The formation constant for the complexation of M⁺ (M=Li, Na, K and Cs) with cryptand, C₂₂₂ follows the order · ABC26IN0113

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

The formation constant for the complexation of $\mathrm{M}^{+}$ $(\mathrm{M}=\mathrm{Li}, \mathrm{Na}, \mathrm{K}$ and $\mathrm{Cs})$ 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 matched to K⁺: Li⁺ < Cs⁺ < Na⁺ < K⁺ (a).

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