NCERT Solutions For Class 12 Chemistry Chapter 8

NCERT Solutions Class 12 Chemistry The d and f Block Elements

NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements are provided here in detail. The chapter 12 of NCERT class 12 chemistry book i.e. The d and f block elements introduces various important topics to the students. In this chapter, students learn the electronic configuration and different properties of the d and f block elements in detail.

NCERT class 12 chemistry solutions for chapter 8 The d and f block elements provided here can help the students clear all their doubts and understand the fundamentals better. These solutions are easy to understand and are given in a simple way. Check the class 12 NCERT solution for chapter 8 i.e. The d and f block elements below. The NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements PDF is also available that the students can download.

Q 8.1:

Write the electronic configuration of:

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

Ans :

(i) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6s6

Or, [Rn]86

(ii) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1

Or, [Xe]54 2f14 3d3

(iii) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6

Or, [Xe]54

(iv) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4

Or, [Xe]54 3d3

(v) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5

Or, [Ar]18 3d5

(vi) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7

Or, [Ar]18 3d7

(vii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10

Or, [Ar]18 3d10

(viii) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3

Or, [Ar]18 3d3

 

 

Q 8.2:

When comparing the oxidation of the +3 state of Fe2+ with Mn2+, why do we find the latter more stable than the former?

Ans :

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

Mn2+ is [Ar]18 3d5.

We know that half and completely filled orbitals are more stable. Hence, Mn with (+2) state has a stable d5 configuration, which is why Mn2+ shows resistance to oxidation to Mn3+. We know that, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe+3 oxidation state.

 

 

Q 8.3:

With increasing atomic number, show how +2 state becomes more stable in the first row of transition elements.

Ans :

In the table below, the oxidation states of the first half of the first row of transition metals are displayed.

3.1

We can see that except, Sc all other metals have a +2 oxidation state. As moving in the increasing order of atomic number from 21 to 25 , that is from Sc to Mn, the number of electrons in 3d orbitals 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).

 

 

Q 8.4:

Explain with examples, how much the electronic configurations of the first series of the transition elements affects the stability of their oxidation states

Ans :

The oxidation states with the maximum number (+2 to +7) are exhibited by the elements of Mn oxidation states that are in the first-half of the transition series. T With the increase in atomic number, the stability of +2 oxidation state increases. Once more electrons are filled in the d-orbital, this happens to a great extent. The +2 oxidation state is not shown by the Sc. Its electronic configuration is 4s2 3d1. 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

 

 

Q 8.5:

Find 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 :

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

 

 

Q 8.6:

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

Ans :

(i) Chromate, \(CrO^{2-}_{4}\)

Oxidation state of Cr is + 6.

(ii) Permanganate, \(MnO^{-}_{4}\)

Oxidation state of Mn is + 7.

(iii) Vanadate, \(VO^-_3\)

Oxidation state of V is + 5.

 

 

Q 8.7:

Explain the lanthanoid contraction. What are the consequences of lanthanoid contraction?

Ans:

On moving along the lanthanoid series, the atomic number gradually increases by one. With the ncrease in atomic number, the number of protons and electrons present in the atom also increases by one. The effective nuclear charge increases as electrons are being added to the same shell . Owing to the proton addition being pronounced more above the interelectronic repulsions resulting from electron addition, nuclear attraction increase happens. 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) 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.)

(ii) There is similarity in the properties of second and third transition series.

(iii) Separation of lanthanoids is possible due to lanthanide contraction

 

 

Q 8.8:

What are the characteristics of the transition elements and why are they called transition elements? Which elements belonging to 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 are found in the d-block and show a transition of properties between s-block and p-block. Hence, these are called transition elements.

Elements such as Cd, Hg and Zn cannot be called as transition elements because they have completely filled d-sub shell.

 

 

Question 8.9:

How does the electronic configuration of non–transition elements vary with that of transition elements?

Ans :

Transition metals have a partially filled d−orbital. Hence, the electronic configuration of transition elements is (n − 1)d1-10 ns0-2.

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 ns2 np1-6.

 

 

Q 8.10:

State the varied kinds of oxidation states shown by the lanthanoids ?

