Maximum number of electrons possible with spin quantum number \[ +\frac{1}{2} \] with principal quantum number $n=4$ in an atom is:

Question

Given below are two statements:
Statement (I): For a given shell, the total number of allowed orbitals is given by $ n^2 $.
Statement (II): For any subshell, the spatial orientation of the orbitals is given by $ -l $ to $ +l $ values including zero. In the light of the above statements, choose the correct answer from the options given below:

Answer 1

Step 1: Identify all subshells in n = 4.
When the principal quantum number $n = 4$, the azimuthal quantum number $l$ can be 0, 1, 2, or 3, giving subshells: 4s, 4p, 4d, 4f.
Step 2: Find the total number of orbitals in n = 4.
Total orbitals = $n^2 = 4^2 = 16$ orbitals.
Step 3: Understand spin quantum number assignment.
Each orbital can hold 2 electrons with opposite spins: one with $m_s = +\frac{1}{2}$ and one with $m_s = -\frac{1}{2}$.
Step 4: Count electrons with $m_s = +\frac{1}{2}$.
Since there are 16 orbitals in $n = 4$, and each orbital can have exactly ONE electron with $m_s = +\frac{1}{2}$, the maximum number of electrons with spin quantum number $+\frac{1}{2}$ is 16.
Step 5: Verify by listing subshell contributions.
4s: 1 orbital, 1 electron with $m_s = +\frac{1}{2}$. 4p: 3 orbitals, 3 electrons. 4d: 5 orbitals, 5 electrons. 4f: 7 orbitals, 7 electrons. Total = $1 + 3 + 5 + 7 = 16$.
Step 6: State the conclusion.
The maximum number of electrons with $m_s = +\frac{1}{2}$ equals the total number of orbitals in the shell, which is $n^2$.
\[ \boxed{16} \]

Maximum number of electrons possible with spin quantum number \[ +\frac{1}{2} \] with principal quantum number \(n=4\) in an atom is:

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The Correct Option is A

Solution and Explanation

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