Coordination Compounds 2 Question 54

####56. Match each set of hybrid orbitals from List-I with complexes given in List-II.

List-I List-II
$P$. $d s p^{2}$ 1. $\left[FeF _6\right]^{4-}$
$Q$. $s p^{3}$ 2. $\left[Ti\left(H _2 O\right) _3 Cl _3\right]$
$R$. $s p^{3} d^{2}$ 3. $\left[Cr\left(NH _3\right) _6\right]^{3+}$
$S$. $d^{2} s p^{3}$ 4. $\left[FeCl _4\right]^{2-}$
5. $\left[Ni(CO) _4\right]$
6. $\left[Ni(CN) _4\right]^{2-}$

The correct option is

(2018 Adv.)

(a) $P \rightarrow 5 ; Q \rightarrow 4,6 ; R \rightarrow 2,3 ; S \rightarrow 1$

(b) $P \rightarrow 5,6 ; Q \rightarrow 4 ; R \rightarrow 3 ; S \rightarrow 1,2$

(c) $P \rightarrow 6 ; Q \rightarrow 4,5 ; R \rightarrow 1 ; S \rightarrow 2,3$

(d) $P \rightarrow 4,6 ; Q \rightarrow 5,6 ; R \rightarrow 1,2 ; S \rightarrow 3$

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Solution:

  1. For, $\boldsymbol{P}$ i.e. $\boldsymbol{d} \boldsymbol{s} \boldsymbol{p}^{2}$, It is seen in $\left[Ni(CN) _4\right]^{2-}$

$$ \begin{aligned} & Ni-[Ar] 3 d^{8} 4 s^{2} \ & Ni^{2+}-[Ar] 3 d^{8} \end{aligned} $$

as $CN^{-}$is a strong ligand so when it approaches towards central metal pairing of unpaired electrons takes place.

Thus, in $\left[Ni(CN) _4\right]^{2-}$

Structure : Square planar

So correct match for $P$ is 6 .

For $Q$ i.e., $s p^{3}$

It is seen in $\left[FeCl _4\right]^{2-}$ and $Ni(CO) _4$

$$ \begin{gathered} Fe-[\operatorname{Ar}] 3 d^{6} 4 s^{2} \ Fe^{2+}-[\operatorname{Ar}] 3 d^{6} \ Fe^{2+}-\begin{array}{|l|l|l|l|l|} \hline 1 & 1 & 1 & 1 & 1 \ \hline \end{array} \quad \begin{array}{|l|l|l|} \hline & & \ \hline \end{array} \end{gathered} $$

$As Cl^{-}$is a weak ligand so when it approaches towards central metal pairing of unpaired electrons does not take place.

Thus, in $\left[FeCl _4\right]^{2-}$

Structure-Tetrahedral

Likewise in $Ni(CO) _4$

$$ \begin{aligned} & Ni-[Ar] 3 d^{8} 4 s^{2} \end{aligned} $$

As CO is a strong ligand, hence when it approaches towards central metal atom pairing of unpaired electron of central atom takes place.

Thus, in $Ni(CO) _4$

Structure Tetrahedral

So, for $Q-4$ and 5 are correct match.

For $R$ i.e., $s p^{3} d^{2}$

It is seen in $\left[FeF _6\right]^{4-}$

$$ \begin{gathered} Fe-[Ar] 3 d^{6} 4 s^{2} \ Fe^{2+}-[Ar] 3 d^{6} \end{gathered} $$

As $F^{-}$is a weak field ligand hence, when it approaches towards central metal atom, pairing of its electrons does not take place. Thus, in $\left[FeF _6\right]^{4-}$

Structure : Octahedral

So, 1 is the correct match for $R$.

For $S$ i.e., $d^{2} s p^{3}$

It is seen in $\left[Ti\left(H _2 O\right) _3 Cl _3\right]$ and $\left[Cr\left(NH _3\right) _6\right]^{3+}$

$$ \begin{aligned} & Ti-[Ar] 3 d^{2} 4 s^{2} \ & Ti^{3+}-[Ar] 3 d^{1} \end{aligned} $$

Here, both $H _2 O$ and $Cl$ are weak ligands

So, in $\left[Ti\left(H _2 O\right) _3 Cl _3\right]$

Structure Octahedral

Likewise in $\left[Cr\left(NH _3\right) _6\right]^{3+}$

$$ \begin{aligned} & Cr-[Ar] 3 d^{5} 4 s^{1} \ & Cr^{3+}-[Ar] 3 d^{3} 4 s^{0} \end{aligned} $$

Here, $NH _3$ is also a weak field ligand so due to its approach no pairing takes place in $Cr$.

Thus, $\operatorname{In}\left[Cr\left(NH _3\right) _6\right]^{3+}$

So for, S-2 and 3 are the correct match.



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