The first ionization energy (in kj/mol) of Na, Mg, Al and Si respectively, are :

Question

The first ionization energy (in kj/mol) of Na, Mg, Al and Si respectively, are : (A) 496, 737, 577, 786 (B) 496, 577, 737, 786 (C) 496, 577, 786, 737 (D) 786, 737, 577, 496

Tardigrade Admin · Accepted Answer

Electronic configuration of Na=[Ne] 3s1 Mg=[Ne] 3s2 Al=[Ne] 3s23p1 Si=[Ne] 3s23p2 So order of first ionisation energy is underset496 Na < underset737 Mg > underset577 Al < underset786 Si kj / mol Na < Al < Mg < Si ( IE 1. order )

Q. The first ionization energy (in kj/mol) of Na, Mg, Al and Si respectively, are :

496, 737, 577, 786

496, 577, 737, 786

496, 577, 786, 737

786, 737, 577, 496

Electronic configuration of $N a = [N e] 3 s^{1}$
$M g = [N e] 3 s^{2}$
$A l = [N e] 3 s^{2} 3 p^{1}$
$S i = [N e] 3 s^{2} 3 p^{2}$
So order of first ionisation energy is
$496 N a < 737 M g > 577 A l < 786 S i kj / m o l$
$N a < A l < M g < S i (I E_{1}$ order $)$

Q. The first ionization energy (in kj/mol) of Na, Mg, Al and Si respectively, are :

496, 737, 577, 786

496, 577, 737, 786

496, 577, 786, 737

786, 737, 577, 496

Electronic configuration of Na=[Ne]3s1 Mg=[Ne]3s2 Al=[Ne]3s23p1 Si=[Ne]3s23p2 So order of first ionisation energy is 496Na​<737Mg​>577Al​<786Si​kj/mol Na<Al<Mg<Si(IE1​ order )

Electronic configuration of $N a = [N e] 3 s^{1}$
$M g = [N e] 3 s^{2}$
$A l = [N e] 3 s^{2} 3 p^{1}$
$S i = [N e] 3 s^{2} 3 p^{2}$
So order of first ionisation energy is
$496 N a < 737 M g > 577 A l < 786 S i kj / m o l$
$N a < A l < M g < S i (I E_{1}$ order $)$