhibits +1 oxidation state very frequently. It is because
Cu ( +1) has an electronic con guration of [Ar]3d 10. The completely lled d-orbital makes it highly
stable.

Page : 242 , Block Name : Exercise

Q8.24 Calculate the number Of unpaired electrons in the following gaseous ions: Mn 3 + , Cr 3 + , V 3 +
and Ti 3 + . Which one of these is the most stable in aqueous solution?

Answer.

3
Cr 3 + is the most stable in aqueous solutions owing to a t 2g

Page : 242 , Block Name : Exercise

Q8.25 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 uorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Answer. (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 nO is basic and Mn 2O 7is acidic.

(ii) Oxygen and uorine 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 uorides. For
example, in OsF 6 and V 2O 5, 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. the So,
oxo-anions of a metal have the highest oxidation state. For example, in MnO 4− oxidation state of
Mn is +7.

Page : 242 , Block Name : Exercise

Q8.26 Indicate the steps in the preparation Of:
(i) K 2Cr 2O 7 from chromite Ore.
(ii) KMnO 4from pyrolusite ore.

Answer. (i)

( ) ( )
Potassium dichromate K 2Cr 2O 7 is prepared from chromite ore FeCr 2O 4 in the

following steps.
Step (1) : Preparation of sodium chromate
4FeCr 2O 4 + 16NaOH + 7O 2 ⟶ 8Na 2CrO 4 + 2Fe 2O 3 + 8H 2O
Step (2) : Conversion of sodium chromate into sodium dichromate
2Na 2CrO 4 + conc. H 2SO 4 ⟶ Na 2Cr 2O 7 + Na 2SO 4 + H 2O
Step (3) : Conversion of sodium dichromate to potassium dichromate
Na 2Cr 2O 7 + 2KCl ⟶ K 2Cr 2O 7 + 2NaCl Potassium chloride being less

soluble than sodium chloride is obtained in the form of orange coloured crystals and can be
removed by ltration.

(
The dichromate ion Cr 2O 27 − ) (
exists in equilibrium with chromate )
CrO 24 −
ion
at pH 4.

However, by changing the pH , they can be interconverted.

2− − 2−
2CrO 4 2HCrO 4 Cr 2O 7
Chromate ⟷ Hydrogen ⟷ Dichromate
(Yellow) chromate (Orange)

(ii)

( )
Potassium permanganate ( KMnO 4) can be prepared from pyrolusite MnO 2 . The ore is

fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such
as KNO 3 or KClO 4 , to give K 2MnO 4 .
heat
2K 2MnO 4 + 2H 2O
2MnO 2 + 4KOH + O 2 ⟶
(Green)
The green mass can be extracted with water and then oxidized either electrolytically or
by passing chlorine/ozone into the solution.
Electrolytic oxidation
2−
K 2MnO 4 ⟷ 2K + + MnO 4

H 2O ⟷ H + + OH −
At anode, manganate ions are oxidized to permanganate ions.

2− −
MnO 4 ⟷ MnO 4
+ e−
Green Purple
Oxidation by chlorine
2K 2MnO 4 + Cl 2 ⟶ 2KMnO 4 + 2KCl
2− −
2MnO 4 + Cl 2 ⟶ 2MnO 4 + 2Cl −
Oxidation by ozone
2−
2K 2MnO 4 + O 3 + H 2O ⟶ 2MnO 4 + 2OH − + O 2
2− 2−
2MnO 4 + O 3 + H 2O ⟶ 2MnO 4 + 2OH − + O 2

Page : 242 , Block Name : Exercise

Q8.27 What are alloys? Name an important alloy which contains some Of the lanthanoid metals.
Mention its uses.

Answer. 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.
(1) Mischmetal is used in cigarettes and gas lighters.
(2) It is used in ame throwing tanks.
(3) It is used in tracer bullets and shells.

Page : 242 , Block Name : Exercise

Q8.28 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.

Answer. 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 lled are called f-block
elements. Among the given atomic numbers, the atomic numbers of the inner transition elements
are 59, 95, and 102.

Page : 242 , Block Name : Exercise

Q8.29 Which is the last element in the series of the actinoids? Write the electronic con guration
of this element. Comment on the possible oxidation state of this element.

Answer. 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.

Page : 242 , Block Name : Exercise

Q8.30 Which is the last element in the series of the actinoids? Write the electronic con guration
of this element. Comment on the possible oxidation state of this element.

Answer.
The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and
its electronic configuration is [Rn]5f 146d 17s 2 . The most common oxidation state
displayed by it is + 3 ; because after losing 3 electrons it attains stable f 4 configuration.

Page : 242 , Block Name : Exercise

Q8.31 use Hund's rule to derive the electronic con guration of Ce 3 + ion and calculate its magnetic
moment on the basis Of ‘spin-only' formula.

Answer.
Ce : 1s 22s 22p 63s 23p 63d 104s 24p 64d 10s 25p 64f 15d ′6s 2
Magnetic moment can be calculated as:
μ = √n(n + 2)
Where,
n = number of unpaired electrons
In Ce, n = 2
Therefore, μ = √2(2 + 2)
= √2 × 4
= √8
= 2√2
= 2.828BM

Page : 243 , Block Name : Exercise

Q8.32 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
con gurations of these elements.

