SATHEE: Coordination Compounds 2 Question 24

Coordination Compounds 2 Question 24

####26. On treatment of $100 m L$ of $0.1 M$ solution of $C o C l_{3} .6 H_{2} O$ with excess of $A g N O_{3}; 1.2 \times 10^{22}$ ions are precipitated. The complex is

(2017 Main)

(a) $[C o {(H_{2} O)}_{4} C l_{2}] C l \cdot 2 H_{2} O$

(b) $[C o {(H_{2} O)}_{3} C l_{3}] \cdot 3 H_{2} O$

(d) $[C o {(H_{2} O)}_{5} C l] C l_{2} \cdot H_{2} O$

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Answer:

Correct Answer: 26. (b, c)

Solution:

Molarity $(M) = \frac{Number of moles of solute}{Volume of solution (in L)}$

$∴$ Number of moles of complex

$\begin{aligned} = \frac{Molarity \times volume (in m L)}{1000} & = \frac{0.1 \times 100}{1000} = 0.01 m o l e \end{aligned}$

Complex ion	Electronic configuration of	Num	nbe elec			paired $(n)$
${[C r {(H_{2} O)}_{6}]}^{2 +}$	$C r^{2 +}; [A r] 3 d^{4}$	1	1	1	1	$; 4$
${[F e {(H_{2} O)}_{6}]}^{2 +}$	$F e^{2 +}; [A r] 3 d^{6}$	11	1	1	1	$1; 4$
${[M n {(H_{2} O)}_{6}]}^{2 +}$	$M n^{2 +}; [A r] 3 d^{5}$	1	1	1	1	$1; 5$
${[C o C l_{4}]}^{2 -}$	$C o^{2 +}; [A r] 3 d^{7}$	11	11	1	1	$1; 3$

Number of moles of ions precipitate

$= \frac{1.2 \times 10^{22}}{6.02 \times 10^{23}} = 0.02 moles$

$∴$ Number of $C l^{-}$ present in ionisation sphere

$= \frac{Number of moles of ions precipitated}{Number of moles of complex} = \frac{0.02}{0.01} = 2$

$∴ 2 C l^{-}$ are present outside the square brackets, i.e. in ionisation sphere. Thus, the formula of complex is

$[C o {(H_{2} O)}_{5} C l] C l_{2} \cdot H_{2} O .$

Coordination Compounds 2 Question 24

Answer:

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Coordination Compounds 2 Question 24

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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