Determine Atomic Number from Successive Ionization Enthalpies

To determine the atomic number $n$, we analyze the trends and sudden jumps in the given ionization enthalpies ($I_1, I_2, I_3$):

1. Analyzing Element $(n + 2)$:

   • $I_1 = 496 \text{ kJ/mol}$
   • $I_2 = 4562 \text{ kJ/mol}$

   The large ratio $\frac{I_2}{I_1} \approx 9.2$ indicates a massive jump between the first and second ionization energies. This signifies that the removal of the second electron involves removing an electron from a stable, fully filled noble gas shell. Therefore, the element with atomic number $(n + 2)$ has $1$ valence electron and belongs to Group 1 (Alkali metals).

2. Analyzing Element $(n + 3)$:

   • $I_1 = 738 \text{ kJ/mol}$
   • $I_2 = 1451 \text{ kJ/mol}$
   • $I_3 = 7733 \text{ kJ/mol}$

   Here, the significant jump occurs between $I_2$ and $I_3$ ($\frac{I_3}{I_2} \approx 5.3$), which indicates that this element has $2$ valence electrons and belongs to Group 2 (Alkaline earth metals).

3. Identifying the specific elements:

   • The first ionization enthalpy of $496 \text{ kJ/mol}$ corresponds specifically to Sodium ($\text{Na}$), which has an atomic number $Z = 11$.
   • Consequently, the element $(n+3)$ with $Z = 12$ is Magnesium ($\text{Mg}$).
   • The elements $n$ and $n+1$ correspond to Fluorine ($\text{F}$, $Z = 9$) and Neon ($\text{Ne}$, $Z = 10$) respectively, which matches the high $I_1$ values of $1681 \text{ kJ/mol}$ and $2081 \text{ kJ/mol}$.

4. Calculating $n$: $n + 2 = 11 \implies n = 9$

This value satisfies the given condition $n < 10$.

Final Answer: The value of $n$ is 9.