How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) $Z r (Z = 40)$ and $H f (Z = 72)$ have almost identical radii.
(iii) Transition metals and their compounds act as a catalyst.

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Solution

(i) There is a very small energy difference in between $(n - 1) d$ and $n s$ orbitals. As a result, both orbitals participate in bond formation, thus transition metals exhibit variable oxidation state.
(ii) $Z r (Z = 40) H f (Z = 72)$ almost have same radii, it is because of lanthanoid contraction.
(iii) Because of their ability of change their oxidation state, transition metals exhibit catalysis.
$\Rightarrow {6 F e}^{2 +} + {C r}_{2} {O_{7}}^{2 -} + 14 H^{+} \to {6 F e}^{3 +} + 2 {C r}^{3 +} + 7 H_{2} O$
$\Rightarrow {2 C r O}_{4}^{-} + 2 H^{+} \to {C r}_{2} {O_{7}}^{2 -} + H_{2} O$
$\Rightarrow {2 M n O}_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{2 -} \to {10 C O}_{2} + 2 {H n}^{2 +} + 8 H_{2} O$

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Z r (Z = 40)

H f (Z = 72)

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Explain giving reasons :
Transition metals and many of their compounds show paramagnetic behaviour.

The enthalpies of atomisation of the transition metals are high.

The transition metals generally form coloured compounds.

Transition metals and their many compounds act as good catalyst.

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.

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