Question

Why does an electron found in a \[2s\] orbital have a lower energy than an electron found in a \[2p\] orbital in multielectron systems?

Answer 1

Hint: Orbital is the region where there is a maximum probability of finding an electron. There are different orbitals like s-orbital, p-orbital, d-orbital, f-orbital. s-orbital has a sphere -like shape whereas p-orbital has barbell shape, which leads to the requirement of more energy in those lobes than the electron in s-orbital.

Complete answer:
Atoms are the smallest particles that consist of subatomic particles. Electrons, protons, and neutrons are the three types of subatomic particles. The region where there is a maximum probability of finding an electron is known as an orbital.
There are mainly four orbitals which are s-orbital, p-orbital, d-orbital, and f-orbital. These four orbitals have different shapes.
s-orbital has spherical shape, and p-orbital has barbell shape with two lobes. Given that the electron present in \[2s\] orbital, the orbital shape is spherical shape, and the electron present in \[2p\] orbitals, whose orbital shape is barbell.
The electron present in \[2s\] orbital has the lower energy as the electron can be easily found within the sphere, but the \[2p\] orbital has two lobes, requiring more energy to find an electron.
Thus, an electron found in a \[2s\] orbital has a lower energy than an electron found in a \[2p\] orbital in multielectron systems due to the shape of the orbital.

Note:
Given orbitals are \[2s\] and \[2p\] which have the same energy level as the principal quantum number, and have the same probability distance from the nucleus. Due to the difference in the shapes, only the probability of finding an electron requires more energy.