[VISUAL] Ionization Energy | (kJ/mol) First | 730 Second | 1450 Third | 7700 Fourth | 10,500 The ion — Atomic Structure Chemistry Question

STEPS:

1. Analyze the successive ionization energies for element X:
The energy required to remove each successive electron from an atom of element X is given as follows:
* First ionization energy ($\text{IE}_1$): $730\text{ kJ/mol}$
* Second ionization energy ($\text{IE}_2$): $1450\text{ kJ/mol}$
* Third ionization energy ($\text{IE}_3$): $7700\text{ kJ/mol}$
* Fourth ionization energy ($\text{IE}_4$): $10,500\text{ kJ/mol}$

2. Identify the core chemical barrier (the "jump" in ionization energy):
* The energy required to remove the second electron ($1450\text{ kJ/mol}$) is roughly double the first ($730\text{ kJ/mol}$), which is a typical increase due to removing an electron from a now positively charged ion.
* However, there is a massive, five-fold jump in energy between the second and third ionization energies ($1450\text{ kJ/mol}$ to $7700\text{ kJ/mol}$).
* This huge jump indicates that the third electron must be removed from a stable, inner-shell core orbital (closer to the nucleus, experiencing a much stronger effective nuclear charge) after all valence electrons have been removed.

3. Determine the valence configuration and common ion charge of X:
* Because only two electrons can be removed relatively easily before encountering the inner-shell barrier, element X must have exactly 2 valence electrons.
* To achieve a stable noble-gas configuration, element X will lose these 2 valence electrons, forming a cation with a $+2$ charge ($\text{X}^{2+}$).

4. Determine the ion charge of phosphorus (P):
* Phosphorus is located in Group 15 of the periodic table.
* Elements in Group 15 possess 5 valence electrons and typically gain 3 electrons to complete their octet.
* Therefore, in an ionic compound, phosphorus forms a phosphide anion with a $-3$ charge ($\text{P}^{3-}$).

5. Balance the charges to find the empirical formula:
* To form a neutral ionic compound, the total positive charge from the $\text{X}^{2+}$ cations must balance the total negative charge from the $\text{P}^{3-}$ anions.
* The least common multiple of 2 and 3 is 6:
* Three $\text{X}^{2+}$ ions provide a total charge of $3 \times (+2) = +6$.
* Two $\text{P}^{3-}$ ions provide a total charge of $2 \times (-3) = -6$.
* Combining these in a $3:2$ ratio yields the empirical formula $\text{X}_3\text{P}_2$, which identifies Option C as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect (XP): This $1:1$ empirical formula would require element X to form a $+3$ cation (to balance $\text{P}^{3-}$). If this were the case, the massive jump in ionization energies would occur between the third and fourth ionization values, rather than between the second and third.
• Option B is incorrect ($\text{X}_3\text{P}$): This $3:1$ empirical formula would imply that element X forms a $+1$ cation (to balance a single $\text{P}^{3-}$ anion). A student might choose this if they mistakenly believed a large ionization energy jump occurred immediately after the first ionization.
• Option D is incorrect ($\text{X}_3\text{P}_4$): This $3:4$ empirical formula would require element X to form a $+4$ cation (since four $\text{P}^{3-}$ anions yield a $-12$ charge, which would require three $+4$ cations to balance). The table shows that removing a third and fourth electron is extremely unfavorable and highly unlikely under typical chemical conditions.