QJEEM22Chba96: JEE Main 2022 Chemistry - Class 11 JEEM22Chba96 Solution: Given below are the quantum numbers for 4 electrons. A. n = 3 , …

The order of increasing energy of electrons in an atom can generally be determined using the principal quantum number ( $n$ ) and the azimuthal quantum number ( $l$ ). The energy can be compared using the rule: the higher the $(n + l)$ value, the higher the energy. If two electrons have the same $(n + l)$ value, the electron with the higher $n$ value has the higher energy.

Analyze each electron's quantum numbers:

A. $$n=3,l=2\Rightarrow n+l=3+2=5$$

B. $$n=4,l=1\Rightarrow n+l=4+1=5$$

C. $$n=4,l=2\Rightarrow n+l=4+2=6$$

D. $$n=3,l=1\Rightarrow n+l=3+1=4$$

Comparing the values of $(n + l)$ and the principal quantum number $n$ :

D: $n+l = 4$ , and $n=3$ .

A and B: $n+l = 5$ , but A has $n=3$ while B has $n=4$ . Therefore, A has a slightly lower energy than B.

C: $n+l = 6$ .

Thus, the correct order of increasing energy is:

$$D<A<B<C$$

Therefore, Option B is correct.