The magnetic moment of a transition metal ion | vedclass

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(B) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M$,where $n$ is the number of unpaired electrons.Given $\mu = 3.87 \

Vedclass · Accepted Answer

$3$

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(B) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M$,where $n$ is the number of unpaired electrons.Given $\mu = 3.87 \, B.M$.Substituting the value: $3.87 = \sqrt{n(n+2)}$.Squaring both sides: $14.9769 \approx n(n+2)$.$n^2 + 2n - 15 \approx 0$.Solving the quadratic equation: $(n+5)(n-3) = 0$.Since $n$ must be positive,$n = 3$.Therefore,there are $3$ unpaired electrons.

The magnetic moment of a transition metal ion is $3.87 \, B.M$. How many unpaired electrons are present in it?

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