JEE Chemistry 2026 Question with Solution

Given: (A) $n=5, m_l=-1$ and (B) $n=3, l=2, m_l=-1, m_s=+1/2$.

Find: The maximum number of electrons that can have these quantum numbers in each case.

For case (A), only $n$ and $m_l$ are fixed. In the shell $n=5$, the possible values of $l$ are $0,1,2,3,4$.

For $l=0$, $m_l=0$ only, so $m_l=-1$ is not possible.

For $l=1,2,3,4$, the value $m_l=-1$ is allowed once in each subshell. Therefore, there are $4$ orbitals corresponding to $m_l=-1$ in the $5p, 5d, 5f,$ and $5g$ subshells.

Each orbital can accommodate $2$ electrons.

For case (B), all four quantum numbers are specified: $n=3, l=2, m_l=-1, m_s=+1/2$.

By Pauli's exclusion principle, only one electron can have one unique set of four quantum numbers.

Therefore, the maximum numbers are $8$ and $1$. The correct option is A.

Given: (A) $n=5, m_l=-1$ and (B) $n=3, l=2, m_l=-1, m_s=+1/2$.

Find: The maximum number of electrons in each case.

Use the shortcut:

• If three quantum numbers are fixed up to $m_l$, one orbital is identified, and that orbital can contain $2$ electrons.
• If only $n$ and $m_l$ are fixed, count how many allowed $l$ values can have that $m_l$.
• If all four quantum numbers are fixed, the answer is always $1$.

For (A), with $n=5$, the allowed $l$ values are $0$ to $4$. The value $m_l=-1$ is possible for $l=1,2,3,4$ only, giving $4$ orbitals.

For (B), all four quantum numbers are fixed, so the maximum number is $1$.

Therefore, the correct option is A.