First slide

Valence bond theory

Question

For the tetrahedral complex ${[{MnBr}_{4}]}^{2 -}$ , the spin only magnetic moment value is: [At. no. of Mn= 25]

Easy

A

1.7
B

5.9
C

4.8
D

2.4

Solution

${[{MnBr}_{4}]}^{- 2}$
Ox.State of Mn
$\begin{array}{l} x + 4 (- 1) = - 2 \\ x = + 2 \end{array}$
$Mn (Z = 25) \dots \dots . [Ar] 4 s^{2} 3 d^{5} {Mn}^{+ 2} \dots \dots . [Ar] 3 d^{5}$
Given that ${[{MnBr}_{4}]}^{- 2}$ is a tetrahedral complex. $∴$ No rearrangement occurs.
Number of unpaired electrons(n) = 5
$\begin{array}{l} μ_{S} = \sqrt{n (n + 2)} B . M \\ μ_{S} = \sqrt{5 (5 + 2)} B . M \\ μ_{S} = 5 .9 B . M \end{array}$
Note:– In the formation of complex only vacant orbitals involve in hybridization.

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