Coordination Compounds 2 Question 28
####30. The equation which is balanced and represents the correct product(s) is
(2014 Main)
(a) $Li _2 O+2 KCl \longrightarrow 2 LiCl+K _2 O$
(b) $\left[CoCl\left(NH _3\right) _5\right]^{+}+5 H^{+} \longrightarrow Co^{2+}+5 NH _4^{+}+Cl^{-}$
(c) $\left[Mg\left(H _2 O\right) _6\right]^{2+}+(\text { EDTA })^{4-} \xrightarrow{\text { Excess } NaOH}$
$[Mg(\text { EDTA })]^{2+}+6 H _2 O$
(d) $CuSO _4+4 KCN \longrightarrow K _2\left[Cu(CN) _4\right]+K _2 SO _4$
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Answer:
Correct Answer: 30. (c)
Solution:
- This problem is based on conceptual mixing of properties of lithium oxide and preparation, properties of coordination compounds. To answer this question, keep in mind that on adding acid, ammine complexes get destroyed.
(a) $Li _2 O+KCl \longrightarrow 2 LiCl+K _2 O$
This is wrong equation, since a stronger base $K _2 O$ cannot be generated by a weaker base $Li _2 O$.
(b) $\left[CoCl\left(NH _3\right) _5\right]^{+}+5 H^{+} \longrightarrow Co^{2+}(a q)+5 NH _4^{+}+Cl^{-}$
This is correct. All ammine complexes can be destroyed by adding $H^{\oplus}$. Hence, on adding acid to $\left[CoCl\left(NH _3\right) _5\right]$, it gets converted to $Co^{2+}(a q)^{+} NH _4^{+}$and $Cl^{-}$.
(c) $\left[Mg\left(H _2 O\right) _6\right]^{2+}+$ EDTA $^{4-} \xrightarrow[\text { Excess }]{OH^{-}}[Mg(\text { EDTA })]^{2+}+6 H _2 O$
This is wrong, since the formula of complex must be $[\operatorname{Mg}(\text { EDTA })]^{2+}$ as EDT.
(d) The 4th reaction is incorrect. It can be correctly represented as
$$ \begin{aligned} 2 CuSO _4+10 KCN \longrightarrow 2 & K _3\left[Cu(CN) _4\right] \ & +2 K _2 SO _4+(CN) _2 \uparrow \end{aligned} $$
