Compare the energies of following sets of quantum numbers for a multielectron system. (A) n =4,l=1 (B) n =4,l=2 (C) n =3,l=1 (D) n =3,l=2 (E) n =4,l=0 Choose the correct answer from the options given below :

Chemistry · Structure of Atom

JEE Main 2024 — 9 April, Shift 1 — Question 71

Compare the energies of following sets of quantum numbers for a multielectron system.

(A) $\mathrm{n}=4,l=1$

(B) $\mathrm{n}=4,l=2$

(D) $\mathrm{n}=3,l=2$

(E) $\mathrm{n}=4,l=0$

Choose the correct answer from the options given below :

Energy level can be determined by comparing $(\mathrm{n}+\ell)$ values

(A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$

(B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$

(D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$

(E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$

For same value of $(\mathrm{n}+\ell$ ), orbital having higher value of n , will have more energy.

Thus, the overall order is $\mathrm{C < E < D < A < B}$ .

Answer key and solution verified before publishing.

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