Question

The ionization energies of an unknown third period element are shown here. Identify the element.

1) Sodium
2) Magnesium
3) Aluminum
4) Silicon

Answer 1

Based on the pattern of ionization energies, the element can be identified as aluminum, which is the only element from group 13 among the provided options.

Step-by-step explanation:

The ionization energies provided for the third period element suggest that this element can lose three electrons at relatively low energies but requires a much higher energy to remove the fourth electron. This high jump after the third ionization energy indicates that the fourth electron is being removed from a closer, more tightly bound energy level. This pattern is characteristic of an element from group 13 in the periodic table, where the first three electrons are valence electrons and the fourth is a core electron. Among the options given, aluminum is the only element from group 13.

Let's write the reactions for each ionization step:

1. Al(g) → Al⁺(g) + e⁻ ; ΔE = 600 kJ/mol
2. Al⁺(g) → Al²⁺(g) + e⁻ ; ΔE = 1,800 kJ/mol
3. Al²⁺(g) → Al³⁺(g) + e⁻ ; ΔE = 2,700 kJ/mol
4. Al³⁺(g) → Al⁴⁺(g) + e⁻ ; ΔE = 11,600 kJ/mol
5. Al⁴⁺(g) → Al⁵⁺(g) + e⁻ ; ΔE = 15,000 kJ/mol

Given the provided ionization energies, the element can be identified as aluminum (Al).