Identify the elements X and Y using the ionization energy values given below: Ionization energy ( $\mathrm{kJ} / \mathrm{mol}$ ) | | $1^{\text {st }}$ | $2^{\text {nd }}$ | |---|---|---| | X | 495 | 4563 | | Y | 731 | 1450 |

(C) $X=\mathrm{Na} ; Y=\mathrm{Mg}$

(A) $\mathrm{X}=\mathrm{Mg} ; \mathrm{Y}=\mathrm{Na}$

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Classification of Elements and Periodicity

Identify the elements X and Y using the ionization energy values given below:
Ionization energy (

kJ / mol

)

	$1^{st}$	$2^{nd}$
X	495	4563
Y	731	1450

(A) $X = Mg; Y = Na$
(B) $X = F; Y = M g$
(C) $X = Na; Y = Mg$ ✓ Correct
(D) $X = M g; Y = F$

Answer: (C)

Solution

A sharp jump between consecutive ionisation energies marks the point at which an electron has to be pulled out of a noble gas core. For X the second value is more than nine times the first, so X has a single valence electron and the second removal already breaks a closed shell; that is sodium. For Y the second value is only about twice the first, so two electrons leave with comparable ease and the large jump would come at the third; that is magnesium. So X is sodium and Y is magnesium.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Classification of Elements and Periodicity

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