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:
- 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.
 |
 |
辛 | 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 $=Q<R<P$
