Magnetic moment

source: thepinsta.com

## Magnetic moment-

### Solution )

Fe – 1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d6

Fe 3+  1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d5

no. of unpaired electrons in Fe3+ = 5

μs = √n (n+2) BM

= √5 (5+2) = √ 35 = 5.92

Mn –  1s2,2s2,2p6, 3s2, 3p 6, 4s 2, 3d 5

Mn2+  1s 2,2s 2,2p 6, 3s 2,3p6,4s0,3d5

no. of unpaired electrons in Mn2+ = 5

μs = √n (n+2) BM

= √5 (5+2) = √ 35 = 5.92

Cr –  1s2,2s2,2p6,3s2, 3p 6, 4s 1, 3d 5

Cr 3+  1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d3

no. of unpaired electrons in Cr3+ = 3

μs = √n (n+2) BM

= √3 (3+2) = √ 15 = 3.87

Zn –  1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d10

Zn 2+  1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d10

no. of unpaired electrons in Zn 2+ = 0

No magnetic moment

### Solution )

μ = 1.73

μs = √n (n+2) BM = 1.73

suppose n = 1, then

μs = √n (n+2) = √1 (1+2)

= √ 3 = 1.73

No. of unpaired electron in Vanadium = 1

23 V  –  1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d3

V4+  1s 2,2s 2,2p 6, 3s 2,3p6,4s0,3d1

### Solution )

μ = 5.92

μs = √n (n+2) BM = 5.92

suppose n = 5, then

μs = √n (n+2) = √5 (5+2)

= √ 35 = 5.92

No. of unpaired electron in manganese = 5

25 Mn  –  1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d5

Mn 2+   1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d5

### Solution )

orbital angular momentum = √ [l ( l + 1)] h / 2π

For ‘d’ orbital , l = 2

orbital angular momentum = √ [2( 2 + 1)] h / 2π = √ (6)  h / 2π Ans.