Question

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

	Column I		Column II
a.	$C{o}^{3+}$	i.	$\sqrt{8}B.M.$
b.	$C{r}^{3+}$	ii.	$\sqrt{35}B.M.$
c.	$F{e}^{3+}$	iii.	$\sqrt{3}B.M.$
d.	$N{i}^{2+}$	iv.	$\sqrt{24}B.M.$
		v.	$\sqrt{15}B.M.$

• A. $a-iv,b-i,c-ii,d-iii$
• B. $a-iv,b-v,c-ii,d-i$
• C. $a-iii,b-v,c-i,d-ii$
• D. $a-i,b-ii,c-iii,d-iv$

Answer 1

Answer: (B)

$a-iv,b-v,c-ii,d-i$

Let us calculate the magnetic moments of the ions given :

The formula of magnetic moment $\left(\mu \right)$ is $\sqrt{n(n+2)}$, where n is the number of unpaired electrons.

(i) ${Co}^{3+}$: It has $d^6$ electrons which means 4 unpaired electrons.

$\mu =\sqrt{4(4+2)}=\sqrt{24}$

(ii) ${Cr}^{3+}$: It has $d^3$ electrons which means 3 unpaired electrons.

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

(iii) ${Fe}^{3+}$: It has $d^5$ electrons which means 5 unpaired electrons.

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

(iv) ${Ni}^{2+}$: It has $d^8$ electrons which means 2 unpaired electrons.

$\mu =\sqrt{2(2+2)}=\sqrt{8}$

Hence, the correct match is given by option B.