Question

Match the species given in Column $I$ with the hybridization given in Column $II$.

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(A\right)& \text{Boron in}\left[B\right(OH{)}_4{]}^{-}& \left(i\right)& s{p}^2\\ \left(B\right)& \text{Aluminium in}\left[Al\right(H_{2}O{)}_6{]}^{3+}& \left(ii\right)& s{p}^3\\ \left(C\right)& \text{Boron in}\left[B_2{H}_6\right]& \left(iii\right)& s{p}^3{d}^2\\ \left(D\right)& \text{Carbon in Buckminsterfullerene}& & \\ \left(E\right)& \text{Silicon in}\left[Si{O}_4^{4-}\right]& & \\ \left(F\right)& \text{Germanium in}[GeC{l}_6{]}^{2-}& & \end{array}$

Answer 1

$\text{Analyzing the options}$

Option (A):

Boron in $\left[B\right(OH{)}_4{]}^{-}$ forms 4 sigma bonds thus it is $s{p}^3$ hybridised. Hence, it is correct match with option $\left(ii\right)$.


Option (B):

Aluminium in $\left[Al\right(H_{2}O){]}^{3+}$

The Electronic configuration of $Al$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$

$Al$ loses its $3$ valence electrons from the $3s$ and $3p$-orbitals to become $A{l}^{3+}$ ion .

Now, it accepts $6$ electron pairs from water molecules after undergoing $s{p}^3{d}^2$ hybridization using it’s one s-orbital, three $p$-orbitals and two $d$-orbitals.


Thus, $Al$ undergo $s{p}^3{d}^2$ hybridisation and correct match will be option $\left(iii\right)$.

Option (C):

Bonding of boron in $B_2{H}_6$ (diborane):


Each $B$ atom uses $s{p}^3$ hybrid orbitals for bonding. Out of the four $s{p}^3$ hybrid orbitals on each $B$ atom, one is without an electron shown in broken lines.

The terminal $B-H$ bonds are normal

$2$-centre-$2$- electron bonds but the two bridge bonds are $3$-centre-$2$-electron bonds. The

$3$-centre-$2$-electron bridge bonds are also referred to as banana bonds.

Hence, due to $s{p}^3$ hybridisation of boron, the correct match will be option $\left(ii\right)$.

Option (D):

Carbon in Buckminsterfullerene:

$C_{60}$ molecule has a shape like soccer ball and called Buckminsterfullerene. It contains $20$ six- membered rings and $12$ five membered rings.

A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo $s{p}^2$ hybridisation.


Each carbon atom forms three sigma bonds with the other three carbon atoms. The remaining electron at each carbon is delocalised in molecular orbitals, which in turn give aromatic character to the molecule.

Hence, correct match is option $\left(i\right)$.

Option (E):


In $Si{O}_4^{4-}$, Silicon atom is bonded to four oxygen atoms in tetrahedral fashion. Due to formation of $4$ sigma bonds and no lone pair on the $Si$ central atom, it has $s{p}^3$ hybridisation.

Hence, correct match is option $\left(ii\right)$.

Option (F):

Germanium in $[GeC{l}_6{]}^{2-}$

The Electronic configuration of $Ge=\left[Ar\right]3d^{10}4s^{2}4p^{2}4d^0$\)

In complex $[GeC{l}_6{]}^{2-}$

$Ge$ have an oxidation state of $+4$.

Electronic configuration of $G{e}^{4+}=\left[Ar\right]3d^{10}4s^{0}4p^{0}4d^0$

Now, $6C{l}^{-}$ ions can donate 6 lone pairs of electrons to $G{e}^{4+}$ which gets accommodated in one $4s$ orbitals, three $4p$ orbitals and two $4d$ orbitals via forming $s{p}^3{d}^2$ hybridisation.
Hence, correct match is option $\left(iii\right)$.

So, the correct match can be given as:

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(A\right)& \text{Boron in}\left[B\right(OH{)}_4{]}^{-}& \left(ii\right)& s{p}^3\\ \left(B\right)& \text{Aluminium in}\left[Al\right(H_{2}O{)}_6{]}^{3+}& \left(iii\right)& s{p}^3{d}^2\\ \left(C\right)& \text{Boron in}\left[B_2{H}_6\right]& \left(ii\right)& s{p}^3\\ \left(D\right)& \text{Carbon in Buckminsterfullerene}& \left(i\right)& s{p}^2\\ \left(E\right)& \text{Silicon in}\left[Si{O}_4^{4-}\right]& \left(ii\right)& s{p}^3\\ \left(F\right)& \text{Germanium in}[GeC{l}_6{]}^{2-}& \left(iii\right)& s{p}^3{d}^2\end{array}$