The d – and f – Block Elements Class 12 Chemistry Exam Questions

Please refer to the below The d – and f – Block Elements important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.

Short Answer Questions :

Question. (i) Write two characteristic of the transition elements.
(ii) Which of the 3d-block elements may not be regarded as the transition elements and why?
Answer : Elements which have incompletely filled d-orbitals in their ground state or in any one of their oxidation states are called transition elements.
Characteristics of transition elements :
1. They show variable oxidation states.
2. They exhibit catalytic properties.
(ii) In the electronic configuration of Zn, Cd and Hg the d-orbitals are completely filled in the ground state as well as in their common oxidation state. So, they are not regarded as transition metals

Question. The elements of 3d transition series are given as Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following :
(i) Write the element which shows maximum number of oxidation states. Given reason.
(ii) Which element has the highest melting point?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidising agent in +3 oxidation state and why?
Answer : (i) Mn shows maximum no. of oxidation states from +2 to +7 because Mn has maximum number of unpaired electrons in 3d sub-shell.
(ii) Cr has maximum melting point, because it has 6 unpaired electrons in the valence shell, hence it has strong interatomic interaction.
(iii) Sc shows only +3 oxidation state because because after losing 3 electrons, it has noble gas electronic configuration.
(iv) Mn is strong oxidising agent in +3 oxidation state because change of Mn³⁺ to Mn²⁺ give stable half filled (d⁵) electronic configuration, E°(Mn³⁺/Mn²⁺) = 1.5 V.

Question. Complete the following equation :
2MnO₄^– + 6H⁺ + 5NO₂^– →
Answer : 2MnO₄^– + 6H⁺ + 5NO₂^– → 2Mn²⁺ + 5NO₃^– +3H₂O

Question. Assign reasons for each of the following :
(i) Transition metals generally form coloured compounds.
(ii) Manganese exhibits the highest oxidation state of +7 among the 3^rd series of transition elements.
Answer : (i) Due to presence of vacant d-orbitals and d-d transitions, compounds of the transition metals are generally coloured.
When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency which generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand.
(ii) As manganese has maximum number of unpaired electrons (5) in 3d subshell in addition to 2 electrons in the 4s subshell, it can use the 7 electrons for bonding purpose.

Question. How would you account for the following?
(i) With the same d-orbital configuration (d4) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidising agent.
(ii) Most of the transition metal ions exhibit characteristic colours in aqueous solutions.
Answer : (i) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺couples are
Cr³⁺_(aq) + e^– → Cr²⁺(aq); E° = –0.41 V
Mn³⁺_(aq) + e^– → Mn²⁺(aq) ; E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising.
(ii) Since, transition elements contain partially filled 5-subshells. Therefore, electrons in these subshells go from lower d-subshells to higher d-subshells. This is called d-d transition. This transition takes place by absorbing energy from the visible light. The mixture of the wavelength which is not absorbed is transmitted out. This accounts for the colour of transition elements.

Question. How is the variability in oxidation states of transition elements different from that of non-transition elements? Illustrate with examples.
Answer : The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals. Their oxidation states differ from each other by unity.
For example, Fe³⁺ and Fe²⁺, Cu²⁺ and Cu⁺, etc.
In case of non transition elements the oxidation states normally differ by units of two. For example Pb²⁺ and Pb⁴⁺, Sn²⁺ and Sn⁴⁺, etc. It arises due to expansion of octet and inert pair effect.

Question. How would you account for the following :
(i) Cr²⁺ is reducing in nature while with the same d-orbital configuration (d4) Mn³⁺ is an oxidising agent.
(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.
Answer : (i) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺couples are
Cr³⁺_(aq) + e^– → Cr²⁺(aq); E° = –0.41 V
Mn³⁺_(aq) + e^– → Mn²⁺(aq) ; E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising.
(ii) Middle of the transition series contains greater number of unpaired electrons in (n –1)d and ns orbitals.

Question. State reasons for the following :
(i) Cu (I) ion is not stable in an aqueous solution.
(ii) Unlike Cr³⁺, Mn²⁺, Fe³⁺ and the subsequent other V²⁺ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species.
Answer : (i) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu²⁺ ion.
2Cu⁺_( aq) → Cu²⁺ _(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.
(ii) This is because due to lanthanoid contraction the expected increase in size does not occur hence they have very high value of ionisation enthalpies.

