[VISUAL] The first five ionization energies of an unknown element are listed in the table above. Whi — Atomic Structure Chemistry Question

STEPS:

1. Understand the concept of successive ionization energies:
Ionization energy (IE) is the energy required to remove an electron from a gaseous atom or ion. Successive ionization energies ($\text{IE}_1$, $\text{IE}_2$, $\text{IE}_3$, etc.) represent the energy required to remove electrons one after another from the same atom.

2. Recall the difference between valence and core electrons:
Valence electrons reside in the outermost occupied shell and are shielded from the full nuclear charge, making them relatively easy to remove. Core electrons are located in fully filled, inner shells closer to the nucleus. Because they experience a much stronger effective nuclear charge and are closer to the nucleus, removing a core electron requires a massive, disproportionate jump in energy.

3. Analyze the given ionization energy data mathematically:
Look at the differences between successive ionization energy values in the table:
* $\text{First } \rightarrow \text{Second}$: $1,817 - 578 = 1,239\text{ kJ/mol}$ (an increase of about 3.1x)
* $\text{Second } \rightarrow \text{Third}$: $2,745 - 1,817 = 928\text{ kJ/mol}$ (an increase of about 1.5x)
* $\text{Third } \rightarrow \text{Fourth}$: $11,577 - 2,745 = \mathbf{8,832\text{ kJ/mol}}$ (an increase of about 4.2x, representing a massive energy jump)
* $\text{Fourth } \rightarrow \text{Fifth}$: $14,842 - 11,577 = 3,265\text{ kJ/mol}$ (an increase of about 1.3x)

4. Determine the number of valence electrons:
Because the colossal jump in ionization energy occurs between the third and fourth ionization energies, the first three electrons are relatively easy to remove, meaning they are valence electrons. The fourth electron is extremely difficult to remove because it is a core electron. Therefore, the unknown element must have exactly three valence electrons.

5. Identify the element:
* Look at the options and find the element with three valence electrons.
* Aluminum ($\text{Al}$) has the ground-state electron configuration $1s^2 2s^2 2p^6 3s^2 3p^1$, which translates to exactly three valence electrons in its outermost shell ($n=3$).
* Removing the fourth electron from $\text{Al}$ requires breaking into the highly stable, lower-energy $2p$ subshell, which accounts for the massive transition from $2,745\text{ kJ/mol}$ to $11,577\text{ kJ/mol}$. This identifies Option B as the correct choice.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Sodium ($\text{Na}$) is an alkali metal with only one valence electron ($[\text{Ne}] 3s^1$). If the element were $\text{Na}$, the massive jump in ionization energy would occur between the *first* and *second* ionization energies ($\text{IE}_1$ to $\text{IE}_2$) as you attempt to remove a core electron from the stable $2p$ shell.
• Option C is incorrect: Although the fifth ionization energy has the absolute greatest value in the table, this is a natural consequence of removing electrons from an increasingly positive cation. It is the *difference* between successive values—representing the transition from valence to core electrons—that identifies the element, not simply which ionization event is the most endothermic.
• Option D is incorrect: While a neutral phosphorus ($\text{P}$) atom does have five valence electrons ($[\text{Ne}] 3s^2 3p^3$), an element with five valence electrons would exhibit its major ionization energy jump between the *fifth* and *sixth* ionization energies ($\text{IE}_5$ to $\text{IE}_6$) when transitioning to removing core electrons.