Ans :

In the case of lanthanoids , +3 oxidation state is most common.

That is, Ln(III) compounds are predominant.

However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

 

 

Q 8.11:

Explain the following while giving reasons:

(i) Transition metals and their many compounds act as good catalyst.

(ii) The transition metals generally form coloured compounds.

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

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

Ans :

(i) The catalytic activity of the transition elements can be explained by two basic facts.

(a) Transition metals provide a suitable surface for the reactions to occur.

(b) 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.

(ii) 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 toanother. 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.

(iii) 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.

(iv) 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.

 

 

Q 8.12:

What are interstitial compounds? Why are they well known for transition metals?

Ans :

Transition metals are large in size and they also contain lots of interstitial sites.

These interstitial sites can be used to trap atoms of other elements (that have small atomic size), such as H, C, N. The compounds resulting are called interstitial compounds.

 

 

Q 8.13:

Comparing transition metals and non transition metals, how does their oxidation states vary ? Explain with examples.

Ans :

In the case of 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.

 

 

Q 8.14:

From iron chromite ore, describe the preparation of potassium dichromate. On increasing pH, what effect will it have on a solution of potassium dichromate?

Ans :

From iron chromite ore, Potassium dichromate is prepared as in the following steps.

Step (1): Preparation of sodium chromate

\(4Fe Cr_2 O_4 +16 Na OH+7 O_2\rightarrow 8 Na_2 Cr O_4 + 2Fe_2 O_3 + 8 H_2 O\)

Step (2): Conversion of sodium chromate into sodium dichromate

\(2Na Cr O_4 + conc. H_2 SO_4 \rightarrow Na_2 Cr O_7 + Na_2 SO_4 + H_2 O\)

Step (3): Conversion of sodium dichromate to potassium dichromate

\(Na Cr_2 O_7 + 2KCl \rightarrow K_2 Cr O_7 + 2Na Cl\)

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

The dichromate ion \((Cr_2 O_7^{2-})\) exists in equilibrium with chromate \((CR O_4^{2-})\) ion at pH 4.

However, by changing the pH, they can be inter converted.

14.1

 

 

Q 8.15:

Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with:

(i) H2S (ii) iodide and (iii) iron(II) solution

Ans :

\(k_2 Cr_2 O_7\) acts as a very strong oxidising agent in the acidic medium.

\(k_2 Cr_2 O_7 + 4 H_2 SO_4\rightarrow K_2 SO_4 + Cr_2(SO_4)_3 + 4 H_2O+3[O]\)

\(k_2 Cr_2 O_7\) takes up electrons to get reduced and acts as an oxidizing agent. The reaction of \(k_2 Cr_2 O_7\) with H2S, iron (II) solution and other iodide are given below.

(i) \(k_2 Cr_2 O_7\) oxidizes H2S to sulphur.

15.3

(ii) \(k_2 Cr_2 O_7\) oxidizes iodide to iodine.

15.1

(iii) \(k_2 Cr_2 O_7\) oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferricions.

15.2

 

 

Q 8.16:

Explain how potassium permanganate is prepared. Also find the reactivity of the potassium permanganate solution with

(i) oxalic acid

(ii) iron(II) ions and

(iii) SO2

Write down the chemical equations for the reactions.

Ans :

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.

\(2 Mn O_2 + 4 KOH + O_2 \overset{heat}{\rightarrow} 2K_2 MnO_4 + 2 H_2 O\)

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

\(K_2 Mn O_4\leftrightarrow 2k^+ +MnO^{2-}_4\) \(H_2 O\leftrightarrow H^+ + OH^{-}\)

At anode, manganate ions are oxidized to permanganate ions.

