Identify which of the following sets of quantum numbers are not possible and explain why.
$(a) n = 0, l = 0, m_l = 0, m_s = +1/2$
$(b) n = 1, l = 0, m_l = 0, m_s = -1/2$
$(c) n = 1, l = 1, m_l = 0, m_s = +1/2$
$(d) n = 2, l = 1, m_l = 0, m_s = -1/2$
$(e) n = 3, l = 3, m_l = 3, m_s = +1/2$
$(f) n = 3, l = 1, m_l = 0, m_s = +1/2$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A, C, E) The following sets of quantum numbers are not possible:
$(a) n = 0$ is not possible because the principal quantum number $n$ must be a positive integer $(n = 1, 2, 3, \dots)$.
$(c) n = 1, l = 1$ is not possible because for a given $n$,the azimuthal quantum number $l$ can only range from $0$ to $n-1$. Thus,if $n = 1$,$l$ must be $0$.
$(e) n = 3, l = 3$ is not possible because $l$ must always be less than $n$ $(l < n)$. Therefore,$l$ can only take values $0, 1, 2$ when $n = 3$.

Explore More

Similar Questions

The four quantum numbers for the valence shell electron or last electron of sodium $(Z = 11)$ are:

The number of angular and radial nodes of $4d$ orbital respectively are

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

The portion of orbital diagrams representing the electronic configuration of certain elements is shown below. Which of them violate Pauli's Exclusion principle?

The maximum number of electrons present in an orbital with $n=4, l=3$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo