Consider the following complex ions P = [ FeF 6 ] 3- Q = [ V ( H 2 O ) 6 ] 2+ R = [ Fe ( H 2 O ) 6 ] 2+ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Chemistry · Coordination Compounds

JEE Main 2024 — 27 January, Shift 1 — Question 62

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 :

$\left[\mathrm{FeF}_{6}\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^{5}$ F :

Weak field Ligand

1	1	1	1	1

No. of unpaired electron's $=5$

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

$\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^{3}$

1	1	1

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

$\uparrow \downarrow$	1	1	1	1

No. of unpaired electron's $=4$

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

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

Answer key and solution verified before publishing.

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