Question. Explain giving a suitable reason for each of the following:
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.
Answer : (i) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.
(ii) The metals of 4d and 5d-series have more frequent metal bonding in their compounds than the 3d-metals because 4d and 5d-orbitals are more exposed in space than the 3d-orbitals. So the valence electrons are less tightly held and form metal-metal bonding more frequently.

Question. How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Transition metals and their compounds act as catalyst.
Answer : (i) Transition elements can use their ns and (n – 1)d orbital electrons for bond formation therefore, they show variable oxidation states.
For example – Sc has ns²(n – 1)d¹ electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.
(ii) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group member of the second (4d) series.
Answer : (i) Transition metals form a large number of interstitial compounds because small atoms of certain non metallic elements (H, B, C, N, etc.) get trapped in voids or vacant spaces of lattices of the transition metals. As a result of filling up of the interstitial spaces such interstitial compounds are hard and rigid.
(ii) This is due to lanthanoid contraction.

Question. Account for the following :
(i) Mn²⁺ compounds are more stable than Fe²⁺compounds towards oxidation to their +3 state.
(ii) Cr²⁺ is reducing and Mn³⁺ oxidizing when both have d⁴ configuration.
Answer : (i) Electronic configuration of Mn²⁺ is 3d⁵which is half filled and hence stable. Therefore, third ionization enthalpy is very high, i.e., 3^rd electron cannot be lost easily. In case of Fe²⁺, electronic configuration is 3d⁶. Hence it can lose one electron easily to give the stable configuration 3d⁵.
(ii) (i) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺couples are
Cr³⁺_(aq) + e^– → Cr²⁺(aq); E° = –0.41 V
Mn³⁺_(aq) + e^– → Mn²⁺(aq) ; E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising.

Question. Explain the following observations :
(i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions.
Answer : (i) As we move along transition metal series from left to right (i.e. Ti to Cu), the atomic radii decrease due to increase in nuclear charge. Hence the atomic volume decreases. At the same time, atomic mass increases. Hence the density from titanium (Ti) to copper (Cu) increases.
(ii) The transition metals and their compounds, are known for their catalytic activity. This activity is
ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. Account for the following :
(i) The lowest oxide of a transition metal is basic, the highest is amphoteric/acidic.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing agents, it is easily oxidised.
Answer : (i) Lowest oxidation compounds of transition metals are basic due to their ability to get oxidised to higher oxidation states. Whereas the higher oxidation state of metal and compounds gets reduced to lower ones and hence acts as acidic in nature.
e.g., MnO is basic whereas Mn₂O₇ is acidic.
(ii) The tendency to form complexes is high for Co(III) as compared to Co(II). Co²⁺ ions are very stable and are difficult to oxidise. Co3+ ions are less stable and are reduced by water. In contrast many Co(II) complexes are readily oxidised to Co(III) complexes and Co(III) complexes are very stable, e.g.,
[Co(NH₃)₆]²⁺ →^Air [Co(NH₃)₆]³⁺
This happens because the crystal field stabilisation energy of Co(III) with a d⁶(t_2g⁶) configuration is higher than for Co(II) with a d⁷ (t_2g⁶ e_g¹) arrangement.

Question. Account for the following :
(i) The transition metals and their compounds act as good catalysts.
(ii) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(iii) A transition metal exhibits higher oxidation states in oxides and fluorides.
Answer : (i) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.
(ii) (i) Electronic configuration of Mn²⁺ is 3d⁵ which is half filled and hence stable. Therefore, third ionization enthalpy is very high, i.e., 3rd electron cannot be lost easily. In case of Fe²⁺, electronic configuration is 3d⁶. Hence it can lose one electron easily to give the stable configuration 3d⁵.
(iii) (i) Manganese can form pπ – dπ bond with oxygen by utilising 2p-orbital of oxygen and 3d-orbital of manganese due to which it can show highest oxidation state of +7. While with fluorine it cannot form such pπ – dπ bond thus, it can show a maximum of +4 oxidation state.

