QJEEAdv18Ch8216: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Ch8216 Solution: Match each set of hybrid orbitals from LIST - A with…

1. $\left[\mathrm{FeF}_6\right]^{4-}$

(A) Let oxidation state of iron (Fe) in $\left[\mathrm{FeF}_6\right]^{4-}$

$$\begin{array}{r}x+6\times (-1)=-4\\ \\ \Rightarrow x=-4+6=+2\end{array}$$

$$\therefore$$ Iron exist as $\mathrm{Fe}^{2+}$


Electron is $3 d$ orbitals remains unpaired as fluorine is weak ligand. One $4 s$ , three $4 p$ and two $4 d$ orbitals undergo hybridisation to form six $s p^3 d^2$ hybrid orbital each of which accumulate a pair of electron from fluorine.


$$\text{ Iron }(\mathrm{Fe})\text{ in }{\left[{\mathrm{FeF}}_6\right]}^{4-}\text{ has }s{p}^3{d}^2\text{ hybridisation. }$$

2. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right]$ : The arrangement of electrons in $\mathrm{Ti}^{3+}\left(3 d^1\right)$ will be


Electron in $3 d$ orbitals remains unpaired as water $\left(\mathrm{H}_2 \mathrm{O}\right)$ and chlorine $(\mathrm{Cl})$ is a weak ligands. Two $3 d$ , one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the chlorine and water molecules.

3. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ : The arrangement of electrons in $\mathrm{Cr}^{3+}\left(3 d^3\right)$ will be


Electrons in $3 d$ orbitals remains paired as ammonia $\left(\mathrm{NH}_3\right)$ is a strong ligand. Two $3 d$ , one $4 s$ and three $4 p$ orbitals undergo hybridisation to form six $d^2 s p^3$ hybrid orbitals. These orbitals accept lone pair of electrons from each of the six ammonia molecules.

4. $${\left[{\mathrm{FeCl}}_4\right]}^{2-}:\text{ The arrangement of electrons in }{\mathrm{Fe}}^{2+}\left(3d^6\right)\text{ will be }$$


Electrons in $3 d$ orbitals remains unpaired as chlorine is a weak ligand. One $4 s$ and three $4 p$ orbitals undergo hybridization to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pair of electrons from each of the chlorine ligand.

5. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ : The arrangement of electrons in $\mathrm{Ni}^0\left(3 d^8 4 s^2\right)$ will be


Electrons in $3 d$ orbitals remains paired and paired electron from $4 \mathrm{~s}$ orbital move to $3 d$ as carbonyl is a strong ligand. One $4 s$ , and three $4 p$ orbitals undergo hybridisation to form four $s p^3$ hybrid orbitals. These $s p^3$ hybrid orbitals accept lone pairs from each of the four carbonyl ligand.

6. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}:$ The arrangement of electrons in $\mathrm{Ni}^{2+}\left(3 d^8\right)$ will be


Electrons in $3 d$ orbitals pair up as cyanide is a strong ligand. One $3 d$ , one $4 s$ and two $4 p$ orbitals undergo hybridisation to form four $d s p^2$ hybrid orbitals. The $d s p^2$ hybrid orbitals accept lone pair of electrons from each of the cyanide ligand.