\(MnO^{2-}_4\leftrightarrow MnO^-_4 + e^-\)

Oxidation by chlorine

\(2 K_2 MnO_4 + Cl_2 \rightarrow 2MnO_4 + 2KCl\) \(2 MnO_4^{2-} + Cl_2 \rightarrow 2MnO_4^- + 2Cl^-\)

Oxidation by ozone

\(2 KMnO_4 + O_3 + H_2O \rightarrow 2KMnO_4 + 2KOH + O_2\) \(2 MnO_4^{2-} + O_3 + H_2O \rightarrow 2MnO_4^{2-} + 2OH^- + O_2\)

(i) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide

16.3

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

16.1

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

16.2

 

 

Q 8.17: For M3+/M2+ and M2+/M systems, the \(E^{\Theta}\) values for some metals are as follows:

Fe2+ /Fe −0.4V

Fe3+/Fe2+ +0.8 V

Cr2+/Cr −0.9V

Cr3/Cr2+ −0.4 V

Mn2+/Mn −1.2V

Mn3+ /Mn2+ +1.5 V

Comment upon:

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

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

Ans :

(i) 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

(ii) The \(E^{\Theta}\) value for Fe3+/Fe2+ 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+

 

 

Q 8.18:

Figure out which of these will be colored in aqueous solution and Give reasons for each. [ Co2+, Fe3+, Mn2+, Sc3+, Cu+, V3+, Ti3+ ]

Ans :

Ions with electrons in d-orbital will be the only ones that will be coloured and the ions with empty d-orbital will be colourless.

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

All other ions, except Sc3+, will be coloured in aqueous solution because of d−d transitions. Not Sc3+ as it has an empty d-orbital.

Q 8.19:

For the elements of the first transition series, compare the stability of +2 oxidation state

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 table above:

Mn shows maximum number of oxidation states, changing between +2 to +7.

The number of oxidation states increases as we move on 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 to bottom from the top, it becomes more and more difficult to remove the third electron from the d-orbital.

Question 8.20:

Note the similarity or dissimilarity between the chemistry of lanthanoids with that of the actinoids with special reference to:

(i) Chemical reactivity

(ii) Atomic and ionic sizes

(iii) Oxidation state

(iv) Electronic configuration

Ans :

(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) 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.

(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.

(iv) Electronic configuration

The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

 

 

Question 8.21:

How would you account for the following?

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

(b) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.

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

Ans:

(i) 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.

(ii) 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 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.

(iii) 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) released in the presence of strong field ligands overcomes this ionization energy.

Q 8.22:

Explain the term ‘disproportionation’? Give examples of disproportionation reaction in aqueous solution.

Ans :

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,

(a) \(3Cr O^{3-}_4 + 8H^+ \rightarrow 2Cr O^{2-}_4 + Cr^{3+} + 4H_2 O\)

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

(b) \(3Mn O^{2-}_4 + 4H^+ \rightarrow 2Mn O^{-}_4 + Mn O_2 + 2H_2 O\)

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

 

 

Question 8.23:

In the first series of transition metals, which metal exhibits +1 oxidation state the most and why?

Ans:

In the first series of transition metals, Cu exhibits +1 oxidation state very frequently. We know Cu has an electronic configuration of [Ar] 3d 10, that is, the completely filled d-orbital makes it highly stable.

 

 

Question 8.24:

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
(i) Mn2+, [Ar] 3d4 4
(ii) Cr3+,[Ar] 3d3 3
(iii) V3+,[Ar] 3d2 2
(iv) Ti3+, [Ar]3d1 1

Cr3+ is the most stable in aqueous solutions owing to a \(t^3_{2g}\) configuration.

 

 

Question 8.25:

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

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

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

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

Ans:

(a) 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 , the oxidation state of Mn is +7.

(b) 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, \(Mn^{II}O\) is basic and \(Mn^{VII}_2O_7\) is acidic.

(c) 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.

Q 8.26:

Explain the procedure involved in the manufacture of :

(i) KMnO4 from pyrolusite ore.

(ii) K2Cr2O7 from chromite ore.

Ans:

(i)

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.

\(2MnO_2 + 4KOH + O_2 \overset{heat}{\rightarrow} 2K_2MnO_4 + 2H_2O\)

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

\(K_2MnO_4\leftrightarrow 2K^+ + MnO^{2-}_4\) \(H_2O\leftrightarrow H^+ + OH^-\)

At anode, manganate ions are oxidized to permanganate ions.

