Question

the ratio of spin only magnetic moments of Fe3+and Co2+ is
a)241/2:151/2
b)7:3
c)351/2:151/2
d)51/2:71/2

Answer 1

the answer of this question is is a

Answer 2

Answer:  a) 241/2:151/2

Explanation:-

1. The metal is Fe with atomic number of 26

The electronic configuration of Fe is, $Ar3d^64s^2$

The oxidation state of $Fe$ is +3.

$Fe^{3+}:23:Ar3d^5$

The number of unpaired electrons in Fe = 5

Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

$\mu=\sqrt{5(5+2)}=\sqrt{35}$

2. The metal is Co with atomic number of 27

The electronic configuration of Fe is, $Ar3d^74s^2$

The oxidation state of $Co$ is +2.

$Fe^{3+}:23:Ar3d^7$

The number of unpaired electrons in Co = 3

Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

$\mu=\sqrt{3(3+2)}=\sqrt{15}BM$

thus ratio of spin only magnetic moments of $Fe^{3+}$ and $Co^{2+}$ is $\frac{\sqrt{35}}{\sqrt{15}}=1.5$