Consider the following complex ions P= [FeF_6 ]^ 3- , Q= [V (H_2 … - Vidyadip

MCQ

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 :

A
$ R < Q < P$
B
$R < P < Q$
✓
$Q < R < P$
D
$ Q < P < R$

✓

Answer

Correct option: C.

$Q < R < P$

c
$\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^5$

$IMAGE$

F : Weak field Ligand 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}$

$IMAGE$

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

$IMAGE$

No. of unpaired electron's$=4$

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

$ \mu=\sqrt{24} \mathrm{BM}$

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