Question

Ionization enthalpies of the elements X, Y, Z having successive atomic numbers respectively are 2080, 496, 737 kJ mol$^{-1}$. X, Y, Z are:

• A. Na, Mg, Al
• B. Ne, Na, Mg
• C. B, C, N
• D. O, N, F

Hint:

Ionization energy spikes indicate noble gas configuration. Compare successive atomic numbers carefully. Low IE → easily lost valence electron. Use IE trend across period/column.

Answer 1

The Correct Option is A

• Ionization enthalpy: energy required to remove an electron.
• X = 2080 kJ/mol → very high, corresponds to noble gas configuration after first electron removed, so Na → Ne configuration.
• Y = 496 kJ/mol → Mg (low, easy to remove electron).
• Z = 737 kJ/mol → Al.
• Hence elements X, Y, Z = Na, Mg, Al.