Structure Of Atom Question 4
Question 4 - 2024 (27 Jan Shift 1)
Consider the following complex ions
$\mathrm{P}=\left[\mathrm{FeF}_{6}\right]^{3-}$
$\mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}{2} \mathrm{O}\right){6}\right]^{2+}$
$\mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}{2} \mathrm{O}\right){6}\right]^{2+}$
The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
(1) $\mathrm{R}<\mathrm{Q}<\mathrm{P}$
(2) $\mathrm{R}<\mathrm{P}<\mathrm{Q}$
(3) $\mathrm{Q}<\mathrm{R}<\mathrm{P}$
(4) $\mathrm{Q}<\mathrm{P}<\mathrm{R}$
Show Answer
Answer (3)
Solution
$\left[\mathrm{FeF}_{6}\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$
F : Weak field Ligand
| 1 | 1 | 1 | 1 | 1 |
|---|
Solutions
No. of unpaired electron’s $=5$
$\mu=\sqrt{5(5+2)}$
$\mu=\sqrt{35} \mathrm{BM}$
$\left[\mathrm{V}\left(\mathrm{H}{2} \mathrm{O}\right){6}\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^{3}$
No. of unpaired electron’s $=3$
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15} \mathrm{BM}$
$\left[\mathrm{Fe}\left(\mathrm{H}{2} \mathrm{O}\right){6}\right]^{+2}: \mathrm{Fe}^{+2}: 3 \mathrm{~d}^{6}$
$\mathrm{H}_{2} \mathrm{O}$ : Weak field Ligand
| 1 | 1 | 1 | 1 | 1 |
|---|
No. of unpaired electron’s $=4$
$\mu=\sqrt{4(4+2)}$
$\mu=\sqrt{24} \mathrm{BM}$
