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Question: Answered & Verified by Expert
Match the following


The correct answer is
ChemistryCoordination CompoundsTS EAMCETTS EAMCET 2023 (12 May Shift 1)
Options:
  • A A-II, B-III, C-IV, D-I
  • B A-III, B-IV, C-II, D-I
  • C A-IV, B-III, C-I, D-II
  • D A-II, B-I, C-IV, D-III
Solution:
1847 Upvotes Verified Answer
The correct answer is: A-II, B-III, C-IV, D-I
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}:-$
$\mathrm{NH}_3$ is a strong field ligand and a neutral ligand. Thus, $\mathrm{Co}$ exists as $\mathrm{Co}^{3+}$ with paired electrons.
Thus, $\mathrm{Co}^{3+}=3 \mathrm{~d}^6$ or $\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^0$
$\left[\mathrm{CoF}_6\right]^{3-}:-$
$\mathrm{F}^{-}$is a weak field ligand and uninegatively charged.
Thus, Co exists as $\mathrm{Co}^{3+}$ with some unpaired electrons.
Thus, $\mathrm{Co}^{3+}=3 \mathrm{~d}^6$ or $\mathrm{t}_{2 \mathrm{~g}}{ }^4 \mathrm{e}_{\mathrm{g}}{ }^2$
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ :-
$\mathrm{CO}$ is a strong field and a neutral ligand.
Thus, $\mathrm{Ni}$ is in 0 oxidation state with all paired electrons.
Thus, $\mathrm{Ni}=3 \mathrm{~d}^8 4 \mathrm{~s}^2$ or $\mathrm{t}_2^4 \mathrm{e}^6$ (outer orbitals)
$\left[\mathrm{Fe}\left(\mathrm{CN}_6\right]^{3-}\right.$ :-
$\mathrm{CN}^{-}$is a strong field ligand with a charge of -1 .
ihus, $\mathrm{Fe}$ exists as $\mathrm{Fe}^{3+}$ here with paired electrons.
Thus, $\mathrm{Fe}^{3+}=3 \mathrm{~d}^5$ or $\mathrm{t}_{2 \mathrm{~g}}{ }^5 \mathrm{e}_{\mathrm{g}}{ }^0$
Therefore, $\mathrm{A} \rightarrow \mathrm{II}, \mathrm{B} \rightarrow \mathrm{III}, \mathrm{C} \rightarrow \mathrm{IV}, \mathrm{D} \rightarrow \mathrm{I}$

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