Question

Which of the following molecules is not linear?
A. $O = C = C = C = O$
B. ${H_2} = C = O$
C. $H \equiv C\,\, - \,\,C \equiv CH$
D. $HC \equiv CH\,\,$

Answer 1

To solve this question, we must first understand how the geometry of a molecule is affected by the hybridization of the atoms in the molecule. Then we must find the hybridization of all the carbon atoms in the given molecules to determine which molecule does not have a linear shape.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
To determine the hybridization and shape of the given molecule, we must follow these steps:
1.Draw the Lewis structure to understand the rough structure of the given molecule and its bonding pattern. We can use the valence electron concept to make the Lewis structure.
2.After that, we must calculate the number of sigma bonds in the given compound. Sigma bonds can be understood as the single bonds which form the basic skeleton while bonding.
3.Following which, we must calculate the number of lone pairs. We can either count the number of lone pairs present in the Lewis structure or we can calculate the number of lone pairs using the formula: $\dfrac{{v - b - c}}{2}$ , where v is the number of valence electrons, b is the total number of bonds and c is the charge on the atom.
4.We must now calculate the steric number. The formula for steric number is:
Steric number = (number of sigma bonds) + (number of lone pairs)
5.Now, the final step includes to correlate the steric number calculated to the following chart:

Steric number	hybridization	Structure
2	sp	Linear
3	$s{p^2}$	Trigonal planar
4	$s{p^3}$	Tetrahedral
5	$s{p^3}d$	Trigonal bipyramidal
6	$s{p^3}{d^2}$	Octahedral
7	$s{p^3}{d^3}$	Pentagonal bipyramidal

The steric numbers for the given compounds can be calculated as follows:
A. $O = C = C = C = O$ : Every carbon in this molecule has no valence electrons, forms 2 sigma bonds in total, and has no charge on it. Hence, all the carbon atoms in this compound are sp hybridised. Hence, this molecule is linear
B. ${H_2}C = C = O$ : There are two carbon atoms present in this molecule. The carbon atom at position 01 has no lone pairs, and forms 3 sigma bonds. Hence, it is $s{p^2}$ hybridised, which means that the given molecule is not linear.
C. $H \equiv C\,\, - \,\,C \equiv CH$ : Every carbon in this molecule has no valence electrons, forms 2 sigma bonds in total, and has no charge on it. Hence, all the carbon atoms in this compound are sp hybridised. Hence, this molecule is linear
D. $HC \equiv CH\,\,$ : Every carbon in this molecule has no valence electrons, forms 2 sigma bonds in total, and has no charge on it. Hence, all the carbon atoms in this compound are sp hybridised. Hence, this molecule is linear

Hence, Option B is the correct option

Note: Hybridization allows for the most stable (and most desirable) structure. When there are hybrid orbitals there are enough electrons to complete the necessary bonds - regardless of whether there is a suitable number of valence electrons.