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) $P \rightarrow 4,6 ; Q \rightarrow 5,6 ; R \rightarrow 1,2 ; S \rightarrow 3$
Show Answer
Solution:
- 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.
