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JEE Mains · Chemistry · STD 12 - 4. d and f- block elements
The magnetic moment of a transition metal compound has been calculated to be \(3.87\) \(B.M.\) The metal ion is
- A \(Cr ^{2+}\)
- B \(Mn ^{2+}\)
- C \(V ^{2+}\)
- D \(Ti ^{2+}\)
Answer & Solution
Correct Answer
(C) \(V ^{2+}\)
Step-by-step Solution
Detailed explanation
\(Cr ^{+2}:[ Ar ], 3 d ^4, 4 s ^0 n =4, \mu=\sqrt{4(4+2)}=\sqrt{24}\) \(=4.89\,BM\) \(Mn ^{+2}:[ Ar ], 3 d ^5, 4 s ^0 n =5, \mu=\sqrt{5(5+2)}=\sqrt{35}\) \(=5.91\,BM\) \(V ^{+2}:[ Ar ], 3 d ^3, 4 s ^0 n =3, \mu=\sqrt{3(3+2)}=\sqrt{15}\) \(=3.87\,BM\)…
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