Question

explain why the zn²⁺ ion has some stability.

a) the zinc ion has achieved a noble gas configuration.

b) the zinc ion has full s, p, and d sublevels to form a pseudo - noble gas configuration.

c) the zinc ion is a transition metal.

d) the zinc ion has bonded with two free electrons.

Explanation:

To determine why \( \text{Zn}^{2+} \) is stable, analyze each option:

• Option A: Zn has an electron configuration of \([Ar] 3d^{10} 4s^2\). When it forms \( \text{Zn}^{2+} \), it loses the 4s electrons, resulting in \([Ar] 3d^{10}\). Noble gas configurations have filled s and p sublevels (e.g., \([Ne]\), \([Ar]\)), but \( \text{Zn}^{2+} \) has a filled d sublevel too, so it's not a typical noble gas configuration. Eliminate A.
• Option B: \( \text{Zn}^{2+} \) has the electron configuration \([Ar] 3d^{10}\). The 3d sublevel is full (10 electrons), the 4s sublevel is empty (after losing electrons), and the 3p sublevel (from Ar) is full. This forms a pseudo - noble gas configuration (filled d, along with filled s and p of the noble gas core), which is a stable electron arrangement. This option is correct.
• Option C: Just being a transition metal does not explain the stability of \( \text{Zn}^{2+} \). Many transition metal ions are stable due to electron configurations, not just the classification. Eliminate C.
• Option D: \( \text{Zn}^{2+} \) is formed by losing electrons, not bonding with free electrons. Eliminate D.

Answer:

B) The zinc ion has full s, p, and d sublevels to form a pseudo - noble gas configuration.