Question

the diagram shows the logarithm of the first 13 ionisation energies of an element.
log
ionisation
energy
number of electrons removed
1 2 3 4 5 6 7 8 9 10 11 12 13
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 an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^1\). The first three ionisation energies remove the 3p and 3s electrons, and then there's a big jump (as inner shell electrons are harder to remove). Looking at the graph, the jump around the 4th - 5th electron removal? Wait, no, let's think again. Aluminium has 13 electrons. The first three are outer (3s²3p¹), then the next 8 are from the 2nd shell (2s²2p⁶), then the last 2 from 1s². Wait, no, the electron configuration of Al is \( [Ne] 3s^2 3p^1 \), so total electrons: 13. The first three ionisation energies correspond to removing the 3p¹ and 3s² electrons. Then, when we try to remove the 4th electron, we're going into the neon core (2nd shell), which has much higher ionisation energy. Wait, the graph: the x - axis is number of electrons removed (1 - 13). Let's see the jumps. The first three electrons are outer (valence), then a jump (since inner shell is harder). Wait, Al has 3 valence electrons. So after removing 3 electrons, the next electron (4th) is from the inner shell (Ne - like), so ionisation energy increases a lot. Then, after that, we are removing electrons from the Ne - like shell (8 electrons: 2s²2p⁶), so 8 electrons, then a jump to remove from 1s². So the number of electrons removed: first 3 (valence), then 8 (inner shell), then 2 (innermost). So the graph should have a jump after 3 electrons (between 3rd and 4th), then a relatively steady increase for 8 electrons, then a jump. Let's check the options. Option B says the element is aluminium. Let's check other options.
• Option C: Silicon (Si) has electron configuration \( [Ne] 3s^2 3p^2 \), so 4 valence electrons. So the jump should be after 4 electrons, not 3. So C is wrong.
• Option D: If the element had only one outer electron, the first ionisation energy would be low, then a big jump (since second electron is inner shell). But the graph shows multiple electrons with relatively low ionisation energy (first 3 or so? Wait, no, the first few have a gradual increase, then a jump, then more. So D is wrong.

So the correct answer is B.

Answer:

B. The element is aluminium.