Question

In second period of the modern periodic table, two elements X and Y have higher first ionization enthalpy values than the preceding and succeeding elements. X and Y are respectively

• A. B, C
• B. Al, S
• C. Be, N
• D. Na, S

Hint:

The two main exceptions to the increasing trend of first ionization enthalpy across a period occur when removing an electron from a fully-filled s-orbital (Group 2, e.g., Be) and a half-filled p-orbital (Group 15, e.g., N). These configurations are extra stable, requiring more energy for ionization.

Answer 1

The Correct Option is C

we are asked to identify the two elements in the second period of the modern periodic table that have higher first ionization enthalpy (IE$_1$) than both their preceding and succeeding elements.
Step 1: Recall the trend of first ionization enthalpy
The first ionization enthalpy generally increases across a period from left to right due to:

• Increasing nuclear charge
• Decreasing atomic size

Step 2: List elements of the second period and their electronic configurations
\begin{tabular}{l l} Li: & [He] $2s^1$
Be: & [He] $2s^2$
B: & [He] $2s^2 2p^1$
C: & [He] $2s^2 2p^2$
N: & [He] $2s^2 2p^3$
O: & [He] $2s^2 2p^4$
F: & [He] $2s^2 2p^5$
Ne: & [He] $2s^2 2p^6$
\end{tabular}
Step 3: Identify anomalies due to electronic configurations

• Beryllium (Be): Fully filled $2s$ orbital ([He] $2s^2$) is stable. Removing an electron from Be requires more energy than from Li or B. $\Rightarrow \text{IE}_1(\text{Be})>\text{IE}_1(\text{Li})$ and $\text{IE}_1(\text{Be})>\text{IE}_1(\text{B})$ Hence, Be is one of the elements (X).
  
• Nitrogen (N): Half-filled $2p$ orbital ([He] $2s^2 2p^3$) is particularly stable. Removing an electron from N is harder than from C or O. $\Rightarrow \text{IE}_1(\text{N})>\text{IE}_1(\text{C})$ and $\text{IE}_1(\text{N})>\text{IE}_1(\text{O})$ Hence, N is the second element (Y).

Step 4: Conclusion The two elements are: \[ \text{X = Be, Y = N} \] \[ \text{Correct Answer: (C) Be, N} \]