Question

Which set of elements shows positive electron gain enthalpy?

• A. $\mathrm{He},\mathrm{N},\mathrm{O}$
• B. $\mathrm{Ne},\mathrm{N},\mathrm{CI}$
• C. $\mathrm{O},\mathrm{CI},\mathrm{F}$
• D. $\mathrm{N},\mathrm{He},\mathrm{Ne}$

Answer 1

The correct option is D

$\mathrm{N},\mathrm{He},\mathrm{Ne}$

Explanation for correct option

(D) $\mathbf{N}\mathbf{,}\mathbf{}\mathbf{He}\mathbf{,}\mathbf{}\mathbf{Ne}$

Electron gain enthalpy

• When one electron is added to a single gaseous atom, a certain amount of energy is released, known as the electron gain enthalpy.
• Energy may be released or used up during the injection of an electron.

$\mathbf{N}\mathbf{,}\mathbf{}\mathbf{He}\mathbf{,}\mathbf{}\mathbf{Ne}$

• For elements in groups $2,15,$ and $18$, where the atoms have symmetrical structures, there exist irregular patterns in the values of electron gain enthalpy.
• They have less need to pick up more electrons as a result of their arrangement being unstable or asymmetric.
• Elements with fully filled orbitals do not accept electrons.
• They are more likely to accept the electron when adding heat energy.
• This shows that the element has a positive enthalpy gain.
• Noble gases like Helium $\left(\mathrm{He}\right)$ and Neon $\left(\mathrm{Ne}\right)$ would be difficult to accommodate an electron.
• Therefore, Helium and Neon have positive electron gain enthalpy.
• Nitrogen is a group $15$ element and shows positive electron gain enthalpy due to half-filled orbital.
• The electronic configuration of nitrogen is as follows:
• $\mathrm{N}=1s^{2}2s^{2}2p^3$

The explanationThe explanation for incorrect options

(A) $\mathbf{He}\mathbf{,}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{O}$

• Atomic number of oxygen $\left(\mathrm{O}\right)$ is $8$.
• The electronic configuration is as follows:

$\mathrm{O}=1s^{2}2s^{2}2p^4$

• Here, Helium and Nitrogen have positive electron gain enthalpy but Oxygen has negative electron gain enthalpy.

(B) $\mathbf{Ne}\mathbf{,}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{CI}$

• Atomic number of Chlorine $\left(\mathrm{Cl}\right)$ is $17$.
• The electronic configuration is as follows:

$\mathrm{Cl}=\left[\mathrm{Ne}\right]3s^{2}3p^5$

• Here, Neon and Nitrogen have positive electron gain enthalpy but Chlorine has negative electron gain enthalpy.

(D) $\mathbf{O}\mathbf{,}\mathbf{}\mathbf{CI}\mathbf{,}\mathbf{}\mathbf{F}$

• Atomic number of Flourine $\left(\mathrm{F}\right)$ is $9$.
• The electronic configuration is as follows:

$\mathrm{F}=1s^{2}2s^{2}2p^5$

• Here, Oxygen, Chlorine and Flourine have negative electron gain enthalpy.

Therefore, the correct option is (D) $\mathbf{N}\mathbf{,}\mathbf{}\mathbf{He}\mathbf{,}\mathbf{}\mathbf{Ne}$.