Question

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

• A. \(\mathrm{Li_2O + 2KCl \rightarrow 2LiCl + K_2O}\)
• B. \(\mathrm{[CoCl(NH_3)_5]^+ + 5H^+ \rightarrow Co^{2+} + 5NH_4^+ + Cl^-}\)
• C. \(\mathrm{[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \xrightarrow{excess\ NaOH} [Mg(EDTA)]^{2+} + 6H_2O}\)
• D. \(\mathrm{CuSO_4 + 4KCN \rightarrow K_2[Cu(CN)_4] + K_2SO_4}\)

Hint:

Check atom balance, charge balance and chemical feasibility of each equation; the Li2O/KCl exchange and the complexes in (3) and (4) are the usual traps.

Answer 1

The Correct Option is B

(1) The exchange \(\mathrm{Li_2O + 2KCl \to 2LiCl + K_2O}\) does not occur: \(\mathrm{K_2O}\) is a far stronger base than \(\mathrm{Li_2O}\) and the reaction is not feasible as the products given.
(3) The charge of \([\mathrm{Mg(EDTA)}]\) is \(+2 - 4 = -2\), not +2, so the equation is not charge-balanced; also excess NaOH would precipitate \(\mathrm{Mg(OH)_2}\).
(4) \(\mathrm{Cu^{2+}}\) is reduced by \(\mathrm{CN^-}\) to Cu(I), giving \(\mathrm{K_3[Cu(CN)_4]}\) and cyanogen; a Cu(II) tetracyanide complex is not formed.
(2) Atoms: Co 1, Cl 1, N 5, H \(15 + 5 = 20\) on both sides; charge: \(+1 + 5 = +6\) on the left and \(+2 + 5 - 1 = +6\) on the right. In acid the ammonia ligands are protonated to \(\mathrm{NH_4^+}\) and the complex breaks down. It is balanced and represents plausible products.
Answer: (2) \(\mathrm{[CoCl(NH_3)_5]^+ + 5H^+ \rightarrow Co^{2+} + 5NH_4^+ + Cl^-}\)