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:

  1. 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

$$ \left[Co\left(H _2 O\right) _5 Cl\right] Cl _2 \cdot H _2 O . $$



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