Structure Of Atom Question 4
Question 4 - 2024 (27 Jan Shift 1)
Consider the following complex ions
$P=\left[FeF _6\right]^{3-}$
$Q=\left[V\left(H _2 O\right) _6\right]^{2+}$
$R=\left[Fe\left(H _2 O\right) _6\right]^{2+}$
The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
(1) $R<Q<P$
(2) $R<P<Q$
(3) $Q<R<P$
(4) $Q<P<R$
Show Answer
Answer (3)
Solution
$\left[FeF _6\right]^{3-}: Fe^{+3}:[Ar] 3 d^{5}$
F : Weak field Ligand
![](https://cdn.mathpix.com/snip/images/SswlRhpSXyFV_mOEFFE2YbzylOretOnOmDbuOkFBmvs.original.fullsize.png)
No. of unpaired electron’s $=5$
$\mu=\sqrt{5(5+2)}$
$\mu=\sqrt{35} BM$
![](https://pristine-salmon.pikapod.net/uploads/original/c1/e0/f85d0f6aefdca2fc63f330c4693c.png)
$\left[V\left(H _2 O\right) _6\right]^{+2}: V^{+2}: 3 d^{3}$
No. of unpaired electron’s $=3$
$\mu=\sqrt{3(3+2)}$
$\mu=\sqrt{15} BM$
$\left[Fe\left(H _2 O\right) _6\right]^{+2}: Fe^{+2}: 3 d^{6}$
$H _2 O$ : Weak field Ligand
![](https://cdn.mathpix.com/snip/images/CKzvaBHF72c2y6Ya5KjVTjOl5IcQnYbWA4OvadDjxyw.original.fullsize.png)
No. of unpaired electron’s $=4$
$\mu=\sqrt{4(4+2)}$
$\mu=\sqrt{24} BM$