Use the pattern of successive ionization enthalpies

The key idea is:

• If there is a very large jump after the 1st ionization enthalpy, the element has 1 valence electron $\Rightarrow$ group 1.
• If there is a very large jump after the 2nd ionization enthalpy, the element has 2 valence electrons $\Rightarrow$ group 2.

Analyze element A

Given for $A$: $I_1 = 899,\quad I_2 = 1757,\quad I_3 = 14847\ \text{kJ mol}^{-1}$

Here, $I_3 - I_2 = 14847 - 1757 = 13090$

There is a huge jump between the 2nd and 3rd ionization enthalpies.

So, after removing 2 electrons, the atom reaches a stable noble gas configuration. Therefore, $A$ has 2 valence electrons.

Hence, A belongs to group 2.

Analyze element B

Given for $B$: $I_1 = 737,\quad I_2 = 1450,\quad I_3 = 7731\ \text{kJ mol}^{-1}$

Here, $I_3 - I_2 = 7731 - 1450 = 6281$

Again, there is a large jump between the 2nd and 3rd ionization enthalpies.

So, $B$ also has 2 valence electrons.

Hence, B also belongs to group 2.

Determine which one is below the other in the group

Down a group, ionization enthalpy generally decreases due to increasing atomic size and shielding.

Compare first ionization enthalpies: $A: 899\ \text{kJ mol}^{-1}, \qquad B: 737\ \text{kJ mol}^{-1}$

Since $B$ has the lower ionization enthalpy, $B$ must lie below $A$ in the same group.

Match with the options

• A: group 1, $B$ below $A$ $\to$ incorrect
• B: group 1, $A$ below $B$ $\to$ incorrect
• C: group 2, $B$ below $A$ $\to$ correct
• D: group 2, $A$ below $B$ $\to$ incorrect

Final answer

The correct option is: $\boxed{\text{C}}$