Coordination Compounds 2 Question 7
8. The calculated spin only magnetic moments (BM) of the anionic and cationic species of $\left[Fe(H_2O)_6\right]_2$ and $\left[Fe(CN)_6\right]$, respectively, are
(a) $ 0$ and $44.9$
(b) $2.84$ and $5.92$
(c) $0$ and $5.92$
(d) $4.9$ and $0$
(2019 Main, 8 April II)
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Answer:
Correct Answer: 8. (a)
Solution:
- $\left[Fe\left(H_2O\right)_6\right]_2 \Rightarrow$ It will form $2$ cationic species. i.e.
I. (i) $As[ \stackrel{II} {Fe } (H _2 O) _6]^{2+} \Rightarrow$ High spin octahedral complex of $Fe^{2+}$.
$Fe^{2+}: 3 d^{6}, x=4$ (unpaired electrons)
$ \mu=\sqrt{4(4+2)} $ $BM=4.9$ $ BM $
or (ii) as $\left[\stackrel{III} {Fe} \left(H _2 O\right) _6\right]^{3+}$ = High spin octahedral complex of $Fe^{3+}$. $Fe^{3+}: 3 d^{5}, x=5, \mu=\sqrt{5(5+2)}=5.92$ $ BM$
[ $H_2O$ is a neutral weak field ligand]
So, $\left[Fe\left(H_2O\right)_6\right]^{2+}$ will be the cationic species, $\mu=5.9 BM$.
$\left[Fe(CN)_6\right]$ will have one anionic complex
II. (i) $\left[\stackrel{II} {Fe}(CN) _6\right]^{4-} \Rightarrow$ Low spin, octahedral complex of $Fe^{2+}$.
As $CN^{-}$ is a strong ligand it will pair up the electrons.
or, (ii) $\left[\stackrel{III} {Fe}(CN)_6\right]^{3-} \Rightarrow$ Low spin octahedral complex of $Fe^{3+}$.
$\left[CN^{-}\right.$is an anionic strong field ligand $]$
So, the anionic species is $\left[Fe(CN) _6\right]^{4-}, \mu=0$
Thus, the calculated spin only magnetic moments (BM) of the anionic and cationic species of $\left[Fe(H_2O)_6\right]^{2+}$ and $\left[Fe(CN)_6\right]^{4-}$ respectively are $5.9$ and $0$.
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