\(MnO^{2-}_4\leftrightarrow MnO^-_4 + e^-\)

Oxidation by chlorine

\(2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl\) \(2MnO^{2-}_4 + Cl_2 \rightarrow 2MnO^-_4 + 2Cl^-\)

Oxidation by ozone

\(2K_2MnO_4 +O_2 + H_2O \rightarrow 2 KMnO_4 + 2KOH + O_2\) \(2MnO^{2-}_4 +O_2 + H_2O \rightarrow 2 MnO^{2-}_4 + 2OH^- + O_2\)

(ii)

Potassium dichromate ( K2Cr2O7 ) is prepared from chromite ore (FeCr2O4) in the following steps.

Step (1): Preparation of sodium chromate

\(4FeCr_2O_4 + 16 NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2 Fe_2O_3 + 8H_2O\)

Step (2): Conversion of sodium chromate into sodium dichromate

\(2Na_2CrO_4 + Conc. H_2SO_4 \rightarrow Na_2Cr_2O_3 + Na_2SO_4 + H_2O\)

Step(3): Conversion of sodium dichromate to potassium dichromate

\(Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl\)

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

The dichromate ion \(Cr_2O^{2-}_7\) exists in equilibrium with chromate ion \(CrO^{2-}_4\) at pH 4.

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

26.1

 

 

Question 8.27:

What are alloys? Find an important alloy containing the alloys some lanthanoid metals and give out a few of its uses.

Ans :

An alloy is a solid solution of two or more elements in a metallic matrix. It can be a partial solid solution or a complete solid solution. Alloys generally are created as they have varied properties compared to their constituent elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of Al, Ca, Si, C, S.

Uses

(a) It is used in tracer bullets and shells.

(b) Mischmetal is used in cigarettes and gas lighters.

(c) It is used in flame throwing tanks.

 

 

Question 8.28:

Give a description of inner transition elements? Figure out, of the given atomic numbers, which are the atomic numbers of the inner transition elements: 104, 102, 95, 74, 59, 29.

Ans :

Inner transition metals : 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.

From the given atomic numbers, the inner transition elements are the ones with atomic numbers 59, 95, and 102.

 

 

Q 8.29:

When comparing the chemistry of lanthanoids and actinoids, the latter’s chemistry is said to be not as smooth as the former. Verify the statement with examples in accordance to the oxidation states of these elements.

Ans :

Lanthanoids project 3 oxidation states (+2, +3, +4). From these, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbital’s is quite large. While, the energy difference between 5f, 6d, and 7s orbital’s is very less. Hence, actinoids display a large number of oxidation states.

For example, neptunium displays +3, +4, +5, and +7 while uranium and plutonium display +3, +4, +5, and +6 oxidation states. The most common oxidation state in case of actinoids is also +3.

Q 8.30 :

In the series of the actinoids, which is the last element ? Give the electronic configuration of the element and give an idea of the possible oxidation state of this element.

Ans :

In the actinoid series, the last element is lawrencium, Lr. The atomic number of the element is 103 and its electronic configuration is [Rn] 5f14 6d1 7s2 . The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.

Q 8.31:

Derive the electronic configuration of Ce3+ ion by using Hund’s rule and also calculate its magnetic moment on the basis of ‘spin-only’ formula.

Ans :

\(Ce \; : \; 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^{10} \; 4s^2 \; 4p^6 \; 4d^{10} \; 5s^2 \; 5p^6 \; 4f^1 \; 5d^1 \; 6s^2\)

Magnetic moment can be calculated as:

\(\nu = \sqrt{n(n+2)}\)

Where,

n = number of unpaired electrons

In Ce, n = 2

\(\nu = \sqrt{2(2+2)}\) \(= \sqrt{2\times 4)}\) \(= \sqrt{2\times 4)}\) \(= 2\sqrt{2)}\)

= 2.828 BM

 

 

Q 8.32:

In the group of lanthanoids, find the elements exhibiting +4 oxidation state and +2 oxidation state. Relate this behavior with the electronic configurations of these elements.

Ans :

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

In the parenthesis are the atomic numbers of the elements are given.

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

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

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

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

 

 

Q 8.33:

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

(i)chemical reactivity.

(ii) electronic configuration

(iii)oxidation states

Ans :

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).

