Question

Out of ${{\text{H}}_{\text{2}}}\text{S}$ , $\text{C}{{\text{O}}_{\text{2}}}$ , $\text{Be}{{\text{F}}_{\text{2}}}$ , and ${{\text{H}}_{\text{2}}}\text{O}$ , the linear molecules are __________________ and _______________.

Answer 1

The shape of the molecules are determined from the hybridization of the central element in the compound and the number of lone pairs and bond pairs present in the central element of the compound. The compound without any lone pairs on the central atom shall be linear.

Formula used: ${H=}\dfrac{V+X-C+A}{\text{2}}$ ,
H is the hybridization of the central element, V is the number of electrons in the valence shell, X is the number of monovalent groups attached to the central atom, C is the cation charge of the compound if it is cationic, A is anion charge of the compound if it is anionic.

Complete Stepwise Solution
Let us see the central elements of the above given compounds:
In hydrogen sulphide or ${{\text{H}}_{\text{2}}}\text{S}$ , the central element is sulphur. There are 16 electrons in sulphur and in the valence shell, there are 6 electrons. The hybridization of sulphur is:
$\text{H=}\dfrac{\text{6 + 2 - 0 + 0}}{\text{2}}$
$\Rightarrow \text{H = }\dfrac{8}{\text{2}}=4$
So the hybridization of sulphur in hydrogen sulphide being $\text{s}{{\text{p}}^{\text{3}}}$ , it will not be linear, but angular.
In carbon dioxide or $\text{C}{{\text{O}}_{\text{2}}}$ , the central element is carbon. There are 6 electrons in carbon and in the valence shell, there are 4 electrons. The hybridization of carbon is:
$\text{H=}\dfrac{\text{4 + 0 - 0 + 0}}{\text{2}}=\dfrac{4}{2}=2$
So the hybridization of carbon in carbon dioxide being $\text{sp}$ , it will be linear.
In Beryllium difluoride or $\text{Be}{{\text{F}}_{\text{2}}}$ , the central element is beryllium. There are 4 electrons in beryllium and in the valence shell, there are 2 electrons. The hybridization of beryllium is:
$\text{H=}\dfrac{\text{2 + 2 - 0 + 0}}{\text{2}}=\dfrac{4}{2}=2$
So the hybridization of beryllium in beryllium difluoride being $\text{sp}$ , it will be linear.
In water or ${{\text{H}}_{\text{2}}}\text{O}$ , the central element is oxygen. There are 8 electrons in oxygen and in the valence shell, there are 6 electrons. The hybridization of oxygen is:
$\text{H=}\dfrac{\text{6 + 2 - 0 + 0}}{\text{2}}=\dfrac{8}{2}=4$
So the hybridization of oxygen in water being $\text{s}{{\text{p}}^{\text{3}}}$ , it will not be linear, but angular.
So, Out of ${{\text{H}}_{\text{2}}}\text{S}$ , $\text{C}{{\text{O}}_{\text{2}}}$ , $\text{Be}{{\text{F}}_{\text{2}}}$ , and ${{\text{H}}_{\text{2}}}\text{O}$ , the linear molecules are “Carbon dioxide” and “Beryllium difluoride”.

Note
The shape of the orbitals is also guided by the VSEPR or Valence Shell Electron Pair Repulsion theory, which states different rules for the prediction of the shape of the molecule and is mainly based on the number of the lone pairs and the bond pairs present in a compound. According to the theory the repulsion between the electron pairs is of the following order:
Lone pair-lone pair> lone pair-bond pair> bond pair-bond pair.