Answer.
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.

Ce after forming Ce 4 + attains a stable electronic configuration of [Xe]
Tb after forming Tb 4 + attains a stable electronic configuration of [Xe]4f 7Tb after forming
Eu 2 + attains a stable electronic configuration of [Xe]4f 7
Yb after forming Yb 2 + attains a stable electronic configuration of [Xe]4f 14

Page : 243 , Block Name : Exercise

Q8.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic con guration
(ii) oxidation states and
(iii) chemical reactivity.

Answer. Electronic con guration
The general electronic con guration for lanthanoids is [Xe] 544f 0 − 145d 0 − 16s 2 and that for actinoids
is [Rn] 865f − 1446d 0 − 17s 2 Unlike 4f orbitals, Sf orbitals are not deeply buried and participate in
bonding to a greater extent.

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- lled and half- lled
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.
Actinoid5 such as lanthanoids have more compounds in +3 state than in +4 state.

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 A1. Actlnoids, on the other hand, are highly reactive metals, especially when they are
nely 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
forrnation of a protective oxide.

Page : 243 , Block Name : Exercise

Q8.34 Write the electronic con gurations Of the elements with the atomic numbers 61, 91, 101,
and 109.

Answer.

Page : 243 , Block Name : Exercise

Q8.35 Compare the general characteristics of the rst 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 con gurations,
(ii) oxidation states,
(iii) ionisation enthalpies, and
(iv) atomic sizes.

Answer. (i) In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbitals are respectively
lled. We know that elements in the same vertical column generally have similar electronic
con gurations.
In the rst transition series, two elements show unusual electronic con gurations:
Cr(24) = 3d 54s 1
Cu(29) = 3d 104s 1

Similarly, there are exceptions in the second transition series. These are:
Mo(42) = 4d 55s 1 Rh(45) = 4d 85s 1
Tc(43) = 4d 65s 1 Pd(46) = 4d 105s 0
Ru(44) = 4d 75s 1 Ag(47) = 4d 105s 1

There are some exceptions in the third transition series as well. These are:
W(74) = 5d 46s 2
Pt(78) = 5d 96s 1
Au(79) = 5d 106s 1
As a result Of these exceptions, it happens many times that the electronic con gurations 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 rst transition
series. All metals present in the rst 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 Fe(Cn) 6 4 − , Co NH 3
6 )] [ (
3 + Ti H O
2 6 )]
3 + are stable complex, 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: WCl 6ReF 7, RuO 4, etc

(iii) In each of the three transition series, the rst ionisation enthalpy increases from left to right.
However, there are some exceptions. The rst ionisation enthalpies of the third transition series
are higher than those of the rst 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 rst ionisation enthalpies than elements corresponding to the same
vertical column in the rst transition series.
There are also elements in the 2nd transition series whose rst 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 rst 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.

Page : 243 , Block Name : Exercise

Q8.36 Write down the number of 3d electrons in each of the following ions:
Ti 2 + , V 2 + , Cr 3 + , Mn 2 + , Fe 2 + , Fe 3 + , CO 2 + , Ni 2 + and Cu 2 +
Indicate how would you expect the ve 3d orbitals to be occupied for these hydrated ions
(octahedral).

Answer.

Page : 243 , Block Name : Exercise

Q8.37 Comment on the statement that elements Of the rst transition series possess many
properties different from those Of heavier transition elements.

Answer. The properties of the elements of the rst transition series differ from those of the
heavier transition elements in many ways.
(i) The atomic sizes of the elements of the rst transition series are smaller than those of the
heavier elements (elements of 2nd and 3'e 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 rst transition series, while
higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomisation of the elements in the rst 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 rst 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 rst transition series form low-spin or high-spin complexes depending
upon the strength of the ligand eld. However, the heavier transition.

Page : 243 , Block Name : Exercise

Q8.38 What can be inferred from the magnetic moment values of the following complex species?
Example Magnetic Moment (BM)

[
K 4 Mn(CN) 6 2.2]
[ (
Fe H 2O )6 ]2 + 5.3
[ ]
K 2 MnCl 4 5.9

Answer. Magnetic movement (μ) is given as μ = √n(n + 2)For value
n = 1, μ = √1(1 + 2) = √3 = 1.732 For value n = 2, μ = √2(2 + 2) = √8 = 2.83
For value n = 3, μ = √3(3 + 2) = √15 = 3.87
For value n = 4, μ = √4(4 + 2) = √24 = 4.899
For value n = 5, μ = √5(5 + 2) = √35 = 5.92

[
(i)K 4 Mn(CN) 6 ]
For in transition metals, the magnetic moment is calculated from the spin-only formula.
Therefore,
√n(n + 2) = 2.2We can see from the above calculation that the given value is closest to n = I . 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 eld ligand that causes the pairing of electrons.

[ (
(ii) Fe H 2O ) 6 ] 2 + We can see from the above calculation that the given value is closest to . Also,
√n(n + 2) = 5.3
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 H 2O is a weak eld ligand and does not cause the pairing of electrons.

[
(iii) K 2 MnCl 4 ]
√n(n + 2) = 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 eld ligand and does not cause the pairing of electrons.

Page : 243 , Block Name : Exercise