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Arrange the following in order of increasing number of unpaired electrons in them
i. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
ii. $\left[\mathrm{MnCl}_6\right]^{3-}$
iii. $\left[\mathrm{FeF}_6\right]^{3-}$
iv. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Options:
i. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
ii. $\left[\mathrm{MnCl}_6\right]^{3-}$
iii. $\left[\mathrm{FeF}_6\right]^{3-}$
iv. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Solution:
1420 Upvotes
Verified Answer
The correct answer is:
iv, i, ii, ii
(i) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}:-$
$\mathrm{Fe}$ is in +3 state and $\mathrm{CN}^{-}$is a strong-field ligand. Thus, $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ or
Thus, there is only one unpaired electron.
(ii) $\left[\mathrm{MnCl}_6\right]^{3-}$ :-
$\mathrm{Mn}$ is in +3 state and $\mathrm{Cl}^{-}$is a weak-field ligand.
Thus, $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4$ or
Thus, there are four unpaired electrons.
(iii) $\left[\mathrm{FeF}_6\right]^{3-}:-$
$\mathrm{Fe}$ is in +3 state and $\mathrm{F}^{-}$is a weak-field ligand. Thus, Fe will be $[\mathrm{Ar}] 3 \mathrm{~d}^5$ with five unpaired electron.
(iv) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ :-
$\mathrm{Co}$ is in +3 state here and $\mathrm{NH}_3$ is a strong-field ligand. Thus, $\mathrm{Co}$ will be $[\mathrm{Ar}] 3 \mathrm{~d}^6$ with zero unpaired electrons.
Thus, the increasing order of the number of unpaired electrons will be :-
(iv) $ < $ (i) $ < $ (ii) < (iii)
$\mathrm{Fe}$ is in +3 state and $\mathrm{CN}^{-}$is a strong-field ligand. Thus, $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ or
Thus, there is only one unpaired electron.
(ii) $\left[\mathrm{MnCl}_6\right]^{3-}$ :-
$\mathrm{Mn}$ is in +3 state and $\mathrm{Cl}^{-}$is a weak-field ligand.
Thus, $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4$ or
Thus, there are four unpaired electrons.
(iii) $\left[\mathrm{FeF}_6\right]^{3-}:-$
$\mathrm{Fe}$ is in +3 state and $\mathrm{F}^{-}$is a weak-field ligand. Thus, Fe will be $[\mathrm{Ar}] 3 \mathrm{~d}^5$ with five unpaired electron.
(iv) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ :-
$\mathrm{Co}$ is in +3 state here and $\mathrm{NH}_3$ is a strong-field ligand. Thus, $\mathrm{Co}$ will be $[\mathrm{Ar}] 3 \mathrm{~d}^6$ with zero unpaired electrons.
Thus, the increasing order of the number of unpaired electrons will be :-
(iv) $ < $ (i) $ < $ (ii) < (iii)
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