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The D-And-f-Block Elements

Question
CBSEENCH12010285

(a) Account for the following:

(i) Zr and Hf have almost similar atomic radii.

(ii) Transition metals show variable oxidation states. 

(iii) Cu+ ion is unstable in aqueous solution.

(b) Complete the following equations:

(i) 2 MnO2 + 4 KOH + O2--->

(ii) 2 Na2CrO4 + 2 H + -->

Solution

(i) Due to lanthanide contraction, Zr and Hf have almost similar atomic radii. It can be explained on the basis of shielding effect. The electrons present in inner shells, shield the outer electrons from the nuclear charge, making them experience a low effective nuclear charge. The shielding effect exerted by the electrons decreases in the orders > p > d > f. The f subshell poorly shields the outer electrons from nuclear attraction, which results in the most attractive pull of nucleus on the outer electron. In the case of post lanthanide elements like Hf, 4f subshell is filled and it is not very effective at shielding the outer shell electrons. Therefore, Zr and Hf have almost similar atomic radii.

(ii) The transition metals have their valence electrons in (n-1)d and ns orbitals. Since there is very little energy difference between these orbitals, both energy levels can be used for bond formation. Thus, transition elements exhibit variable oxidation states.

(iii) Cu2+ is more stable than Cu+ in an aqueous medium. Cu2+ has high hydration energy that compensates for the energy required to remove one electron from Cu+to form Cu2+. So in an aqueous medium, Cu+ gives Cu2+ and Cu.
Chemical reaction: 2 Cu+(aq) à Cu2+(aq) + Cu(s)

(b)

(i) 2 MnO2 + 4 KOH + O2 ---> 2 K2MnO4 + 2 H2O
(ii) 2 Na2CrO4 + 2 H+ ----> Na2Cr2O7 + 2 Na+ + H2O