Question

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(Compound)}& \text{(Hybridisation)}\\ \text{(I)}\left[Fe\right(EDTA){]}^{-}& \text{(P)}s{p}^3{d}^2\\ \text{(II)}\left[Ni\right(CN{)}_4{]}^{2-}& \text{(Q)}s{p}^3\\ \text{(III)}[CuC{l}_4{]}^{2-}& \text{(R)}{d}^{3}s\\ \text{(IV)}\left[Fe\right(H_{2}O{)}_6{]}^{2-}& \text{(S)}{d}^{2}s{p}^3\\ & \text{(T)}ds{p}^2\\ & \text{(U)}s{p}^{3}d\end{array}$

Which of the following option has correct combination considering List-I and List-II?

• A. $I\to S;III\to Q$
• B. $I\to P;III\to Q$
• C. $I\to P;III\to T$
• D. $I\to S;III\to T$

Answer 1

The correct option is A $I\to S;III\to Q$

(I) Here $Fe$ is in $+3$ oxidation state having $\left[Ar\right]3d^5$ configuration.
EDTA is a moderate strong field ligand which causes pairing of electrons.


(III) Here $Cu$ is in $+2$ oxidation state having $\left[Ar\right]3d^9$ configuration.
Since $Cl$ is a weak field ligand so it does not cause pairing of electrons.