Easy MCQ +4 / -1 PYQ · JEE Mains 2021

Identify the elements X and Y using the ionisation energy values given below :

Ionization energy (kJ/mol)
1${st}$ 2${nd}$
X 495 4563
Y 731 1450

  1. A X = F; Y = Mg
  2. B X = Mg; Y = F
  3. C X = Na; Y = Mg Correct answer
  4. D X = Mg; Y = Na

Solution

Due to 2p<sup>6</sup>, noble gas electronic configuration, the second ionisation enthalpy of Na is very high. That’s why has large difference between IE<sub>1</sub>, and IE<sub>2</sub><br><br> Mg<sup>+</sup> is 2p<sup>6</sup>, 3s<sup>1</sup>.<br><br> After the loss of one electron, Mg<sup>+</sup> will be formed with noble gas electronic configuration. That’s why has less difference between I.E<sub>1</sub> and I.E<sub>2</sub>, but it's I.E<sub>3</sub> is very high.

About this question

Subject: Chemistry · Chapter: Periodic Table and Periodicity · Topic: Periodic Trends

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