(a) Account for the following :

(i) Transition metals show variable oxidation states.

(ii) Zn, cd and Hg are soft metals.

(iii) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+ 1.57 V) as compared to Cr³⁺/Cr²⁺.

(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

OR

(a) Following are the transiton metal ions of 3d series :

Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺

(Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24)

Answer the following :

(i) Which ion is most stable in an aqueous solution and why?

(ii) Which ion is a strong oxidizing agent and why?

(iii) Which ion is colourless and why?

(b) Complete the following equations :

(i)

(ii)

(a) (i) Transition metals show variable oxidation states.

The transition elements have their valence electrons in two different sets of orbitals, i.e., (n-1) and ns. Since there is very little difference in the energies of these orbitals, both energy levels can be used for the bond formation. In simple compounds, we may presume that the two electrons from ns orbital of a transition elements are used to give an oxidation state of +2 and the (n-1)d electrons remain unaffected. But the higher oxidation states like +3, +4, +5, +6, +7 correspond to the use of all 4s and 3d electrons in the transition series of elements. In the excited state, the (n-1)d electrons become bonding and thus give variable oxidation states to the atoms. In short, the energies of (n-1)d orbitals and ns orbitals are very close. Hence, electrons from both can participae in bonding and thus give variable oxidation states.

(ii) Zn, Cd and Hg are soft metals.

The melting point decreases with increase in the number of pairing electrons (decreasing no. of unpaired electrons).

The electronic configuration of Zn = [Ar] 3d¹⁰ 4s²

The electronic configuration of Cd = [Kr] 4d¹⁰ 5s²

The electronic configuration of Hg = [Xe] 4f¹⁴ 5d¹⁰ 6s²

Here, we can see that, in Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, absence of unpaired electrons cause the weak metallic bonds present in them. As a result, they are soft (low melting point).

(iii) E^o value for the (Mn³⁺/Mn²⁺) couple is highly positive (+ 1.57 V) as compared to Cr³⁺/Cr²⁺ is due to the following reason:

The comparatively high value of E° (Mn³⁺/Mn²⁺) shows that Mn²⁺ is very stable which is on account of stable d⁵ configuration of Mn²⁺ whereas Cr³⁺ is more stable due to stable t³_2g.

(b) One similarity and one difference between the chemistry of

lanthanoid and actinoid elements.

Similarities: As both lanthanoids and actinoids involve filling of f-orbitals, they show similarities in many respects as follows:

⇒ Both show an oxidation state of +3.

⇒ Both are electropositive and very reactive.

⇒ Both exhibit magnetic and spectral properties.

Differences:

OR

(a) (i) Cr³⁺ is the most stable ion in aqueous solution because it has half-filled t³_2g

(ii) Mn³⁺ is strong oxidizing agent which is on account of stable d⁵ configuration of Mn²⁺

(iii) Ti⁴⁺ is colorless because it has no unpaired electrons.

Color is due to the presence of incomplete d-subshell. As the electronic configuration of Ti = [Ar] 3d² 4s²

And, the electronic configuration of Ti⁴⁺ = [Ar] 3d⁰ 4s⁰

As we can see that, Ti⁴⁺ has completely empty d-orbitals. Hence it is colorless.

(b) (i) 2MnO₄^- + 16H⁺ + 5 S^2-→ 2Mn²⁺ + 8H₂O + 5S

This is the oxidizing property of KMNO₄ in the acidic medium

2KMnO₄ + 3H₂SO₄→ K₂SO₄ + 2MnSO₄ + 2H₂O + 5O

H₂S + O → H₂O + S

By multiplying the second reaction by 5, we get

(ii) When KMNO₄ is heated(513K), it readily decomposes giving oxygen.