Magnetic moment-numerical

Magnetic moment

source: thepinsta.com

Magnetic moment-

Question 1 ) Calculate the magnetic moment of Fe ³⁺ , Mn⁺⁺, Cr³⁺ , Zn ⁺⁺ ?

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

Question 2) Magnetic moment of a compound of vanadium is 1.73 BM. Write the electronic configuration of Vanadium in this compound.

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

V⁴⁺ ion has one unpaired electron in ‘d’  sub-shell.

Question 3) Magnetic moment of a compound manganese is 5.92 BM. Find the charge on metal cation in the compound.

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

Mn²⁺ ion has five  unpaired electron in ‘d’ sub-shell. So charge on metal cation is +2

Question 4) Calculate the orbital angular momentum for a ‘d’ electron ?

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.