Electronic configuration

The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried andparticipate 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-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.

Q 8.34 :

Of the elements with the atomic numbers 61, 91, 101, and 109, write down the electronic configurations

Ans :

Atomic number Electronic configuration
61 [Xe]544f55d06s2
91 [Rn]565f26d17s2
101 [Rn]565f135d07s2
109 [Rn]565f146d77s2

 

Q 8.35:

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) atomic sizes.

(ii) ionisation enthalpies

(iii) oxidation states

(iv) electronic configurations,

Ans :

(i) 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.

(ii) 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.

(iii) 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.

Some stable complexes are :

\({\left [ Fe^{II}(Cn)_6 \right ]}^{4-}\),

\({\left [ Co^{III}(NH_3)_6 \right ]}^{3+}\),

\({\left [ Ti(H_2O)_6 \right ]}^{3+}\)

The issue is that 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.

(iv) 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:

Cu (29) = 3d104s1

Cr (24) = 3d54s1

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

Ag(47) = 4d105s1

Pd(46) = 4d10 5s0

Rh(45) = 4d85s1

Ru (44) = 4d75s1

Tc (43) = 4d65s1

Mo (42) = 4d55s1

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

Au (79 ) = 5d10 6s1

Pt (78) = 5d9 6s1

W (74) = 5d4 6s2

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

Q 8.36:

Find the number of 3d electrons in each of the following ions:

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

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

Ans :

Metal ion Number of d-electrons Filling of d-orbitals

Cu2+ 9 \(t^{6}_{2g}e^{3}_{g}\)
Ni2+ 8 \(t^{6}_{2g}e^{2}_{g}\)
CO2+ 7 \(t^{5}_{2g}e^{2}_{g}\)
Fe3+ 5 \(t^{3}_{2g}e^{2}_{g}\)
Fe2+ 6 \(t^{4}_{2g}e^{2}_{g}\)
Mn2+ 5 \(t^{3}_{2g}e^{2}_{g}\)
Cr3+ 3 \(t^{3}_{2g}\)
V2+ 3 \(t^{3}_{2g}\)
Ti2+ 2 \(t^{2}_{2g}\)

 

 

 

Q 8.37:

Compared to heavier transition elements, first transition series elements vary in a lot of ways properties vise. Comment on the statement.

Ans :

By many ways, the properties of elements of heavier transition elements differs from those of first transition series.

 

(a) 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.

(b) 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.

(c) The meltingand 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).

(d) For heavier elements , the higher oxidation states are more common whereas for first transition series elements the +2 and +3 oxidation states are more common.

(e) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

 

 

Q 8.38:

From the below complex species, what could be inferred from the magnetic moment values?

Example Magnetic Moment (BM)

K2[MnCl4] = 5.9

K4[Mn(CN)6] = 2.2

[Fe(H2O)6]2+ = 5.3

Ans :

Magnetic moment ( . \(\nu = \sqrt{n(n+2)}\) )

For value n = 1, \(\nu = \sqrt{1(1+2)} = \sqrt{3} = 1.732\)

For value n = 2, \(\nu = \sqrt{2(2+2)} = \sqrt{8} = 2.83\)

For value n = 3, \(\nu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\)

For value n = 4, \(\nu = \sqrt{4(4+2)} = \sqrt{24} = 4.899\)

For value n = 5, \(\nu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\)

(a) K2[MnCl4]

\(\sqrt{n(n+2)} = 2.2\)

What we could see from the calculation above is that n=5 would be the closest value given to the term.Here the +2 oxidation state in the complex is Mn. This would mean that in the d-orbital, the Mn has 5 electrons.

Hence, we can say that Cl is a weak field ligand and does not cause the pairing of electrons.

(b) K4[Mn(CN)6]

\(\sqrt{n(n+2)} = 2.2\)

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.

(c) [Fe(H2O)6]2+

\(\sqrt{n(n+2)} = 5.3\)

Thus the calculation given above proves that the closest value to n= 4 . In addition to this, the +2 oxidation state in this complex is none other than Fe. The d-orbital for Fe would have 6 electrons.

 

Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.