Among the following, the energy of
Question:

Among the following, the energy of $2 \mathrm{~s}$ orbital is lowest in :

1. K

2. H

3. Li

4. Na

Correct Option: 1

Solution:

As the value of $Z$ (atomic number) increases, energy of orbitals decreases (becomes more -ve value)

$\therefore$ Order of energy of $2 s$ orbital is $\mathrm{H}>\mathrm{Li}>\mathrm{Na}>\mathrm{K}$.