Question

1. what is the lewis dot structure for an element that will form an ion with a charge of 2+?

Explanation:

Step1: Recall Lewis Dot Structure Rules

Elements forming \(2^+\) ions lose 2 valence electrons. So the neutral atom should have 2 valence electrons (since losing 2 gives a stable octet or duplet, like alkaline earth metals: Group 2, 2 valence electrons). Lewis dot structures show valence electrons as dots around the symbol.

Step2: Analyze Each Option

• First option (\(X^\bullet\)): 1 valence electron (forms \(1^+\) ion, e.g., Group 1).
• Second option (\(\bullet X^\bullet\)): 2 valence electrons (matches Group 2, loses 2 to form \(2^+\) ion).
• Third option (\(\ddot{\underset{\bullet}{X}}\ddot{}\) – Wait, no, let's count dots. Wait, the third and fourth: Wait, no, re - check. Wait, the second option has two dots (one on left, one on right? Wait, no, the second option is \(\bullet X^\bullet\) – two valence electrons. The third and fourth have more (like 6 or 7? Wait, no, let's re - express. Wait, the second option: the dots are two (one on left, one on right of X), so valence electrons = 2. So when it loses 2, it forms \(X^{2+}\). The other options: first has 1, third and fourth have more (like 6 or 7, which would gain electrons, forming negative ions).

Answer:

\(\bullet X^\bullet\) (the second option)