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) $\left[Co\left(H _2 O\right) _5 Cl\right] Cl _2 \cdot H _2 O$
Show Answer
Answer:
Correct Answer: 26. (d)
Solution:
- Molarity $(M)=\dfrac{\text { Number of moles of solute }}{\text { Volume of solution (in L) }}$
$\therefore$ Number of moles of complex
$ \begin{aligned} & =\dfrac{\text { Molarity } \times \text { volume }(\text { in } mL)}{1000} \\ & =\dfrac{0.1 \times 100}{1000}=0.01 mole \end{aligned} $
Number of moles of ions precipitate
$ =\dfrac{1.2 \times 10^{22}}{6.02 \times 10^{23}}=0.02 \text { moles } $
$\therefore$ Number of $Cl^{-}$present in ionisation sphere
$ =\dfrac{\text { Number of moles of ions precipitated }}{\text { Number of moles of complex }}=\dfrac{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 . $
BETA
