Question

The first ionization enthalpies of $Be , B , N$ and $O$ follow the order

• A. $O < N < B < Be$
• B. $Be < B < N < O$
• C. $B < Be < N < O$
• D. $B < Be < O < N$

Answer 1

The Correct Option is D

The problem is to find the correct order of the first ionization enthalpies for the elements Beryllium (Be), Boron (B), Nitrogen (N), and Oxygen (O). To solve this, let's understand the concept of ionization enthalpy.

Ionization enthalpy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms or ions. Generally, it increases across a period due to an increase in nuclear charge, making it more difficult to remove an electron.

However, there are exceptions owing to electronic configurations:

• Beryllium (Be): Electronic configuration: \(1s^2 2s^2\)
• Boron (B): Electronic configuration: \(1s^2 2s^2 2p^1\)
• Nitrogen (N): Electronic configuration: \(1s^2 2s^2 2p^3\)
• Oxygen (O): Electronic configuration: \(1s^2 2s^2 2p^4\)

 

Let's consider these electronic configurations and the associated stability:

• Among Be and B, Be has a completely filled 2s orbital making it relatively more stable. Thus, B has a lower ionization enthalpy than Be.
• N has a half-filled 2p subshell which is stable due to exchange energy. Thus, the ionization enthalpy of N is higher than O.

 

Based on the above reasoning and configurations, the order is:

• Be has a higher ionization enthalpy than B.
• N has a higher ionization enthalpy than O.

Thus, the correct order is \(B < Be < O < N\).

In summary, the anomalies in the periodic trend are due to the extra stability of half-filled and completely filled orbitals, which explains why the correct answer is the fourth option: \(B < Be < O < N\).