First slide

Aplicaitons of VBT

Question

The magnetic moment of K₃[Fe(CN)₆] is found to be 1.73 BM. How many unpaired electron(s) is/are present per molecule?

Moderate

A

2
B

3
C

4
D

1

Solution

$\begin{array}{l} μ = \sqrt{n (n + 2)} \\ 1 .73 = \sqrt{n (n + 2)} \\ So, n = 1 \end{array}$

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