Question

In an atom having magnetic moment 1.73 BM, the number of unpaired electron is

Answer 1

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Magnetic moment(μ) is related with the no. of unpaired electron

μ=n(n+2)→1 Bohr Magneton (BM)-Units

The given magnetic moment in the question is 1.73BM.

Putting that value (μ) in equation 1; we get:-

(1.73)2=n(n+2)

On solving,

⇒n=1

Thus, the complex/compound having one unpaired electrons exhibit a magnetic moment of 1.73BM

a. In [Cu(NH3)4]2+

Cu2+=[Ar]3d9, n=1

b. In [Ni(CN)4]2−

Ni2+=[Ar]3d8 , n=2

c. In [TiCl4]

Ti4+=[Ar]3d0, n=0

d. In [CoCl6]4−

Co2+=[Ar]3d7 , n=3

In above all configuration only [Cu(NH3)4]2+ has one unpaired electron.

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