Question

a lewis structure with placeholder elements is shown. if the formal charge of the central atom is +1, choose the possible identity or identities of the central atom.

Explanation:

Step1: Recall the formula for formal charge

The formula for formal charge is \(FC = V - N - \frac{B}{2}\), where \(V\) is the number of valence electrons, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.

Step2: Analyze the Lewis structure

In the given Lewis structure, the central atom has 4 bonding electrons (two double bonds). Let's assume the central atom has \(x\) valence electrons and \(y\) non - bonding electrons. The formal charge \(FC=x - y-\frac{4}{2}=x - y - 2\). Given \(FC = + 1\), so \(x-y=3\).

Step3: Check each option

• Option A (F): Valence electrons \(V = 7\). If \(x = 7\), then \(y=x - 3=4\). But F usually forms one bond (not two double bonds), so it's not likely.
• Option B (O, S, Se): Valence electrons \(V = 6\). If \(x = 6\), then \(y=x - 3=3\). But O, S, Se usually have even numbers of non - bonding electrons in common bonding situations.
• Option C (B or Al): Valence electrons \(V = 3\) (for B) or \(V = 3\) (for Al). If \(x = 3\), then \(y=x - 3=0\). B and Al can form compounds with double - bond - like resonance structures (e.g., in some boranes and aluminates), and having \(y = 0\) (no non - bonding electrons) for a + 1 formal charge is possible.
• Option D (N, P, As): Valence electrons \(V = 5\). If \(x = 5\), then \(y=x - 3=2\). But N, P, As usually form three bonds (not two double bonds) for a more stable octet.
• Option E (Cl, Br, I): Valence electrons \(V = 7\). If \(x = 7\), then \(y=x - 3=4\). But Cl, Br, I usually form one bond (not two double bonds) in simple Lewis structures.

Answer:

C. B or Al