JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 16)
The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is _________ BM.
Answer
2
Explanation
Z = 29 [Cu] $$\buildrel { - 2{e^ - }} \over \longrightarrow $$ Cu2+ = [Ar] 3d9
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Number of unpaired electron, n = 1
$$\therefore$$ Spin only magnetic moment,
$$\mu = \sqrt {n(n + 2)} BM = \sqrt {1(1 + 2)} BM = \sqrt 3 BM$$
= 1.73 BM $$ \simeq $$ 2 BM
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