JEE MAIN - Chemistry (2021 - 17th March Morning Shift - No. 2)
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?
zero
5.26
5.0
5.92
Explanation
25Mn = [Ar]3d5 4s2
Mn2+ = [Ar]3d5 4s0
No. of unpaired electron (n) = 5
$$\mu = \sqrt {n\left( {n + 2} \right)} \,BM$$
$$ = \sqrt {5 \times 7} {\rm{ }} = \sqrt {35} {\rm{ }} = {\rm{ }}5.92{\rm{ }}$$ BM
Mn2+ = [Ar]3d5 4s0
No. of unpaired electron (n) = 5
$$\mu = \sqrt {n\left( {n + 2} \right)} \,BM$$
$$ = \sqrt {5 \times 7} {\rm{ }} = \sqrt {35} {\rm{ }} = {\rm{ }}5.92{\rm{ }}$$ BM
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