Question

Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II.
The correct option is:

LIST-I	LIST-II
$ds{p^2}$	${\left[ {Fe{F_6}} \right]^{4 - }}$
$s{p^3}$	$\left[ {Ti{{\left( {{H_2}O} \right)}_3}C{l_3}} \right]$
$s{p^3}{d^2}$	${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$
${d^2}s{p^3}$	${\left[ {FeC{l_4}} \right]^{2 - }}$
$\left[ {Ni{{\left( {CO} \right)}_4}} \right]$
${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$

A. $P \to 5;Q \to 4;R \to 2,3;S \to 1$
B. $P \to 5,6;Q \to 4;R \to 3;S \to 1,2$
C. $P \to 6;Q \to 4,5;R \to 1;S \to 2,3$
D. $P \to 4,6;Q \to 5,6;R \to 1,2;S \to 3$

Answer 1

To solve this question, you must recall the electronic configurations of the metal cations present in the given coordination compounds. Coordination compounds are additional compounds that do not lose their identity in solution form.

Complete step by step solution:
${\left[ {Fe{F_6}} \right]^{4 - }}$: Since fluoride ion is a negative ligand, we can see that the oxidation state of iron is $+ 2$.
$F{e^{2 + }}:\left[ {Ar} \right]3{d^6}$; As fluoride ion is a weak field ligand; it does not force pairing of unpaired electrons in the d orbital of iron.
Since coordination no. of iron is 6, we can say that the hybridisation is $s{p^3}{d^2}$.
$\left[ {Ti{{\left( {{H_2}O} \right)}_3}C{l_3}} \right]$: we know that, chloride is a negative ligand and water is neutral. Thus, the oxidation state of titanium is $+ 3$.
$T{i^{3 + }}:\left[ {Ar} \right]3{d^1}4{s^0}$; the $3d,4s,4p$ orbitals are available for bonding.
Since coordination no. is 6, we can infer that the hybridisation of titanium ion is ${d^2}s{p^3}$.
${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$: Oxidation state of chromium is $+ 3$.
$C{r^{3 + }}:\left[ {Ar} \right]3{d^3}4{s^0}$; It is clear from the electronic configuration that the available orbitals for ligand electron pairs are $3d,4s,4p$orbitals.
Since coordination no. is 6, we can infer that the hybridisation of chromium ion is ${d^2}s{p^3}$.
${\left[ {FeC{l_4}} \right]^{2 - }}$: Oxidation state of iron is $+ 2$.
$F{e^{2 + }}:\left[ {Ar} \right]3{d^6}$; As chloride ion is a weak field ligand; it does not force the pairing of unpaired electrons in the d orbital of iron.
Since coordination number is 4, it can be concluded that the hybridisation of ferrous ion is $s{p^3}$.
$\left[ {Ni{{\left( {CO} \right)}_4}} \right]$: Oxidation state of nickel is 0.
$Ni:\left[ {Ar} \right]3{d^8}4{s^2}$; we know that the carbonyl ligand is a strong field ligand and forces the back pairing of electrons.
Thus, the electronic configuration becomes, $Ni:\left[ {Ar} \right]3{d^{10}}$.
4s and 4p orbitals are available for bonding.
Since, coordination number is 4, thus, the hybridisation of nickel is $s{p^3}$.
${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$
Oxidation state of nickel is $+ 2$.
$Ni:\left[ {Ar} \right]3{d^8}$; cyanide ion is a strong field ligand and forces the back pairing of unpaired electrons.
Thus, a $d$ subshell is vacated.
Since, coordination number is 4, the hybridisation of nickel ions is $ds{p^2}$.
The matching is as follows; $P \to 6;Q \to 4,5;R \to 1;S \to 2,3$

Thus, the correct option is C.

Note:
$CO,{\left( {CN} \right)^ - },N{O_2}^ - ,phen,dipy,en,N{H_3}$are strong field ligands. These ligands cause the pairing of unpaired electrons in the atom with which they form a coordination bond.
Generally, the ligands which donate through their $\pi$ orbitals are weak (known as weak field ligands) while those which donate through sigma orbitals are comparatively stronger and those which donate through sigma and accept through $\pi$orbitals are even stronger (known as strong field ligands)