The 1 st , 2 nd , and the 3 rd ionization enthalpies, I 1 , I 2 , and I 3 , of four atoms with atomic numbers n, n+1 , n+2 , and n+3 , where n<10 , are tabulated below. What is the value of n ? Atomic Number Ionization Enthalpy (kJ/mol) --- --- --- --- I 1 I 2 I 3 n 1681 3374 6050 n+1 2081 3952 6122 n+2 496 4562 6910 n+3 738 1451 7733

Chemistry · Periodicity of Elements and Periodic Properties

JEE Advanced 2020 — Paper 2 — Question 17

The $1^{\text {st }}, 2^{\text {nd }}$ , and the $3^{\text {rd }}$ ionization enthalpies, $I_{1}, I_{2}$ , and $I_{3}$ , of four atoms with atomic numbers $n, n+1$ , $n+2$ , and $n+3$ , where $n<10$ , are tabulated below. What is the value of $n$ ?

Atomic Number	{Ionization Enthalpy (kJ/mol)}
$I_{1}$	$I_{2}$	$I_{3}$
$n$	1681	3374	6050
$n+1$	2081	3952	6122
$n+2$	496	4562	6910
$n+3$	738	1451	7733

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Since I.E. of Na is $495.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , so $(\mathrm{n}+2)$ should be 11 .

$n+2=11$ , So $n=9$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Chemistry
Chapter: Periodicity of Elements and Periodic Properties
Topic: Periodicity in Chemical Properties (Ionization Energy, Electron Affinity, Electronegativity)