The spin only magnetic moment of - ZCl 4-2- is 3-87 BM where Z isA- MnB- NiC- CoD- Cu

The correct option is C CoThe spin only magnetic moment is given in quantum mechanics as µ = √(n(n+2)) BM where n is the number of unpaired e and BM (Bohr mag ...

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Standard XII

Chemistry

Crystal Field Split in Octahedral Complexes

The spin only...

Question

The spin only magnetic moment of $[Z C l_{4}]^{2 -}$ is 3.87 BM where Z is

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Solution

The correct option is C Co
The spin only magnetic moment is given in quantum mechanics as µ = $\sqrt{n (n + 2)}$ BM where n is the number of unpaired e- and BM (Bohr magnetons)
So, $\sqrt{n (n + 2)} = 3.87$
⟹ $n (n + 2) = (3.87)^{2} = 15$
⟹ $n^{2} + 2 n - 15 = 0$
⟹ n = -5 or 3.

So, No. of unpaired electrons are 3
Hence $Z^{2 +} = d^{7}$

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The spin only magnetic moment of [ZCl4]2− is 3.87 BM where Z is

The correct option is C CoThe spin only magnetic moment is given in quantum mechanics as µ = √n(n+2) BM where n is the number of unpaired e- and BM (Bohr magnetons) So, √n(n+2)=3.87 ⟹ n(n+2)=(3.87)2=15 ⟹ n2+2n−15=0 ⟹ n = -5 or 3. So, No. of unpaired electrons are 3 Hence Z2+=d7

The spin only magnetic moment of $[Z C l_{4}]^{2 -}$ is 3.87 BM where Z is