Coordination Compounds 2 Question 30

####32. Consider the following complex ions, $P, Q$ and $R$. $P=\left[FeF _6\right]^{3-}, Q=\left[V\left(H _2 O\right) _6\right]^{2+}$ and $R=\left[Fe\left(H _2 O\right) _6\right]^{2+}$

The correct order of the complex ions, according to their spin-only magnetic moment values (in BM) is

(a) $R<Q<P$

(b) $Q<R<P$

(c) $R<P<Q$

(d) $Q<P<R$

(2013 Adv.)

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Answer:

Correct Answer: 32. (b)

Solution:

  1. PLAN Spin only magnetic moment have the formula $\sqrt{n(n+2)} B M$, where $N$ is the number of unpaired electrons. In the presence of weak ligand (as $H _2 O, Cl^{-}, F^{-}$) there is no pairing of electrons, and electrons donated by ligands are filled in outer vacant orbitals.
ن்
$P:\left[FeF _6\right]^{3-}$
weak ligand
26 +3 $[Ar] 3 d^{5}$ 5 $\sqrt{35} BM$
$Q:\left[V\left(H _2 O\right) _6\right]^{2+}$
weak ligand
23 +2 [Ar] 3 $\sqrt{15} BM$
$R:\left[Fe\left(H _2 O\right) _6\right]^{2+}$ 26 +2 $[Ar] 3 d^{6}$ 4 $\sqrt{24} BM$

In the presence of strong ligand (as $CN^{-}, CO, NH _3$, en) electrons are paired and electrons from ligands are filled in available inner orbitals

Thus, order of spin-only magnetic moment $=Q<R<P$



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