d and f Block Elements — CBSE Class 12 · 2009 · Delhi Board Question with Solution

Complete the following chemical equations for reactions:
a) i) MnO₄⁻(aq) + S₂O₃²⁻(aq) + H₂O(l) →
ii) Cr₂O₇²⁻(aq) + H₂S(g) + H⁺(aq) →
b) Give an explanation for each of the following observations:
i) The gradual decrease in size (actinoid contraction) from left to right is greater among the actinoids than among the lanthanoids.
ii) The greater number of oxidation states is exhibited by the members in the middle of a transition series.
iii) With the same d-orbital configuration (d⁴), Cr²⁺ is a reducing agent but Mn³⁺ is an oxidising agent.

Solution

(a)(i)

8Mn O_{4}^{-} + 3 S_{2} O_{3}^{2 -} + H_{2} O \to 8Mn O_{2} + 6S O_{4}^{2 -} + 2O H^{-}

(ii)

C r_{2} O_{7}^{2 -} + 3 H_{2} S + 8 H^{+} \to 2C r^{3 +} + 3S + 7 H_{2} O

(b)(i) 5f orbitals are more diffuse and shield the nuclear charge even less effectively than 4f; each successive actinoid feels a bigger rise in effective nuclear charge, so the contraction per element is larger.
(ii) Mid-series members (Cr, Mn) have the greatest number of unpaired d electrons in addition to the ns pair; with the most electrons available for loss or sharing, they display the widest spread of states.
(iii) Cr

^{2 +}

^{4}

) is oxidised to Cr

^{3 +}

^{3}

— stable t

_{2 g}^{3}

), so it donates an electron: reducing. Mn

^{3 +}

^{4}

) is reduced to Mn

^{2 +}

^{5}

— half-filled, extra stable), so it accepts an electron: oxidising.

CBSE Class 12 Board · 2009 · Delhi · Subjective

CBSE Class 12 · 2009 · Delhi · d and f Block ElementsLong Answer

Solution

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