The first five ionization energies of a second-period element are listed in the table. [VISUAL] Whic — Atomic Structure Chemistry Question

STEPS:

1. Understand successive ionization energy trends: Ionization energy is the energy required to remove an electron from a gaseous atom or ion. Successive ionization energies (first, second, third, etc.) always increase because each subsequent electron is being removed from a more positively charged ion.
2. Identify the massive jump in the data: Look at the ratios or differences between successive ionization energies in the provided table:
* First to Second: $801\text{ kJ/mol} \rightarrow 2,430\text{ kJ/mol}$ (approx. 3-fold increase)
* Second to Third: $2,430\text{ kJ/mol} \rightarrow 3,660\text{ kJ/mol}$ (approx. 1.5-fold increase)
* Third to Fourth: $3,660\text{ kJ/mol} \rightarrow \mathbf{25,000\text{ kJ/mol}}$ (approx. 7-fold increase!)
* Fourth to Fifth: $25,000\text{ kJ/mol} \rightarrow 32,820\text{ kJ/mol}$ (approx. 1.3-fold increase)
3. Relate the jump to electron shielding and core vs. valence electrons:
* The first three electrons are relatively easy to remove because they are valence electrons located in the outermost shell ($n=2$ for a second-period element).
* The fourth electron requires a massive, disproportionate amount of energy to remove because it is a core electron located in an inner shell ($n=1$). Core electrons are much closer to the nucleus, experience a much stronger effective nuclear charge ($Z_{\text{eff}}$), and experience negligible shielding compared to valence electrons.
* This huge energy gap between $IE_3$ and $IE_4$ reveals that the unknown element has exactly three valence electrons.
4. Identify the element:
* A second-period element with exactly three valence electrons has the valence electron configuration $2s^2 2p^1$.
* This element has a total of 5 electrons (3 valence + 2 core), which is Boron (B) (atomic number 5).
5. Select the correct option: Option B correctly identifies the element as Boron (B) and explains that it has three valence electrons.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While the element is indeed Boron (B), it does not have five core electrons. Boron has 5 *total* electrons, of which only two are core electrons ($1s^2$) and three are valence electrons ($2s^2 2p^1$).
• Option C is incorrect: Nitrogen (N) has 5 valence electrons ($2s^2 2p^3$). If N were the element in the table, the massive jump in ionization energy would occur between the fifth and sixth ionization energies (when the first core electron from the $1s$ subshell is removed), not between the third and fourth.
• Option D is incorrect: Although Nitrogen (N) does indeed have three electrons in its $p$ sublevel ($2s^2 2p^3$), this does not explain the dramatic jump between the third and fourth ionization energies. A jump at $IE_4$ indicates the removal of a core electron, which points to Boron rather than Nitrogen.