Question

$CO_{ 2 }$ is isostructural with:
A.) $HgCl_{ 2 }$
B.) $H_{ 2 }O$
C.) $SnCl_{ 2 }$
D.) $NO_{ 2 }$

Answer 1

Hint: To find a compound to be isostructural, first match its hybridization of the central atom of species in question with the central atom of species given in options. Also, $CO_{ 2 }$ has a linear structure. Now you can easily find the answer.

Complete step by step answer:

Carbon dioxide is a molecule composed of one carbon atom bonded to two oxygen atoms via two double bonds and there are no lone pairs on C atoms. Therefore, hybridization is sp hybridization. $CO_{ 2 }$ is a linear molecule with a C-atom in the middle and two O-atoms bonded at $180^{ 0 }$.
Now, we will discuss all the options one by one -
In option A, the molecular geometry of mercuric chloride ($HgCl_{ 2 }$) is Linear. As there are no lone pairs with Hg and the number of bonding pairs is two. Therefore, hybridization is sp hybridization and the structure is linear.

In option B, Water has 4 regions of electron density around the central oxygen atom (2 bonds and 2 lone pairs). These are arranged in a tetrahedral ($sp^{ 3 }$) shape. The resulting molecular shape is bent with an H-O-H angle of $104.5^{ 0 }$.

In option C, tin (Sn) is a group 14 element, it shows $sp^{ 3 }$ hybridization with two bond pairs between $sp^{ 3 }$ - p orbital of two chlorine atoms, and the remaining two lone pairs are present. The shape of the molecule is bent or V-shaped.

In option D, $NO_{ 2 }$ is bent because of the presence of an extra electron. This electron introduces repulsion in the molecule. The hybridization of $NO_{ 2 }$ is $sp^{ 2 }$. Hence, the shape is somewhat bent but not linear.

Therefore, the correct answer to this question is option A.

Note: You should know that in chemistry, hybridization is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.