Question. Explain the following observations :
(i) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
(ii) Cu⁺ ion is not known in equeous solutions.
Answer : (i) This is due to presence of maximum number of unpaired electrons in Mn²⁺ in (3d⁵).
(ii) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu2+ ion.
2Cu⁺_( aq) → Cu²⁺ _(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.

Question. How would you account for the following :
(i) Metal-metal bonding is more extensive in the 4d and 5d series of transition elements than the 3d series.
(ii) Mn (III) undergoes disproportionation reaction easily.
(iii) Co (II) is easily oxidised in the presence of strong ligands.
Answer : (i) The metals of 4d and 5d-series have more frequent metal bonding in their compounds than the 3d-metals because 4d and 5d-orbitals are more exposed in space than the 3d-orbitals. So the valence electrons are less tightly held and form metal-metal bonding more frequently.
(ii) Mn³⁺ is less stable and changes to Mn²⁺ which is more stable due to half-filled d-orbital configuration . That is why, Mn³⁺ undergoes disproportionation reaction.
(iii) The tendency to form complexes is high for Co(III) as compared to Co(II). Co²⁺ ions are very stable and are difficult to oxidise. Co³⁺ ions are less stable and are reduced by water. In contrast many Co(II) complexes are readily oxidised to Co(III) complexes and Co(III) complexes are very stable, e.g.,
[Co(NH₃)₆]²⁺ →^Air [Co(NH₃)₆]³⁺
This happens because the crystal field stabilisation energy of Co(III) with a d⁶(t⁶_2g) configuration is higher than for Co(II) with a d⁷(t⁶_2ge¹_g) arrangement.

Question. Why do transition elements show variable oxidation states?
Answer : Transition elements can use their ns and (n – 1)d orbital electrons for bond formation therefore, they show variable oxidation states.
For example – Sc has ns²(n – 1)d¹ electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.

Question. Complete the following equation :
3MnO₄^2– + 4H⁺ →
Answer : 3MnO₄^2–+ 4H⁺ → 2Mn₄^– + MnO₂ + 2H₂O

Question. Complete the following equations.
(i) 2MnO₂ + 4KOH + O₂→^Δ
(ii) Cr₂O₇^2– + 14H⁺ + 6I^– →
Answer :

The d – and f – Block Elements Class 12 Chemistry Exam Questions

Question. How would you account for the following :
(i) The oxidising power of oxoanions are in the order
VO⁺₂ < Cr₂O₇^2– < MnO₄^–
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
(iii) Cr²⁺ is a stronger reducing agent than Fe²⁺.
Answer : (i) Change in Cr₂O₇^2– to Cr(III) is 3 and in MnO₄^– to Mn (II) is 5.
Change in oxidation state is large and the stability of reduced products in V(III) < Cr(III) < Mn(II). This is why oxidising power of VO₂⁺ < Cr₂O₇^2– < MnO₄^–.
(ii) Third ionization enthalpy of Mn is very high because the third electron has to be removed from the stable half-filled 3d-orbitals [Mn²⁺ (Z = 25) = 3d⁵].
(iii) Zn²⁺ ion has completely filled d-subshell and no d-d transition is possible. So zinc salts are white.
Configuration of Cu²⁺ is [Ar] 3d⁹. It has partly filled d-subshell and hence it is coloured due to d-d transition.

Question. Account for the following :
(i) Cu⁺ ions are not stable in aqueous solution.
(ii) Most of the transition metal ions exhibit paramagnetic behaviour.
Answer : (i) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu²⁺ ion.
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu(_s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.
(ii) Transition metals and most of their compounds contain unpaired electrons in the (n – 1)d orbitals hence show paramagnetic behaviour.

Question. What is meant by ‘disproportionation? Give an example of a disproportionation reaction in aqueous solution.
Answer : Disproportionation reaction involves the oxidation and reduction of the same substance. The two examples of disproportionation reaction are :
(i) Aqueous NH₃ when treated with Hg₂Cl₂ (solid) forms mercury aminochloride disproportionatively.
Hg₂Cl₂+ 2NH₃ → Hg + Hg(NH₂)Cl + NH₄Cl
(ii) 2Cu⁺ → Cu + Cu²⁺

