Question

Which molecule is linear? Give an example.

Answer 1

VSEPR theory is used in the determination of linearity of a molecule. Usually, linear geometry occurs at central atoms with two bonded atoms and zero or three lone pairs. Here, the polarity of the dipole moment is also $0$.

Complete answer:
Linear molecules are those molecules in which two atoms are situated at a straight line or a bond angle of $180^\circ$ around a central atom.
Linearity of molecules is determined by their molecular geometry using the Valence Shell Electron Pair Repulsion Theory or VSEPR theory.
It involves the $3 -$ dimensional arrangement of substrates attached (bonded) to a 'central' element of a 'binary' structure. Linear molecular geometry occurs at central atoms of molecules having $A{X_n}$ molecular formula with $A$ being the central element, $X$ being the attached substrates and $n$ being the total number of electron pairs.
To get the structural geometry of compounds we have to determine the number of bonded pairs of electrons and number of lone pairs of electrons. The number of bonded pairs is equal to the number of substrates attached to the central element of the compound and represents the number of 'Bonded Pairs' of electrons.
For example, in $BeC{l_2}$, there are $2$ substrates attached to the central element Beryllium. Hence the number of bonded pairs is $2$.
To calculate the number of lone pairs, we will use the following expression:
Number of lone pairs $= \dfrac{{{\text{Valence electrons}} - {\text{Substrate electrons}}}}{2}$
In $BeC{l_2}$, there are $16$ valence electrons and $16$ substrate electrons.
Hence, Number of lone pairs $= \dfrac{{16 - 16}}{2} = 0$
Total number of electron pairs $=$ number of bonded pairs $+$ number of lone pairs $= 2 + 0 = 2$
Hence geometry is $X - A - X$, thus $BeC{l_2}$ is a linear molecule.
Let’s consider another example of $SnC{l_2}$.
Here, the number of bonded pairs is again $2$ while there are $18$ valence electrons and $16$ substrate electrons.
Number of lone pairs $= \dfrac{{{\text{Valence electrons}} - {\text{Substrate electrons}}}}{2}$
Number of lone pairs $= \dfrac{{18 - 16}}{2} = \dfrac{2}{2} = 1$
Total number of electron pairs $=$ number of bonded pairs $+$ number of lone pairs $= 2 + 1 = 3$
Hence the parent geometrical structure is $A{X_3}$, a trigonal planar geometry on the central $Sn$ atom. The resulting geometry is $A{X_2}E$ bent angular structure. Hence, it is a non-linear molecule.
Some other examples of linear compounds are $C{O_2},Xe{F_2}$ and ${C_2}{H_2}$.

Note:
A simpler intuition to understand this is to simply remember that lone pairs repel each other. They try to get as far as possible from each other and hence create a linear geometry of the compound. For example, in $Be{H_2}$, there are $0$ lone pairs and it is a linear compound as the hydrogen atom repels each as much as possible hence making a straight line.
You might consider ${H_2}O$ as a linear compound, but it actually has a tetrahedral bent shape. The $2$ hydrogens only use $4$ electrons out of $8$ electrons of oxygen. Hence, there are $2$ lone pairs left on oxygen. These lone pairs repel each other and give ${H_2}O$ its tetrahedral bent shape.