The order of increasing energies of the orbit | vedclass

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

Question

(A) The energy of an orbital is determined by the $(n + l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.For

Vedclass · Accepted Answer

$3s < 3p < 4s < 3d < 4p$

Vedclass · Answer

(A) The energy of an orbital is determined by the $(n + l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.For $3s$: $n=3, l=0$,so $(n+l) = 3+0 = 3$.For $3p$: $n=3, l=1$,so $(n+l) = 3+1 = 4$.For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.For $4p$: $n=4, l=1$,so $(n+l) = 4+1 = 5$.If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Comparing $3p$ $(n=3)$ and $4s$ $(n=4)$,$3p Comparing $3d$ $(n=3)$ and $4p$ $(n=4)$,$3d Thus,the increasing order of energy is $3s < 3p < 4s < 3d < 4p$.

The order of increasing energies of the orbitals follows:

Similar Questions

The atomic number of an element having the valence shell electronic configuration $4s^2 4p^6$ is

The minimum number of quantum numbers required to specify an orbital in an atom is

Which among the following species has the same number of electrons in its outermost as well as penultimate shell?

$4s^2$ is the configuration of the outermost orbit of an element. Its atomic number would be $:-$

If the atomic number of element $X$ is $7,$ the best electron dot symbol for the element is

Vedclass Test Series

Exam Paper Generator

Online Exam Module