Class 12 - Chemistry - D and F Block Elements

Question 8.1.

Write down the electronic configuration of:

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

Answer:

  • Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or, [Ar]18 3d3
  • Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4

Or, [Xe] 54 3d3

  • Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10
  • Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or, [Xe]54
  • Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar]18 3d7
  • Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1

Or, [Xe] 54 2f14 3d3

  • Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5
  • Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d1 5s2 5p6 5d10 6s2 6s6 Or, [Rn]86

 

 

Question 8.2.

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

Answer:

Electronic configuration of Mn2+ is [Ar] 18 3d5 and Electronic configuration of Fe2+ is [Ar] 18 3d6.

It is known that half-filled and fully-filled orbitals are more stable.

Therefore, Mn in (+2) state has a half-filled stable configuration, whereas the Fe in +3 oxidation state has partially filled subshells,

which are relatively unstable. This is the reason Mn2+ shows resistance to oxidation to Mn3+.

Also, Fe2+ has 3d6 configurations and by losing one electron, it attains half-filled stable configuration.

Hence, Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state.

 

 

Question 8.3.

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

Answer:
As the atomic number increases from 21 to 25, the number of electrons in the 3d-orbital also increases from 1 to 5, +2

oxidation state is attained by the loss of the loss of the two of the two 4s electrons by these metals. Sc, does not exhibit +2 oxidations state.

As the number of d-electrons in +2 state increases from Ti to Mn, the stability of +2 state increases (d –orbital gradually becoming half filled).

Mn (+2) has d5 electrons which is highly stable.

 

Question 8.4.

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

Answer:

The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting a 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 ([Ar] 3d14s2) does not show +2 oxidation states, instead, it loses all the three valence electrons to form Sc3+.

The +3 oxidation state of Sc is very stable as it attains stable configuration.

For Mn ([Ar] 3d54s2), +2 oxidation state is very stable because after losing two electrons,

it attains stable half-filled structure.

 

Question 8.5.

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?

Answer:

The stable oxidation state of transition element with the above electron configuration in ground state of atoms is as follows:-

3d3 (Vanadium): +2, +3, +4, and +5

3d5 (chromium): +3, +4, +6,

3d5 (manganese):  +2, 4, +6 and +7

3d8 (cobalt): +2, +3(in complexes)

3d4: There is no d4 configuration in the ground state.

Lower the stable oxidation state generally leads to ionic bond and higher oxidation state corresponds to covalent bond.

 

Question 8.6.

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.

Answer:

(i) Vanadate (VO3-)-Oxidation state of V is + 5.

(ii) (CrO7)2- and (CrO4 2- )-Oxidation state of Cr is + 6.

(iii) Permanganate (MnO4 -) Oxidation state of Mn is + 7.

 

Question 8.7.

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

  • There is similarity in the properties of second and third transition series.
  • Separation of lanthanoids is possible due to lanthanide contraction.
  • 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.)

 

Question 8.8.

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 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.

 

  

Question 8.9.

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

Answer:

Transition metals have a partially filled d−orbital.

Therefore, the electronic configuration of transition elements is (n−1) d1-10 ns0-2 .

The non-transition elements either do not have a partially filled d−orbital.

Therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np1-6.

 

Question 8.10.

What are the different oxidation states exhibited by the lanthanoids?

Answer:

In the lanthanide series, +3 oxidation state is most common i.e., Ln (III) compounds are predominant.

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

 

Question 8.11.

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:

  • 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.
  • 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.
  • 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.
  • 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.

 

Question 8.12.

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

Answer:

Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice.

Interstitial compounds are well known for transition metals because small-sized atoms of H, B, C, N etc.

can easily occupy positions in the voids present in the crystal lattices of transition metals.

 

Question 8.13.

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 (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.

 

Question 8.14.

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 (FeCr2O4) by the following steps:

Step1: Preparation of sodium dichromate in the reaction of Chromite ore with sodium hydroxide and oxygen gas.

4FeCr2O4 + 16NaOH + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8H2O

Step2: Conversion of Sodium Chromate on reaction with concentrated Sulfuric acid gives Sodium dichromate as a product.

2Na2CrO4 + conc. H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

Step3: Sodium dichromate on reaction with potassium chloride converts to potassium dichromate as a product.

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

As potassium dichromate is less soluble than sodium chloride so, potassium dichromate is obtained in form of orange crystals.

Dichromate ion exists in equilibrium with chromate ion at around pH. However, by changing the pH they can be interconverted.

