Question

the diagram shows the logarithm of the first 13 ionisation energies of an element.
chart: y - axis labeled log ionisation energy, x - axis labeled number of electrons removed (ranging from 1 to 13) with x marks plotted
which statement is correct?
a a proton is lost for each successive ionisation energy.
b the element is aluminium.
c the element is silicon.
d the element has only one outer electron.

Explanation:

• Option A: Ionisation energy involves removing electrons, not protons. Protons are in the nucleus and don't get lost during ionisation, so A is wrong.
• Option B: Aluminium (Al) has electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^1\). The first three ionisation energies remove the 3p, 3s electrons (outer shell), then the next 8 remove from the second shell, and the 13th from the first. The graph shows a big jump after 3 electrons (removing outer shell), then a smaller jump after 10 (removing second shell), matching Al's electron removal pattern.
• Option C: Silicon (Si) has electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^2\). The big jump should be after 4 electrons (removing outer shell), but the graph has a big jump after 3, so C is wrong.
• Option D: If the element had one outer electron, the big jump would be after 1 electron (removing the only outer electron), but the graph's big jump is after 3, so D is wrong.

Answer:

B. The element is aluminium.