Identify the orbital having lowest energy from following.

The energy of an orbital in an atom depends on both the principal quantum number \((n)\) and the azimuthal quantum number \((l)\). According to the general trend, for orbitals with different principal quantum numbers, the one with the lower \(n\) typically has lower energy. For orbitals with the same \(n\), the one with the lower \(l\) value will have lower energy.
To identify the orbital with the lowest energy from the given options, consider:
The principal quantum number (\(n\)):
\(\begin{aligned}
& \text {2p: } n=2 \\
& 3 \mathrm{~s}: n=3 \\
& 3 \mathrm{~d}: n=3 \\
& \text {4p: } n=4
\end{aligned}\)
The azimuthal quantum number \((l)\) values for the different orbitals are:
\(\begin{aligned}
& 2 \mathrm{p}: l=1 \\
& 3 \mathrm{~s}: l=0 \\
& 3 \mathrm{~d}: l=2 \\
& 4 \mathrm{p}: l=1
\end{aligned}\)
The energy order for orbitals generally increases with increasing values of \((n+l)\), and for orbitals with the same \((n+l)\) value, the one with the smaller \(n\) has the lower energy.
Calculating \((n+l)\) :
\(2 \mathrm{p}: 2+1=3\)
\(3 \mathrm{~s}: 3+0=3\)
3d: \(3+2=5\)
\(4 \mathrm{p}: 4+1=5\)
Among the options with the same \((n+l)\) value (\(2 p\) and \(3 s\)), \(2 p\) has the lower principal quantum number (\(n=2\) vs. \(n=3\)). Thus, the orbital with the lowest energy is:
Option A: 2p