Periodic TableMedium

Question

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The first ionization enthalpy of O is lower than that of N and F.

Reason R: The loss of an electron from O leads to stable half-filled $p$ orbital.

In light of the above statements, choose the most appropriate answer from the options given below:

Options

A.$\mathbf{A}$ is correct but $\mathbf{R}$ is not correct.
B.$\mathbf{A}$ is not correct but $\mathbf{R}$ is correct
C.Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
D.Both A and R are correct and R is NOT the correct explanation of A

Solution

Sol. [Key Concept: Assertion–Reason evaluation]

$N = 1{\text{ }s}^{2}2{\text{ }s}^{2}2p^{3}$

$$F:1s^{2}2s^{2}2p^{5} $$First ionization enthalpy ( $\Delta_{i}H$ )

$${\begin{matrix} & \left( \Delta_{i}H \right)\text{~}\text{for}\text{~}N = 1402\text{ }kJ{\text{ }mol}^{- 1} \\ & \Delta_{i}H\text{~}\text{for}\text{~}O = 1314\text{ }kJ{\text{ }mol}^{- 1} \\ & \Delta_{i}H\text{~}\text{for}\text{~}F = 1681\text{ }kJ/{mol}^{-} \end{matrix} } {Due\ to\ half\ filled\ stable\ electronic\ configuration\ of\ nitrogen,\ it's\ first\ ionization\ enthalpy\ is\ higher\ than\ that\ of\ oxygen } {\begin{matrix} & O(\text{ }g) \longrightarrow O^{+}(g) + e^{-} \\ & 2{\text{ }s}^{2}2p^{4} \\ & 2{\text{ }s}^{2}2p^{3} \end{matrix} } {(half\ filled\ p\ subshell)}$$

Create a free account to view solution

View Solution Free
Topic: Periodic Table·Practice all Periodic Table questions

More Periodic Table Questions