Magnetic moment is given by µ = √n(n+2) ; Here, n = number of unpaired electrons ⇒ 2.83 = √n(n+2) ⇒ (2.83)2 = n (n + 2) ⇒ 8.00 = n2+2n ⇒ n2+2n−8 = 0 ⇒ n2+4x−2n−8 = 0 ⇒ (n + 4) (n - 2) = 0 n = 2 [Since m ≠- 4] Among the given ions
‌
22
Ti3+ → +3d1 →
↿
⇒ 1 unpaired electron
‌
28
Ni+ → +3d8 →
↿⇂
↿⇂
↿⇂
↿
↿
⇒ 2 unpaired electrons
‌
24
Cr3+ → +3d3 →
↿⇂
↿
↿
⇒ 3 unpaired electrons
‌
25
Mn2+ → +3d5
↿⇂
↿
↿
↿
↿
⇒ 5 unpaired electrons Hence, Ni2+ possesses a magnetic moment of 2.83BM.