KCET · Chemistry · Coordination Compounds
Which one of the following is wrongly matched?
- A \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} \quad-\) Square planar
- B \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \quad-\) Neutral ligand
- C \(\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)^{3-}\right] \quad-\mathrm{sp}^{3} \mathrm{~d}^{2}\)
- D \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+} \quad-\) Follows EAN rule
Answer & Solution
Correct Answer
(C) \(\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)^{3-}\right] \quad-\mathrm{sp}^{3} \mathrm{~d}^{2}\)
Step-by-step Solution
Detailed explanation
(a) In \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}, \mathrm{Cu}\) is present as \(\mathrm{Cu}^{2+}\)
\(
\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9} 4 \mathrm{~s}^{0}
\)

\(\left(\mathrm{NH}_{3}\right.\) being a strong field ligand shifts one electron from \(3 \mathrm{~d}\) orbital to \(4 \mathrm{p}\) orbital.)
(b) \(\mathrm{In}\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}\) is a neutral ligand.
(c) \(\mathrm{In}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\), Fe is present as \(\mathrm{Fe}^{3+}\).
\(
\begin{aligned}
&\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} \\
&{\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}=[\mathrm{Ar}]}
\end{aligned}
\)

Thus, its hybridisation is \(\mathrm{d}^{2} \mathrm{sp}^{3}\) not \(\mathrm{sp}^{3} \mathrm{~d}^{2}\), ie, it is an inner orbital complex.
(d) \(\left[\operatorname{Co}(\mathrm{en})_{3}\right]^{3+}\) contains total 36 electrons, ie, follows EAN rule.
\(
\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{9} 4 \mathrm{~s}^{0}
\)

\(\left(\mathrm{NH}_{3}\right.\) being a strong field ligand shifts one electron from \(3 \mathrm{~d}\) orbital to \(4 \mathrm{p}\) orbital.)
(b) \(\mathrm{In}\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \mathrm{CO}\) is a neutral ligand.
(c) \(\mathrm{In}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\), Fe is present as \(\mathrm{Fe}^{3+}\).
\(
\begin{aligned}
&\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} \\
&{\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}=[\mathrm{Ar}]}
\end{aligned}
\)

Thus, its hybridisation is \(\mathrm{d}^{2} \mathrm{sp}^{3}\) not \(\mathrm{sp}^{3} \mathrm{~d}^{2}\), ie, it is an inner orbital complex.
(d) \(\left[\operatorname{Co}(\mathrm{en})_{3}\right]^{3+}\) contains total 36 electrons, ie, follows EAN rule.
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