The equation, which is balanced and represents the correct product(s) is

A
${Li}_{2} O + 2 KCl \to 2 LiCl + K_{2} O$
B
${[CoCl {({NH}_{3})}_{5}]}^{+} + 5 H^{+} \to {Co}^{2 +} + 5 {NH}_{4}^{+} + {Cl}^{-}$
C
${[Mg {(H_{2} O)}_{6}]}^{2 +} + (EDTA)^{4 -} \overset{excess NaOH}{⟶} [Mg (EDTA)]^{2 +} + 6 H_{2} O$
D
${CuSO}_{4} + 4 KCN \to K_{2} [Cu (CN)_{4}] + K_{2} {SO}_{4}$

Solution:

$(a) {Li}_{2} O + KCl ⟶ 2 LiCl + K_{2} O$
wrong equation, since a stronger base (K₂O) cannot be generated by a weaker base Li₂O
$(b) [CoCl {({NH}_{3})}_{5}] + 5 H^{\oplus} ⟶ {CO}^{2 +} (aq) + 5 {NH}_{4}^{\oplus} + {Cl}^{-}$
This is correct. All ammine complexes can be destroyed by adding H⁺
$(c) {[Mg {(H_{2} O)}_{6}]}^{2 +} {EDTA}^{4 -} \frac{{OH}^{-}}{excess} [Mg (EDTA)]^{2 +} + 6 H_{2} O$
This is wrong, since the formula of complex must be $[Mg (edta)]^{2 -}$
$(d) 2 {CuSO}_{4} + 10 KCN ⟶ 2 K_{3} [Cu (CN)_{4}] + 2 K_{2} {SO}_{4} + (CN)_{2} ↑$
It is a wrong option.

Solution:

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