Question

Given, (A) n=5, m___l = -1; (B) n=3, l=2, m___l = -1, m___s = +1/2. The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :

• A. 8 and 1
• B. 26 and 1
• C. 2 and 4
• D. 4 and 1

Hint:

If all four quantum numbers are specified, the answer is always 1 electron. If only three are specified ($n, l, m_l$), the answer is 2 electrons.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Quantum numbers define the state of an electron. $n$ is the shell, $l$ is the subshell, $m_l$ is the specific orbital, and $m_s$ is the spin.
Step 2: Detailed Explanation:
Case (A): \(n=5, m_l = -1\)
In the 5th shell ($n=5$), the possible values for $l$ are $0, 1, 2, 3, 4$.
- For $l=0$ (s), $m_l$ can only be 0.
- For $l=1$ (p), $l=2$ (d), $l=3$ (f), and $l=4$ (g), each has exactly one orbital where $m_l = -1$.
There are 4 such orbitals ($5p, 5d, 5f, 5g$). Each orbital holds 2 electrons.
Total electrons $= 4 \times 2 = 8$.
Case (B): \(n=3, l=2, m_l = -1, m_s = +1/2\)
This specifies a single shell ($n=3$), a single subshell ($l=2$), a single orbital ($m_l = -1$), and a specific spin ($m_s = +1/2$).
According to Pauli's Exclusion Principle, only 1 electron can have this unique set of four quantum numbers.
Step 3: Final Answer:
The maximum numbers are 8 and 1.