Coordination Compounds 2 Question 24
####26. On treatment of $100 mL$ of $0.1 M$ solution of $CoCl _3 .6 H _2 O$ with excess of $AgNO _3 ; 1.2 \times 10^{22}$ ions are precipitated. The complex is
(2017 Main)
(a) $\left[Co\left(H _2 O\right) _4 Cl _2\right] Cl \cdot 2 H _2 O$
(b) $\left[Co\left(H _2 O\right) _3 Cl _3\right] \cdot 3 H _2 O$
(c) $\left[Co\left(H _2 O\right) _6\right] Cl _3$
(d) $\left[Co\left(H _2 O\right) _5 Cl\right] Cl _2 \cdot H _2 O$
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Answer:
Correct Answer: 26. (b, c)
Solution:
- Molarity $(M)=\frac{\text { Number of moles of solute }}{\text { Volume of solution (in L) }}$
$\therefore$ Number of moles of complex
$$ \begin{aligned} & =\frac{\text { Molarity } \times \text { volume }(\text { in } mL)}{1000} \ & =\frac{0.1 \times 100}{1000}=0.01 mole \end{aligned} $$
| Complex ion | Electronic configuration of |
Num | nbe elec |
paired $(n)$ |
||
|---|---|---|---|---|---|---|
| $\left[Cr\left(H _2 O\right) _6\right]^{2+}$ | $Cr^{2+} ;[Ar] 3 d^{4}$ | 1 | 1 | 1 | 1 | $; 4$ |
| $\left[Fe\left(H _2 O\right) _6\right]^{2+}$ | $Fe^{2+} ;[Ar] 3 d^{6}$ | 11 | 1 | 1 | 1 | $1 ; 4$ |
| $\left[Mn\left(H _2 O\right) _6\right]^{2+}$ | $Mn^{2+} ;[Ar] 3 d^{5}$ | 1 | 1 | 1 | 1 | $1 ; 5$ |
| $\left[CoCl _4\right]^{2-}$ | $Co^{2+} ;[Ar] 3 d^{7}$ | 11 | 11 | 1 | 1 | $1 ; 3$ |
Number of moles of ions precipitate
$$ =\frac{1.2 \times 10^{22}}{6.02 \times 10^{23}}=0.02 \text { moles } $$
$\therefore$ Number of $Cl^{-}$present in ionisation sphere
$$ =\frac{\text { Number of moles of ions precipitated }}{\text { Number of moles of complex }}=\frac{0.02}{0.01}=2 $$
$\therefore 2 Cl^{-}$are present outside the square brackets, i.e. in ionisation sphere. Thus, the formula of complex is
$$ \left[Co\left(H _2 O\right) _5 Cl\right] Cl _2 \cdot H _2 O . $$
