On treatment of 100 mL of 0.1 M solution of $\mathrm{CoCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ with excess $\mathrm{AgNO}_{3}$; $1.2 \times 10^{22}$ ions are precipitated. The complex is :

(D) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

(A) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot 2 \mathrm{H}_{2} \mathrm{O}$

(B) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$

(C) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$

JEE Main · 2017 · 02 Apr 2017 · Offline · Coordination Compounds

On treatment of 100 mL of 0.1 M solution of

CoCl_{3} \cdot 6 H_{2} O

with excess

AgNO_{3}

;

1.2 \times 1 0^{22}

ions are precipitated. The complex is :

Answer: (D)

The complex holds

0.01

mol, since

100

mL of

0.1

M is

0.01

mol. The precipitated ions number

1.2 \times 1 0^{22}

, which is

\frac{1.2 \times 1 0 ^{22}}{6.02 \times 1 0 ^{23}} = 0.02

mol — exactly twice the complex. So two chlorides per formula unit are free to drop out as AgCl while the third stays bound to cobalt, which fits

[Co (H X_{2} O) X_{5} Cl] Cl X_{2} \cdot H X_{2} O

: it ionises to give two

Cl X^{-}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option