Magnetic moment

source: thepinsta.com

## Magnetic moment-

### Question 1 ) Calculate the magnetic moment of Fe ^{3+} , Mn^{++}, Cr^{3+} , Zn ^{++} ?

### Solution )

Fe – 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{2},3d^{6}

Fe ^{3+} 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{0},3d^{5}

no. of unpaired electrons in Fe^{3+} = 5

**μ**s = **√n** (n+2) BM

= **√5** (5+2) = **√ 35 = 5.92**

Mn – 1s^{2},2s^{2},2p^{6}, 3s^{2}, 3p ^{6}, 4s ^{2}, 3d ^{5}

Mn^{2+} 1s ^{2},2s ^{2},2p ^{6}, 3s ^{2},3p^{6},4s^{0},3d^{5}

no. of unpaired electrons in Mn^{2+} = 5

**μ**s = **√n** (n+2) BM

= **√5** (5+2) = **√ 35 = 5.92**

Cr – 1s^{2},2s^{2},2p^{6},3s^{2}, 3p ^{6}, 4s ^{1}, 3d ^{5}

Cr ^{3+} 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{0},3d^{3}

no. of unpaired electrons in Cr^{3+} = 3

**μ**s = **√n** (n+2) BM

= **√3** (3+2) = **√ 15 = 3.87**

Zn – 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{2},3d^{10}

Zn ^{2+} 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{0},3d^{10}

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**

**μ**s = **√n** (n+2) BM = 1.73

suppose n = 1, then

**μ**s = **√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},3p^{6},4s^{2},3d^{3}

V^{4+} 1s ^{2},2s ^{2},2p ^{6}, 3s ^{2},3p^{6},4s^{0},3d^{1}

#### V^{4+} 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**

**μ**s = **√n** (n+2) BM = 5.92

suppose n = 5, then

**μ**s = **√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},3p^{6},4s^{2},3d^{5}

Mn ^{2+} 1s ^{2},2s ^{2},2p ^{6},3s ^{2},3p^{6},4s^{0},3d^{5}

#### Mn^{2+} 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.**