Question

Among the compounds, ${\text{B}}{{\text{F}}_{\text{3}}}$ , ${\text{NC}}{{\text{l}}_{\text{3}}}$ , ${{\text{H}}_{\text{2}}}{\text{O}}$ , ${\text{S}}{{\text{F}}_{\text{4}}}$ and ${\text{BeC}}{{\text{l}}_{\text{2}}}$ , identify the one in which the central atom has the same hybridization.
(A) ${\text{B}}{{\text{F}}_{\text{3}}}$ and ${\text{NC}}{{\text{l}}_{\text{3}}}$
(B) ${\text{NC}}{{\text{l}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$
(C) ${\text{B}}{{\text{F}}_{\text{3}}}$ , ${\text{NC}}{{\text{l}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$
(D) ${\text{S}}{{\text{F}}_{\text{4}}}$ and ${\text{BeC}}{{\text{l}}_{\text{2}}}$

Answer 1

In the above question, we have to find out the individual hybridisation and then we can compare each one of them to get the correct answer. Basically here, we have to find which compound is ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridised or which is ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridized and so on.

Formula Used:
${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}$
where H= hybrid orbitals
ve = number of valence electrons
sa= number of side atoms which are attached to central atom
c = charge on the compound.

Complete Step by step solution:
As we have to find the hybridisation on the central atom, we can use the formula
${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}$
Here, we will get the values in numerical form, so we should have knowledge about how to relate this number with hybridization. We can refer to table given below for the same:

VALUE OF H	HYBRIDISATION
2	${\text{sp}}$
3	${\text{s}}{{\text{p}}^2}$
4	${\text{s}}{{\text{p}}^3}$
5	${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$
6	${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$

For ${\text{B}}{{\text{F}}_{\text{3}}}$ :
The number of electron in the outermost shell is 3, so ve=3
The number of fluorine atom attach to it, sa = 3
Charge on the compound, c = 0
Hence, ${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}{\text{ = }}\dfrac{{3 + 3 + 0}}{2}{\text{ = 3}}$
$\therefore$ It is ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridized.
For ${\text{NC}}{{\text{l}}_{\text{3}}}$ :
The number of electron in the outermost shell is 5, so ve=5
The number of chlorine atom attach to it, sa = 3
Charge on the compound, c = 0
Hence, ${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}{\text{ = }}\dfrac{{5 + 3 + 0}}{2}{\text{ = 4}}$
$\therefore$ It is ${\text{s}}{{\text{p}}^3}$ hybridized.
For ${{\text{H}}_{\text{2}}}{\text{O}}$ :
The number of electron in the outermost shell is 6, so ve=6
The number of hydrogen atom attached to it, sa = 2
Charge on the compound, c = 0
Hence, ${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}{\text{ = }}\dfrac{{6 + 2 + 0}}{2}{\text{ = 4}}$
$\therefore$ It is ${\text{s}}{{\text{p}}^3}$ hybridized.
For ${\text{S}}{{\text{F}}_{\text{4}}}$ :
The number of electron in the outermost shell is 6, so ve=6
The number of fluorine atom attach to it, sa = 4
Charge on the compound, c = 0
Hence, ${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}{\text{ = }}\dfrac{{6 + 4 + 0}}{2}{\text{ = 5}}$
$\therefore$ It is ${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$ hybridized.
For ${\text{BeC}}{{\text{l}}_{\text{2}}}$ :
The number of electron in the outermost shell is 2, so ve=2
The number of fluorine atom attach to it, sa = 2
Charge on the compound, c = 0
Hence, ${\text{H = }}\dfrac{{{\text{ve + sa + c}}}}{{\text{2}}}{\text{ = }}\dfrac{{2 + 2 + 0}}{2}{\text{ = 2}}$
$\therefore$ It is ${\text{sp}}$ hybridized.
Since, ${\text{NC}}{{\text{l}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ are ${\text{s}}{{\text{p}}^2}$ hybridized.
So, option B is correct.

Note:
In these types of questions, we can easily find out the hybridisation of the atom if we have knowledge about how h is related with the hybridisation.
The side atoms and charge on the compound can easily be found out. So, we must have ideas about valence electrons of the central atom.