Question

Give reasons for the following:
(i) $\left(C H_{3}\right)_{3} P=O$ exists but $\left(C H_{3}\right)_{3} N=O$ does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) $H_{3} P O_{2}$ is a stronger reducing agent than $H_{3} P O_{3}$

Answer 1

Hint: In order to solve this question we are going to use electronic configuration of atoms.The concepts of periodic properties and redox also has to be used in this question.

Complete step by step answer:-
(i) $\left(C H_{3}\right)_{3} P=O$ In this compound phosphorus (p) can expand its covalency beyond four due to the presence of ‘d’ orbital and form $d \pi-p \pi$ bonds, therefore the $\left(C H_{3}\right)_{3} P=O$, whereas in case of $\left(C H_{3}\right)_{3} N=O$ Nitrogen cannot expand its covalency due to absence of ‘d’ subshell. Hence the $d \pi-p \pi$ bonds cannot be formed. Hence $\left(C H_{3}\right)_{3} N=O$ doesn’t exist.

(ii) As we move across the period from left to right the atom's size decreases, due to this oxygen having a compact atomic size and presence of negative charge makes it difficult for the accommodation of incoming electrons, hence oxygen has less electron gain enthalpy compared to sulphur.

(iii) $H_{3} P O_{2}$ is a stronger reducing agent than $H_{3} P O_{3}$ because a compound is considered a good reducing agent when it is easily able to give away hydrogen in large numbers. In this question $H_{3} P O_{2}$ has 2 hydrogen bonds as compared to one hydrogen bond in $H_{3} P O_{3}$. Hence $H_{3} P O_{2}$ is a good reducing agent.

Note:
Students generally make mistakes in understanding the characteristics of atoms across the period and down the group. They will also make mistakes in atoms' electronic nature and find it difficult to determine why electron affinity is less in some atoms and more in some atoms. They might make mistakes in understanding the meaning of reduction and oxidation process.