(a) Account for the following:

(i) Ce⁴⁺ is a strong oxidizing agent in aqueous solution.

(ii) Transition metals have high enthalpy of atomization.

(iii) Mn shows maximum number of oxidation states in 3d series.

(b) Complete the following equations:

(i)

(ii)

(a)

(i) Ce⁴⁺ is a strong oxidizing agent in aqueous solution.

The atomic number for Cerium is 58 and its electronic configuration is 4f¹ 5d¹ 6s², thus it has a tendency to give 4 electrons to achieve the +4 oxidation state. But this configuration is unstable as it has 4 empty orbitals at higher energy state. Thus it can accept one electron and thus acts as a strong oxidizing agent, which is contrary to other metals of the group.

(ii) Transition metals have higher enthalpy of atomization

Transition metals all are d-block elements and have outermost shells as d-orbital. These metals are called as transition metals as they contain atleast one unpaired electron in d-orbital in their ground state or oxidation state. As this fact implies that they consist of higher number of unpaired electrons and they have a higher interatomic attraction due to higher inter-nuclear attraction. This increases the bond strength resulting in higher enthalpies of atomization.

(iii) Mn shows maximum number of oxidation states in 3d series.

The electronic configuration of Mn is 4s² 3d⁵, which suggests that among the 3d series it is half-filled and thus have maximum number of unpaired electrons. The more unpaired electrons an atom has the more increasingly it can give those electrons as compared to paired electrons. So it can donate 2 electrons from s orbital and 5 electrons from d-orbital. Thus it shows maximum oxidation states in 3d series from +2 to +7.

(b)

Step (i) we generate the unbalanced skeleton:

Here Mn undergoes reduction and S oxidation.

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since the atoms are balanced the reaction is same as previous step.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in reduction reaction, so add 4 water molecule to RHS of the equation to balance oxygen and 1 H₂O molecule on LHS of oxidation reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Equal charges on both sides imply balanced equation.

Option (ii)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Verify whether all charges are balanced.

LHS = 6×+2 + 1× -2 + 14 = 24

RHS = 6×+3 + 2×+3 + 7×0 = 24

Thus the sum total is same the solve reaction is correct.