The $$1^{st}, 2^{nd}$$ and the $$3^{rd}$$ ionization enthalpies, $$I_1, I_2$$ and $$I_3$$, of four atoms with atomic numbers 𝑛, 𝑛 + 1, 𝑛 + 2, and 𝑛 + 3 , where 𝑛 < 10, are tabulated below. What is the value of 𝑛?

The large increase ( “jump” ) that first appears in the series of ionisation enthalpies pin-points the moment an electron is removed from a closed-shell (noble-gas) configuration. The position of that jump tells us the number of valence electrons and therefore the group of the element.

• For an element of Group 1 (configuration …s1) the first electron is easy to remove, but removal of the second electron breaks into a noble-gas core, so $$I_2 \gg I_1$$.
• For Group 2 (…s2) the big jump is between $$I_2$$ and $$I_3$$.
• For Group 13 (…s2\!}p^{1}$$) the jump is between $$I_3$$ and $$I_4$$, and so on.

The given table (atoms with atomic numbers $$n,\;n+1,\;n+2,\;n+3$$, with $$n\lt10$$) shows
1. for the first atom a huge increase from $$I_1$$ to $$I_2$$,
2. for the second atom a huge increase from $$I_2$$ to $$I_3$$,
3. for the third and fourth atoms no such enormous jump in the first three values.

Interpretation:
Case 1: first atom - jump after the first electron ⇒ Group 1.
Therefore the first atom is Li (atomic number 3) because it is the only Group 1 element with $$Z\lt10$$ that can start a block of four consecutive elements.
Case 2: second atom - jump after the second electron ⇒ Group 2, matching Be (atomic number 4), which indeed follows Li.
The remaining two atoms are B ($$Z=5$$) and C ($$Z=6$$); their first three ionisation enthalpies show no dramatic jump, just as listed in the table.

Thus the only self-consistent assignment is

$$n = 3$$.

Hence, the value of $$n$$ is $$\mathbf{3}$$.