 

Question 8.15.

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

Answer:

K2Cr2O7 acts as a very strong oxidizing agent in acidic medium.

K2Cr2O7 gets reduced and acts as an oxidizing agent by oxidizing Iodide to iodine

(i) K2Cr2O7 oxidizes iodide to iodine

Cr2O72-+ 14H++ 6e-→ 2Cr3+ + 7H2O

2I-→ (I2 + 2e-} x3

Overall: Cr2O72-+ 14H++ 6I-→ 2Cr3+ + 7H2O+ 3I2

(ii) K2Cr2O7 oxidizes iron (II) to iron (III)

Cr2O72-+ 14H++ 6e-→ 2Cr3+ + 7H2O

Fe2+→ {Fe3+ + e-} x6

Overall: Cr2O72-+ 14H++ 6Fe2+→ 2Cr3+ + 7H2O + 6Fe3+

(iii) K2Cr2O7 oxidized H2S to sulphur

Cr2O72-+ 14H++ 6e-→ 2Cr3+ + 7H2O

H2S → {S + 2H+ + 2e-} x3

Overall: Cr2O72-+ 8H++ 3H2S → 2Cr3+ + 7H2O +3S

  

Question 8.16.

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.

Answer:

Potassium permanganate can be prepared from MnO2.

It can be done by fusing the ore with KOH in presence of an oxidizing agent like atmospheric oxygen/KNO3 etc. to give green coloured K2MnO4 as the product.

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

The Product K2MnO4 is extracted with water and then oxidised by passing ozone/chlorine into the solution or electrolytically.

Electrolytic oxidation:

K2MnO42- ßà 2K+ + MnO42-

H2O ßà H+ + OH-

At Anode, manganite ions are oxidized to permanganate ions.

MnO42- ßà MnO4- + e-

(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions and water as product

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Fe2+→ {Fe3++ e-} x5

Overall: MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid as follows:

MnO4- + 6H+ + 5e- → {Mn2+ + 3H2O} x2

2H2O + 2SO2 + O2→ {4H+ + 2SO42- + 2e-} x5

Overall: MnO4- + 4H2O + 10SO2 + 5O2→ 8H+ + 10SO42- + 2H+

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

MnO4- + 8H+ + 5e- → {Mn2+ + 4H2O} x2

C2O42-→ 2CO2 + 2e-} x5

Overall: 2MnO4- + 16H++ 5C2O42- → 2Mn2+ + 10CO2+ 8H2O

  

Question 8.17.

For M2+/M and M3+/M2+ systems the EV values for some metals are as follows:

Cr2+/Cr        -0.9V           Cr3/Cr2+ -0.4 V

Mn2+/Mn     -1.2V          Mn3+/Mn2+ +1.5 V

Fe2+/Fe       -0.4V         Fe3+/Fe2+ +0.8 V

Use this data to comment upon:

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

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

 

Answer:

(i) The value E(-)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+

(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

 

 

Question 8.18.

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

Answer:

Metal ions which have valence electrons in d-orbital and in which d-d transition can take place will be coloured and the metal ions which have completely filled orbital or have d-orbital 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.

All other ions, except Sc3+, will be coloured in aqueous solution because of d−d transitions.

 

Question 8.19.

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

Answer:

Sc

 

 

+3

 

 

 

 

Ti

+1

+2

+3

+4

 

 

 

V

+1

+2

+3

+4

+5

 

 

Cr

+1

+2

+3

+4

+5

+6

+7

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

 

 

 

 

 

 

It can be observed from the above table that in the starting of 3d transition series elements like Sc, Ti, V, Cr in +2 state are not that stable in their elements in the +3 state.

In the middle Mn2+, Fe2+, Co2+are quite known. In fact, Mn2+ and Mn7+ are most stable states in Mn.

Fe2+ is less stable when compared to Fe3+ which is due to fact that Fe3+ is able to lose one electron to acquire d5 state which has extra stability.

Co2+ is less stable as compared to Co3+. Ni2+ is the most common and stable among its +2, +3, +4 states.

Cu2+ is more stable and is quite common as compared to Cu+.

Towards the end, Zn forms only Zn2+ which is highly stable as it has 3d10 states.

Note: As it becomes difficult to remove the third electron from d-orbital, the stability of +2 oxidation state increases from top to bottom

 

Question 8.20.

