Question

$1s<2s=2p<3s=3p=3d<4s=4p=4d=4f<$ __________
For multi electron atoms, the energy of the orbitals?

Answer 1

Hint: We can explain this question by referring with Aufbau’s principle. Keep in mind, the energy of an orbital will depend on its principle quantum number (n) and the azimuthal quantum number (l) and the energy is to be correlated with the $(n+l)$ rule.

Complete step by step solution:
We know that an atom is composed of a nucleus containing neutrons and protons with electrons dispersed in space. Electrons are not simply floating within the atom. Instead, they are fixed within electronic orbitals. Electronic orbitals are regions within the atom in which electrons have the highest probability to be found.
The energy of the atomic orbitals increases as the principal quantum number (denoted by ‘n’) increases. As the principal quantum number increases, the size of the orbitals increases and the electrons spend more time farther from the centre of the atom. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher i.e. less stabilised. Thus, we can apply the $(n+l)$ rule to find out which orbital will have the highest energy.
The azimuthal quantum number (i.e.) for s-orbital is $0$, for p-orbital is $1$, for d-orbital is $2$ and for f-orbital is $3$.
The energy of the orbitals should be in increasing principal quantum number.

So, the order for the multi electron atoms should be:
 $1s<2s<2p<3s<3p<3d<4s<4p<4d<4f<5s<5p<5d<5f$ and so on.

Note: It is important to note that the $n+l$ rule only works up to the atomic number having $20$. It can predict the orbital energy levels and the order of the occupancy. But above this atomic number, it fails to correctly predict the orbital energy levels. The physical meaning of the $n+l$ rule is related to the size and shape of a given orbital.