Elements involved

We compare first ionization enthalpies of Group 13 elements: $\mathrm{B},\ \mathrm{Al},\ \mathrm{Ga},\ \mathrm{Tl}$

General trend down a group

Ionization enthalpy generally decreases down a group because:

• atomic size increases,
• outer electron is farther from nucleus,
• shielding increases.

So we may first expect: $\mathrm{B} > \mathrm{Al} > \mathrm{Ga} > \mathrm{Tl} \quad \text{(rough expectation)}$

Important exceptions in Group 13

In Group 13, the trend is not perfectly regular because of poor shielding by inner $d$ and $f$ electrons.

• From $\mathrm{Al}$ to $\mathrm{Ga}$: Ga has intervening $3d$ electrons, which shield poorly.
• Hence effective nuclear charge on the valence electron in Ga increases, making Ga slightly smaller and its first ionization enthalpy slightly higher than Al.

Therefore: $\mathrm{Ga} > \mathrm{Al}$

• From $\mathrm{Ga}$ to $\mathrm{Tl}$: due to poor shielding by $4f$ and $5d$ electrons and inert pair effect, Tl has relatively high ionization enthalpy.
• In fact, $\mathrm{Tl}$ has higher first ionization enthalpy than Ga.

Therefore: $\mathrm{Tl} > \mathrm{Ga}$

Comparison with boron

Boron is the smallest in the group, so its first ionization enthalpy is the highest among these lighter members.

Thus the actual order among the listed relevant elements is: $\mathrm{B} > \mathrm{Tl} > \mathrm{Ga} > \mathrm{Al}$

Check options

• A: $\mathrm{Ga} > \mathrm{Al} > \mathrm{B}$ ❌ since $\mathrm{B}$ should be highest.
• B: $\mathrm{Tl} > \mathrm{Ga} > \mathrm{Al}$ ✅ correct relative order among these three.
• C: $\mathrm{Al} > \mathrm{Ga} > \mathrm{Tl}$ ❌ wrong because $\mathrm{Ga} > \mathrm{Al}$ and $\mathrm{Tl}$ is not lowest.
• D: $\mathrm{B} > \mathrm{Al} > \mathrm{Ga}$ ❌ wrong because $\mathrm{Ga} > \mathrm{Al}$.

Final answer

The correct option is: $\boxed{\text{B: } \mathrm{Tl} > \mathrm{Ga} > \mathrm{Al}}$