The incorrect statement is:

🧠 Zn ($Z=30$) has filled $3d^{10}\,4s^2$ — that gives Zn an unusually high $IE_1$. Ga ($Z=31$) removes a loose $4p$ electron, so Ga's $IE_1$ is LOWER than Zn's — option (d) reverses this Going from Zn to Ga, you cross from a "filled-shell stable" element to one with a single $4p$ electron. The $4p$ electron is shielded by the full $3d^{10}$ and the $4s^2$, so it is easy to pull. Result: $IE_1(\ce{Ga}) < IE_1(\ce{Zn})$.

🗺️ Check each statement (a) $IE_1(\ce{K}) < IE_1(\ce{Na}) < IE_1(\ce{Li})$: Group 1 IE rises as you go up. K (Period 4) has the lowest. True. (b) Xe is not the lowest in Group 18 — Rn is below Xe and has even lower $IE_1$. True (Xe is not the lowest). (c) $Z=37$ is Rb (Group 1, Period 5), $Z=38$ is Sr (Group 2, Period 5). Group 1 < Group 2 in $IE_1$. True. (d) Ga ($IE_1 \approx 579$) vs Zn ($IE_1 \approx 906$). Ga is lower, not higher. The statement claims Ga is higher — False. This is the incorrect option.

⚠️ The trap Students assume "higher $Z$ means higher $IE_1$" within a period. But Zn's filled $3d^{10}\,4s^2$ is unusually stable, giving it an anomalously high $IE_1$ that the next element Ga drops below. Always remember: filled subshells (and especially $d^{10}$) cause IE bumps that break the smooth trend.

$\boxed{\text{Answer: (d)}}$