JEE Mains 2020 Model Paper 3 New Pattern

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.