The energy is lowest for the orbital :

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number

Vedclass · Accepted Answer

$4s$

Vedclass · Answer

(A) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number $(l)$.For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.For $4p$: $n=4, l=1$,so $(n+l) = 4+1 = 5$.For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.For $4f$: $n=4, l=3$,so $(n+l) = 4+3 = 7$.Comparing the $(n+l)$ values,$4s$ has the lowest value of $4$. If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy. Since $4s$ has the lowest $(n+l)$ value,it has the lowest energy.

Similar Questions

Maximum number of electrons that can be accommodated in the subshell with azimuthal quantum number $l=4$ is

$A$ certain orbital has no angular nodes and two radial nodes. The orbital is

$Be$'s $4^{th}$ electron will have four quantum numbers $(n, l, m, s)$:

What is the number of electrons in a $Na$ atom for which $l + m = 0$?

Which of the following orbitals has a dumbbell shape?

Vedclass Test Series

Exam Paper Generator

Online Exam Module