First slide

Aplicaitons of VBT

Question

Consider the following complex ions, P, Q, and R. $P= {[{FeF}_{6}]}^{3 -}, Q = {[V {(H_{2} O)}_{6}]}^{2 +} and R = {[Fe {(H_{2} O)}_{6}]}^{2 +}$ , the correct order of the complex ions, according to their spin only magnetic moment values (in B. M.) is:

Easy

A

Q < R < P
B

R < Q < P
C

R < P < Q
D

Q < P < R

Solution

$\underset{(P)}{{[{FeF}_{6}]}^{3 -}} \to {Fe}^{3 +} \to d^{5}$ configuration,
$F^{-}$ is weak field ligand
Unpaired electrons = 5; $∴ μ= \sqrt{5 (5 + 2)} = 5 .95 B . M .$
$\underset{(Q)}{{[V {(H_{2} O)}_{6}]}^{2 +}} \to V^{+ 2} \to d^{3}$ configuration,
$H_{2} O$ is weak field ligand
Unpaired electron = 3; $∴ μ = \sqrt{3 (3 + 2)} = 3.87 B . M .$
$\underset{(R)}{{[Fe {(H_{2} O)}_{6}]}^{2 +}} \to {Fe}^{+ 2} \to d^{6}$ configuration,
$H_{2} O$ is weak field ligand
Unpaired electron = 4; $∴ μ = \sqrt{4 (4 + 2)} = 4.90 B . M .$
Therefore only spin magnetic moment $(μ) = \sqrt{n (n + 2)}$
B.M. (n = unpaired electron)
So, $μ$ of P > R > Q.

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