The spin only magnetic moment of [Cu(NH3)4]2+ complex is:-

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Question

The spin only magnetic moment of [Cu(NH₃)₄]²⁺ complex is:-

1.732 B.M.

3.87 B.M.

2.2 B.M.

$5.96 B.M$

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In [Cu(NH₃)₄]²⁺ complex Cu is present in +2 oxidation state.

(Cu) ⇒ [Ar]4 s²,3d⁹

(Cu⁺²) ⇒ [Ar]3d⁹

No. of (Cu⁺²) s unpaired electron is 1 . So, spin only magnetic moment $= \sqrt (1 (1 + 2)) = \sqrt 3 = 1.73 BM$
Final Answer:- option (A)