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

(i) Electronic configuration

(ii) Atomic and ionic sizes and (iii) oxidation state (iv) chemical reactivity.

Answer:

(i) Electronic configuration

The general electronic configuration for lanthanoids is [Xe] 54 4f0-14 5d 0-1 6s2 and that for actinoids is [Rn] 86 5f1-14 6d 0-1 7s2. Unlike 4f orbitals,

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

(ii) Oxidation states

The principal oxidation state of lanthanoids is (+3).

However, sometimes we also encounter oxidation states of + 2 and + 4.

This is because of extra stability of fully-filled and half-filled orbitals.

Actinoids exhibit a greater range of oxidation states.

This is because the 5f, 6d, and 7s levels are of comparable energies.

Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 states than in +4 states.

(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.

  1. 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).

 

Question 8.21.

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

Answer:

  • Cr2+ is strongly reducing in nature. It has a d4 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.
  • 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.
  • The ions in d1 configuration tend to lose one more electron to get into stable d0
  • 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.

 

Question 8.22.

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.

Example:

2MnO42- + 4H+→ 2MnO4- + MnO2 + 2H2O

Mn (VI) is oxidised to Mn (VII) and also reduced to Mn (IV).

2CrO43- +2H+→ CrO42- +Cr3+ + 4H2O

Cr (V) is oxidised to Cr (VI) and also reduced to Cr (III).

 

Question 8.23.

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

Answer:

Copper (29) has electronic configuration 1s22s22p63s23p63d104s1.

It can easily lose one electron to give stable configuration as it has completely filled d-orbital.

 

 

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?

Answer:

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 t3 2g configuration.

 

Question 8.25.

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.

 

Answer:

(i) The lower oxide have low oxidation state while the higher oxide has high oxidation state, example MnO is basic and Mn2O7 is acidic.

(ii) Oxygen and fluorine have a small size and high electronegativity and can easily oxidize metals, example V2O5.

(iii) Oxoanions of metals have higher oxidation states because of

high electronegativity of oxygen and highly oxidizing property example, Cr in CrO72- has an oxidation state of +6

 

Question 8.26.

Indicate the steps in the preparation of:

(i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore

Answer:

Preparation of K2Cr2O7 is prepared from chromite ore (FeCr2O4) in the following 3 steps:-

Step 1:- Conversion of chromite ore into sodium chromate

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

Step 2:- Conversion of sodium chromate into sodium dichromate

2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

Step 3:- Conversion of sodium dichromate into potassium dichromate

Na2Cr2O7 + 2KCl → K2Cr2O7 + 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 (Cr2O7)2- exists in equilibrium with chromate ion (Cr2O4)2- at pH 4.

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

In presence of Acid and Alkali

Cr2O42-            à         2HCrO4-                    ßCr2O72-

Chromate (Yellow)       Hydrogen Chromate   Dichromate (Orange)

(ii) KMnO4 from pyrolusite ore -

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.

                              (Heat)

2MnO2 + 4KOH + O2 à 2K2MnO4 + 2H2O

                                       (Green)

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

Electrolytic –oxidation

K2MnO4 à 2K+ + MnO42-

H2O à H+ + OH-

At anode, manganate ions are oxidized to permanganate ions.

MnO42- ßà MnO4- + 2e-

 Green         Purple

Oxidation by chlorine

2K2MnO4 + Cl2 --> 2KMnO4 + 2KCl

2MnO42- + Cl2 --> 2MnO4- + 2Cl-

Oxidation by ozone

2K2MnO4 + O3 + H2O --> 2KMnO4 + 2KOH + O2

2MnO4- + O3 + H2O --> 2MnO42- + 2OH- + O2

 

Question 8.27.

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.

Uses

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

(2) It is used in flame throwing tanks.

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

 

Question 8.28.

What are inner transition elements?

Decide which of the following atomic numbers the atomic numbers of the inner transition elements are: 29, 59, 74, 95, 102, and 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 filled are called f-block elements.

Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.

 

Question 8.29.

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.

Answer:

Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 states are 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.

 

Question 8.30.

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.

Answer:

The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is [Rn]5 f146d17s2.

The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.

 

Question 8.31.

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

Answer:

Ce (Z = 58) = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f1 5d1 6s2

Magnetic moment µ = √n (n + 2)

Where n = number of unpaired electrons.

In Ce, n=2

Therefore, µ = √2 (2 + 2)

= √2x4

=√8

=2√2

= 2.828 BM.

  

Question 8.32.

Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states.

Try to correlate this type of behaviour with the electronic configurations 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.

+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.

 

 Question 8.33.

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

(i) Electronic configuration (ii) oxidation states and (iii) chemical reactivity

Answer:

Electronic Configuration

The general electronic configuration for lanthanoids is [Xe] 54 4f0-145d0-1 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.

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 states than in +4 states.

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, is 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).

 

Question 8.34.

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

Answer:

Z = 61 (Promethium, Pm) ⇒ [Xe] 54 4f5 5d° 6s2

 Z = 91 (Protactium, Pa) ⇒ [Xe] 86 4f2 5d1 7s2

 Z = 101 (Mendelevium, Md) ⇒ [Xe] 86 4f13 5d° 7s2

 Z = 109 (Meitnerium, Mt) ⇒ [Xe] 86 4f14 5d7 7s2

 

Question 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) Electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes.

 

Answer:

(a) Electronic configuration: The elements in the same vertical column generally have similar electronic configuration.

First transition series shows only two exceptions (i.e. Cr = 3d5 4s1 and Cu = 3d10 4s1).

But second transition series shows more exceptions (i.e. Mo = 4d5 5s1Tc = 4d6 5s1, Ru = 4d7 5s1, Rh = 4d8 5s1 Pd = 4d10 5s0, Ag = 4d105s1).

In third transition, there are many exceptions (i.e. W = 5d4 6s2, Pt = 5d9 6s1 and Au = 5d10 6s1).

Thus in the same vertical column, in a number of cases, the electronic configuration of the three series are not similar.

(b) Oxidation states: The elements in the same vertical column generally show similar oxidation states.

The number of oxidation states shown by the elements in the middle of each series is maximum and 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 [Fe(CN)6]4- , [Co(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.

(c) Ionisation enthalpies:

The first ionisation enthalpies in each series generally increases gradually as we move from left to right

though some exceptions are observed in each series.

The first ionisation enthalpies of some elements in the second (4d) series are higher while some of them have

lower value than the elements of 3d series in the same vertical column.

However, the first ionisation enthalpies of third (5d) series are higher than those of 3d and 4d series.

This is because of weak shielding of nucleus by 4f-electrons in the 5d series.

(d) Atomic sizes: In general, ions of the same charge or atoms in a given series show

progressively decrease in radius with increasing atomic number though the decrease is quite small.

But the size of the atoms of the 4d series is larger than the corresponding elements of the 3d series,

whereas those of corresponding elements of the 5d-series nearly the same as those of 4d series because of lanthanoid contraction.

(e) Enthalpies of atomisation:

The metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series.

This is due to much more frequent metal bonding in compounds of heavy transition metals.

 

Question 8.36.

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 you would expect the five3d orbitals to be occupied for these hydrated ions (octahedral).

Answer:

Metal ion

Number of d-electrons

Filling of d-orbitals

Ti2+

2

t22g

V2+

3

t32g

Cr3+

3

t32g

Mn2+

5

t32g e2g

Fe2+

6

t42g e2g

Fe3+

5

t32g e2g

Co2+

7

t52g e2g

Ni2+

8

t62g e2g

Cu2+

9

t62g e3g

 

Question 8.37.

Comment on the statement that elements of the first transition series possess

many properties different from those of heavier transition elements.

Answer:

The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.

  • 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.

  • +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.
  • 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.
  • 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).
  • 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.

 

Question 8.38.

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

Answer:

Magnetic Moment 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) K4 [Mn (CN) 6]

For in transition metals, the magnetic moment is calculated from the spin-only formula.

Therefore,

μ = √n (n+2) = 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 and μ = 2.2BM shows that it has only one unpaired electron.

Hence, we can say that CN is a strong field ligand that causes the pairing of electrons.

The hybridisation is d2 sp3 forming inner-orbital octahedral complex.

(ii) [Fe (H2O) 6]2+: In this complex, Fe is in +2 oxidation state and m = 5×3.

It means that there are four unpaired electrons in 3d. Also, the 3d electrons do not pair up when the H2O molecules approach.

 Thus, H2O is a weak ligand. The hybridisation involved is sp3d2, forming an outer-orbital octahedral complex.

(iii) K2 [MnCl4]: In this complex, Mn is in +2 oxidation state and m = 5 × 92.

It means that there are five unpaired electrons.

The hybridisation involved is sp3, forming a tetrahedral complex. Cl is a weak ligand

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