Element $A$ has $Δ_{i} H_{1} = 496$ and $Δ_{i} H_{2} = 4562 kJ mo l^{- 1}$ . Element $Z$ (period 3) has a first electron gain enthalpy of $- 200 kJ mo l^{- 1}$ and a very positive second value. The most probable formula of the ionic compound formed between them is

Answer: B

The huge

Δ_{i} H_{2}

shows A has one valence electron and forms

A X^{+}

(Na). Z needs two electrons to reach octet (S). Charge balance gives

A X_{2} Z

, i.e.

Na X_{2} S