$1 \mathrm{~L}$, $0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}$ is mixed with $1 \mathrm{~L}$, $0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$. The resulting solution is divided into two equal parts (X) and treated with excess $\mathrm{AgNO}_{3}$ solution and $\mathrm{BaCl}_{2}$ solution respectively as shown below: 1 L Solution $(\mathrm{X})+\mathrm{AgNO}_{3}$ solution (excess) $\rightarrow \mathrm{Y}$ 1 L Solution (X) $+\mathrm{BaCl}_{2}$ solution (excess) $\rightarrow \mathrm{Z}$ The number of moles of Y and Z respectively are

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Coordination Compounds

1 L

0.02 M

solution of

[Co (NH_{3})_{5} SO_{4}] Br

is mixed with

1 L

0.02 M

solution of

[Co (NH_{3})_{5} Br] SO_{4}

. The resulting solution is divided into two equal parts (X) and treated with excess

AgNO_{3}

solution and

BaCl_{2}

solution respectively as shown below:
1 L Solution

(X) + AgNO_{3}

solution (excess)

\to Y

1 L Solution (X)

+ BaCl_{2}

solution (excess)

\to Z

The number of moles of Y and Z respectively are

Answer: (B)

The two complexes are ionisation isomers that differ in which ion is held inside the coordination sphere and which is free in solution. In

[Co (N H_{3})_{5} S O_{4}] Br

the sulphate is bound to cobalt and the bromide is the free counter-ion, so only bromide can meet silver ion. In

[Co (N H_{3})_{5} Br] S O_{4}

the bromide is bound and the sulphate is free, so only sulphate can meet barium ion. Each one-litre portion holds

0.01 mol

of each complex. Silver nitrate therefore precipitates the

0.01 mol

of free bromide as silver bromide, and barium chloride precipitates the

0.01 mol

of free sulphate as barium sulphate, so the two precipitates amount to

0.01 mol

each.

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option