Question. Give reason :
Orange solution of potassium dichromate turns yellow on adding sodium hydroxide to it.
Answer : When the pH of the solution of potassium dichromate is decreased, the colour of the solution changes from yellow to orange due to the conversion of CrO42– ions into Cr₂O₇^2– ions.
2CrO₄^2– + 2H⁺ ⇌ Cr₂O₇^2– + H₂O
Yellow Orange

Question. Account for the following :
(i) In the series Sc to Zn, the enthalpy of atomisation of zinc is the lowest.
(ii) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺.
Answer : (i) Electronic configuration of Mn²⁺ is 3d⁵ which is half filled and hence stable. Therefore, third ionization enthalpy is very high, i.e., 3rd electron cannot be lost easily. In case of Fe²⁺, electronic configuration is 3d⁶. Hence it can lose one electron easily to give the stable configuration 3d⁵.
(ii) Much larger third ionisation energy of Mn(where change is d⁵ to d⁴) is mainly responsible for this.
This also explains that +3 state of Mn is of little importance.

Question. When chromite ore FeCr₂O₄ is fused with NaOH in presence of air, a yellow coloured compound
(A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCl forms a orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C).
Answer : (i)

(ii) Potassium dichromate is used as a powerful oxidising agent in industries and for staining and tanning of leather.

Question. Give reasons for the following :
(i) Mn²⁺ is a good oxidising agent.
(ii) E°_M²⁺/M values are not regular for first row transition metals (3d-series).
(iii) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF₄, whereas the highest oxide is Mn₂O₇.
Answer : (i) Mn²⁺ is more stable due to half filled d⁵ configuration and Mn³⁺ easily changes to Mn²⁺ hence, it is oxidising.
(ii) The E°_M²⁺/M values are not regular which can be explained from the irregular variation of ionisation enthalpes i.e., IE₁ + E₂ and also the sublimation enthalpies which are relatively much less for manganese and vanadium.
(iii) (i) Manganese can form pπ – dπ bond with oxygen by utilising 2π-orbital of oxygen and 3π-orbital of manganese due to which it can show highest oxidation state of +7. While with fluorine it cannot form such pπ – dπ bond thus, it can show a maximum of +4 oxidation state.

Question. How would you account for the following :
(i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
(ii) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ couple or Fe³⁺/Fe²⁺ couple.
(iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
Answer : (i) This is due to lanthanoid contraction.
(ii) Much larger third ionisation energy of Mn(where change is d⁵ to d⁴) is mainly responsible for this.
This also explains that +3 state of Mn is of little importance.
(iii) All transition elements except the first and the last member in each series show a large number of variable oxidation states. This is because difference of energy in the (n – 1)d and ns orbitals is very little.
Hence, electrons from both the energy levels can be used for bond formation.

Question. Give reasons :
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
Answer : (i) Manganese can form pπ – dπ bond with oxygen by utilising 2p-orbital of oxygen and 3d-orbital of manganese due to which it can show highest oxidation state of +7. While with fluorine it cannot form such pπ – dπ bond thus, it can show a maximum of +4 oxidation state.
(ii) Transition elements can use their ns and (n – 1)d orbital electrons for bond formation therefore, they show variable oxidation states.
For example – Sc has ns²(n – 1)d¹ electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.

Question. How would you account for the following :
(i) Many of the transition elements and their compounds can act as good catalysts.
(ii) The metallic radii of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second series.
Answer : (i) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.
(ii) This is due to lanthanoid contraction.

Question. How would you account for the following :
(i) Transition elements have high enthalpies of atomisation.
(ii) The transition metals and their compounds are found to be good catalysts in many processes?
Answer : (i) As transition metals have a large number of unpaired electrons in the d-orbitals of their atoms they have strong interatomic attraction or metallic bonds. Hence, they have high enthalpy of atomization.
(ii) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. How would you account for the following :
(i) Highest fluoride of Mn is MnF₄ whereas the highest oxide is Mn₂O₇.
(ii) Transition metals and their compounds show catalytic properties.
Answer : (i) Manganese can form pπ – dπ bond with oxygen by utilising 2p-orbital of oxygen and 3d-orbital of manganese due to which it can show highest oxidation state of +7. While with fluorine it cannot form such pπ – dπ bond thus, it can show a maximum of +4 oxidation state.
(ii) The transition metals and their compounds, are known for their catalytic activity. This activity is
ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. Complete the following chemical equation :
Cr₂O^2–_7(aq) + H₂S_(g) + H⁺_(aq) →
Answer : Cr₂O^2–_7(aq) + 3H₂S_(g) + 8H⁺_(aq) → 2Cr³⁺_(aq)+ 7H₂O_(l) + 3S_(s)

