SATHEE: Coordination Compounds - Result Question 41

Coordination Compounds - Result Question 41

####43. Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : [2018]

Column I

a. $C o^{3 +}$

b. $C r^{3 +}$

c. $F e^{3 +}$

d. $N i^{2 +}$ Column II i. $\sqrt{8} B M$

ii. $\sqrt{35} B M$

iii. $\sqrt{3} B M$

iv. $\sqrt{24} B M$ v. $\sqrt{15} B M$

\begin{tabular}{lllll} & a & b & c & d \

(a) & iv & v & ii & i \

(b) & i & ii & iii & iv \

(d) & iv & i & ii & iii \end{tabular}

Show Answer

Solution:

(a) $C o^{3 +} = [A r] 3 d^{6}$ , unpaired $e^{-} (n) = 4$

Spin magnetic moment

$= \sqrt{4 (4 + 2)} = \sqrt{24}$ B.M.

$C r^{3 +} = [A r] 3 d^{3}$ , unpaired $e^{-} (n) = 3$

Spin magnetic moment

$= \sqrt{3 (3 + 2)} = \sqrt{15}$ B.M.

$F e^{3 +} = [A r] 3 d^{5}$ , unpaired $e^{-} (n) = 5$

Spin magnetic moment

$= \sqrt{5 (5 + 2)} = \sqrt{35}$ B.M.

$N i^{2 +} = [A r] 3 d^{8}$ , unpaired $e^{-} (n) = 2$

Spin magnetic moment

$= \sqrt{2 (2 + 2)} = \sqrt{8}$ B.M.

44 (b) In a solution containing $H g C l_{2}, I_{2}$ and $I^{-}$ , both $H g C l_{2}$ and $I_{2}$ compete for $I^{-}$ .

Since formation constant of $[H g I_{4}]^{2 -}$ is very large $(1.9 \times 10^{30})$ as compared with $Γ_{3}$ $(K_{f} = 700)$ .

$∴ I^{-}$ will preferentially combine with $H g C l_{2}$ .

$\begin{matrix} H g C l_{2} + 2 I^{-} \to H g I_{2} ↓ + 2 C l^{-} \\ R e d p p t^{H g I_{2} + 2 I^{-} \to [H g I_{4}]^{2 -}} \\ soluble \end{matrix}$

Coordination Compounds - Result Question 41

Column I

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Coordination Compounds - Result Question 41

Column I

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu