If there are three possible values (-1/2, 0, +1/2) for the spin quantum number, then the maximum cap — Atomic Structure Chemistry Question

💡 Solution & Explanation

**Step 1 - Determine the Number of Orbitals in the Second Orbit** The second orbit corresponds to the principal quantum number $n = 2$. The total number of spatial orbitals ($N_{\text{orbitals}}$) in a shell with principal quantum number $n$ is given by the formula: $$N_{\text{orbitals}} = n^2$$ Substituting the value $n = 2$ into the formula: $$N_{\text{orbitals}} = 2^2 = 4\text{ orbitals}$$ These $4$ spatial orbitals are: * One $\ce{2s}$ orbital ($l = 0$, $m_l = 0$) * Three $\ce{2p}$ orbitals ($l = 1$, $m_l \in \{-1, 0, +1\}$) --- **Step 2 - Determine the Electron Capacity per Orbital** In standard quantum mechanics, the spin quantum number ($m_s$) has only $2$ possible values ($+1/2$ and $-1/2$), meaning each spatial orbital can accommodate a maximum of $2$ electrons. However, this question presents a hypothetical scenario where there are $3$ possible values for the spin quantum number: $$m_s \in \left\{-\frac{1}{2}, 0, +\frac{1}{2}\right\}$$ Consequently, each orbital can now accommodate a maximum of $3$ electrons. --- **Step 3 - Calculate the Maximum Electronic Capacity of the Second Orbit** $$N_{\text{electrons}} = N_{\text{orbitals}} \times (\text{capacity per orbital})$$ $$N_{\text{electrons}} = 4 \times 3 = \boxed{12\text{ electrons}}$$ --- **Step 4 - Evaluation of the Options** * **Option (A) is incorrect:** $8$ electrons is the standard real-world maximum ($2n^2 = 8$), which assumes only 2 spin states. * **Option (B) is incorrect:** $6$ electrons is the capacity of the three $\ce{2p}$ orbitals alone. * **Option (C) is correct:** Under 3 spin states, $4 \times 3 = 12$ electrons. * **Option (D) is incorrect:** $27$ electrons is mathematically incorrect.