Chemistry · Coordination Compounds

JEE Advanced 2021 — Paper 2 — Question 13

The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of $\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni}$ and Cu are 26, 27, 28 and 29, respectively)

Option A:
$\left[\mathrm{FeCl}_{4}\right]^{-}$ and $\left[\mathrm{Fe}(\mathrm{CO})_{4}\right]^{2-}$
Correct
Option B:
$\left[\mathrm{Co}(\mathrm{CO})_{4}\right]^{-}$ and $\left[\mathrm{CoCl}_{4}\right]^{2-}$
Correct
Option C:
$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Option D:
$\left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{+}$ and $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}$
Correct

Answer: A, B, D

Step-by-step solution

(A) $\left[\mathrm{FeCl}_{4}\right]$

\begin{aligned} & \mathrm{Fe}^{+3}=3 \mathrm{~d}^{5}, \mathrm{Cl}^{-} \text {weak field liganol } \mathrm{sp}^{3} & {\left[\mathrm{Fe}(\mathrm{CO})_{4}\right]^{-2}} & \mathrm{Fe}^{-2}=3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2} \end{aligned}

CO strong field liganol pairing occurs $\mathrm{Fe}^{-2}=3 \mathrm{~d}^{10}$ hence $\mathrm{sp}^{3}$ (B) $\left[\mathrm{Co}(\mathrm{CO})_{4}\right]$ $\mathrm{Co}=3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2} \quad$ due to CO pairing occurs Hence $=3 d^{10}$ $\left[\mathrm{CoCl}_{4}\right]^{-2}$ $\mathrm{Co}^{+2}=3 \mathrm{~d}^{7}, \quad \mathrm{Cl}^{-} \quad$ weak field liganol $\mathrm{sp}^{3}$ (D) $\left[\mathrm{Cu}(\mathrm{Py})_{4}\right]^{+1}$ $\mathrm{Cu}^{+1}=3 \mathrm{~d}^{10}, \quad \mathrm{sp}^{3}$ $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{-3}$ $\mathrm{Cu}^{+1}=3 \mathrm{~d}^{10}, \mathrm{sp}^{3}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Chemistry
Chapter: Coordination Compounds
Topic: Isomerism in Coordination Compounds