Question. Give reasons :
(i) d-block elements exhibit more oxidation states than f-block elements.
(ii) The enthalpies of atomization of the transition metals are high.
(iii) The variation in oxidation states of transition metals is of different type from that of the non-transition metals.
Answer : (i) All transition elements except the first and the
last member in each series show a large number of variable oxidation states. This is because difference of energy in the (n – 1)d and ns orbitals is very little.
Hence, electrons from both the energy levels can be used for bond formation.
(ii) (i) As transition metals have a large number of unpaired electrons in the d-orbitals of their atoms they have strong interatomic attraction or metallic bonds. Hence, they have high enthalpy of atomization.
(iii) The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals. Their oxidation states differ from each other by unity.
For example, Fe³⁺ and Fe²⁺, Cu²⁺ and Cu⁺, etc.
In case of non transition elements the oxidation states normally differ by units of two. For example Pb²⁺ and Pb⁴⁺, Sn²⁺ and Sn⁴⁺, etc. It arises due to expansion of octet and inert pair effect.

Question. Explain the following :
(i) Copper (I) ion is not stable in an aqueous solution.
(ii) With same (d⁴) configuration Cr (II) is reducing whereas Mn (III) is oxidising.
(iii) Transition metals in general act as good catalysts.
Answer : (i) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu2+ ion.
2Cu⁺_( aq) → Cu²⁺ _(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.
(ii) E° values for the Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺couples are
Cr³⁺_(aq) + e^– → Cr²⁺(aq); E° = –0.41 V
Mn³⁺_(aq) + e^– → Mn²⁺(aq) ; E° = +1.551 V
These E° values indicate that Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising.
(iii) The transition metals and their compounds, are known for their catalytic activity. This activity is
ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. Account for the following :
(i) Transition metals show variable oxidation states.
(ii) Cu⁺ ion is unstable in aqueous solution.
Answer : (i) Transition elements can use their ns and (n – 1)d orbital electrons for bond formation therefore, they show variable oxidation states.
For example – Sc has ns²(n – 1)d¹ electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation state = +2. When it utilizes both the electrons then its oxidation state = +3.
(ii) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu2+ ion.
2Cu⁺_( aq) → Cu²⁺ _(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.

Question. Complete the following chemical equations :
(i) 8Mn₄^– + 3S₂O₃^2– + H₂O →
(ii) Cr₂O^2–₇ + 3Sn2⁺ + 14H⁺ →
Answer :

Question. Explain giving reasons :
Transition metals and their compounds generally exhibit a paramagnetic behaviour.
Answer : Transition metals and most of their compounds contain unpaired electrons in the (n – 1)d orbitals hence show paramagnetic behaviour.

Question. Give reason for the following :
(i) Transition metals have high enthalpies of atomisation.
(ii) Fe³⁺|Fe²⁺ redox couple has less positive electrode potential than Mn³⁺|Mn²⁺ couple.
(iii) Copper (I) has d¹⁰ configuration, while copper (II) has d⁹ configuration, still copper
(II) is more stable in aqueous solution than copper (I).
Answer : (i) As transition metals have a large number of unpaired electrons in the d-orbitals of their atoms they have strong interatomic attraction or metallic bonds. Hence, they have high enthalpy of atomization.
(ii) Electronic configuration of Mn²⁺ is 3d⁵which is half filled and hence stable. Therefore, third ionization enthalpy is very high, i.e., 3^rd electron cannot be lost easily. In case of Fe²⁺, electronic configuration is 3d⁶. Hence it can lose one electron easily to give the stable configuration 3d⁵.
(iii) In aqueous solutions, Cu⁺ undergoes disproportionation to form a more stable Cu2+ ion.
2Cu⁺_( aq) → Cu²⁺ _(aq) + Cu_(s)
Cu²⁺ in aqueous solutions is more stable than Cu⁺ ion because hydration enthalpy of Cu²⁺ is higher than that of Cu⁺ . It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.