Question

which of the following has the greatest number of unshared electrons around the central atom?
nh₄⁺
ch₄
clf₃
bcl₃
icl₅

Explanation:

Step1: Calculate valence electrons for each central atom

• $\text{NH}_4^+$ central N: valence e⁻ = 5 - 1 (charge) = 4; bonding pairs = 4; lone pairs = $\frac{4 - 4}{2}$ = 0 → 0 unshared e⁻.
• $\text{CH}_4$ central C: valence e⁻ = 4; bonding pairs = 4; lone pairs = $\frac{4 - 4}{2}$ = 0 → 0 unshared e⁻.
• $\text{ClF}_3$ central Cl: valence e⁻ = 7; bonding pairs = 3; lone pairs = $\frac{7 - 3}{2}$ = 2 → 2×2 = 4 unshared e⁻.
• $\text{BCl}_3$ central B: valence e⁻ = 3; bonding pairs = 3; lone pairs = $\frac{3 - 3}{2}$ = 0 → 0 unshared e⁻.
• $\text{ICl}_5$ central I: valence e⁻ = 7; bonding pairs = 5; lone pairs = $\frac{7 - 5}{2}$ = 1 → 2 unshared e⁻.

Answer:

C. $\text{ClF}_3$