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Question: Which one of the following gas is most soluble in water? A. \({\rm{CO}}\) B. \({{\rm{O}}_{\rm{2...

Which one of the following gas is most soluble in water?
A. CO{\rm{CO}}
B. O2{{\rm{O}}_{\rm{2}}}
C. NO{\rm{NO}}
D. CO2{\rm{C}}{{\rm{O}}_{\rm{2}}}

Explanation

Solution

The effect of the inter-molecular interaction for a gas and solvent on its solubility in water can be used to deduce one by using another.
We can define solubility as the maximum amount of a given solute that we can dissolve in a given solvent at given temperature. Its value varies with the nature of the solute, solvent and conditions of temperature and pressure.

Complete step by step solution
Here, we are given different solutes and water as a solvent so we will have a look at the effect of nature of solute on the solubility.
We can understand the nature of a gas in terms of its structural properties as it is the inter-molecular attraction between the solute and solvent that would decide whether the gas is soluble or not. Here, we have CO,O2,NO{\rm{CO,}}{{\rm{O}}_{\rm{2}}},{\rm{NO}} and CO2{\rm{C}}{{\rm{O}}_{\rm{2}}} gases as solutes. So, let’s have a look at their structures:

We can go through the charge separation and bond polarity in these molecules to have a better understanding of the bonding and their nature. We can find out the formal charge (FC)\left( {FC} \right) on an atom by the following formula:
FC=ntotalvalenceelectrons−nnon−bondingelectrons−nbondingelectrons2FC = {n_{{\rm{total valence electrons}}}} - {n_{{\rm{non - bonding electrons}}}} - \dfrac{{{n_{{\rm{bonding electrons}}}}}}{2}
Let’s do this for all the molecules one by one as follows:
carbon monoxide:
FC(C)=4−2−62 =−1 FC(O)=6−2−62 =+1 FC\left( C \right) = 4 - 2 - \dfrac{6}{2}\\\ = - 1\\\ FC\left( O \right) = 6 - 2 - \dfrac{6}{2}\\\ = + 1
Nitric oxide:
FC(N)=5−3−42 =0 FC(O)=6−4−42 =0 FC\left( N \right) = 5 - 3 - \dfrac{4}{2}\\\ = 0\\\ FC\left( O \right) = 6 - 4 - \dfrac{4}{2}\\\ = 0
carbon dioxide:
FC(C)=4−0−82 =0 FC(O)=6−4−42 =0 FC\left( C \right) = 4 - 0 - \dfrac{8}{2}\\\ = 0\\\ FC\left( O \right) = 6 - 4 - \dfrac{4}{2}\\\ = 0
Among these, in oxygen gas, the charge would be zero for both the atoms as they have complete octet and are the same atoms with equal number of electrons and there would be no bond polarity as well. In nitric oxide, there is no charge separation but there would be bond polarity for electronegativity difference between nitrogen and oxygen yet t would be lesser than in case of carbon and oxygen. Now for carbon monoxide and carbon dioxide, we have charge separation in the first but there is a positive charge on oxygen which is not favorable. In case of carbon dioxide, we have two polar bonds and it can also react with water to form carbonic acid that facilitates its solubility.
Hence, we can say that **the correct option is D.

Note: **
The molecule of carbon dioxide is non-polar as the two equal bond dipoles are directed in opposite directions but the bonds are polar and can interact with the water molecules.