Q55 · CSIR-NET Chemistry, December 2015

Paper: CSIR-NET December 2015 · Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Coordination Chemistry – General · Marks: 2 · Difficulty: Medium

The formation constant for the complexation of $\mathrm{M}^{+}(\mathrm{M}=\mathrm{Li}, \mathrm{Na}, \mathrm{K}$ and Cs$)$ with cryptand, $\mathrm{C}_{222}$ follows the order
(a)$\mathrm{Li}^{+}<\mathrm{Cs}^{+}<\mathrm{Na}^{+}<\mathrm{K}^{+}$
(b)$\mathrm{Li}^{+}<\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Cs}^{+}$
(c)$\mathrm{K}^{+}<\mathrm{Cs}^{+}<\mathrm{Li}^{+}<\mathrm{Na}^{+}$
(d)$\mathrm{Cs}^{+}<\mathrm{K}^{+}<\mathrm{Li}^{+}<\mathrm{Na}^{+}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Cryptand-2.2.2 has a cavity (≈1.4 Å) matched to K⁺, so K_f peaks at K⁺; Na⁺ fits next best, Cs⁺ is too big and Li⁺ far too small: Li⁺ < Cs⁺ < Na⁺ < K⁺.

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