Question

Maximum possible electrons in Na, for which n + I + m = 3 is/are
0 6
O 3
O 4
o5​

Answer 1

Sodium Electronic Configuration:

$\sf \: Na_{11} : {1s}^{2} \: {2s}^{2} \: {2p}^{6} \: {3s}^{1}$

So, the possible combinations for n + l + m = 3 are as follows:

• 1 + 1 + 1 : 0 electrons
• 1 + 0 + 2 :0 electrons
• 2 + 0 + 1 : 2 electrons (2s²)
• 2 + 1 + 0 : 2 electrons $(2p_{z}^{2})$
• 3 + 0 + 0 : 1 electrons (3s¹)

So, total possible electrons are :

• TOTAL 5 ELECTRONS FOLLOWING THE GIVEN EQUATION.

$\star$ See diagram (tick marks) for better clarification !