When the value of azimuthal quantum number is 3, the maximum and minimum values of spin multiplicity — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Identify the Subshell and Orbital Count For $l = 3$, the electron is in an $f$-subshell. The number of degenerate orbitals: $$N_{\text{orbitals}} = 2l + 1 = 2(3) + 1 = 7 \text{ orbitals}$$ ### Step 2 - Calculate Maximum Spin Multiplicity Spin multiplicity is given by: $$\text{Spin Multiplicity} = 2S + 1$$ Where $S = \left|\sum m_s\right|$ is the total spin. According to **Hund's Rule**, maximum spin occurs when all 7 orbitals are singly occupied with parallel spins: $$S_{\text{max}} = 7 \times \frac{1}{2} = \frac{7}{2}$$ $$\text{Maximum Spin Multiplicity} = 2\left(\frac{7}{2}\right) + 1 = 7 + 1 = \boxed{8}$$ ### Step 3 - Calculate Minimum Spin Multiplicity Minimum spin occurs when electrons are completely paired ($S_{\text{min}} = 0$): $$\text{Minimum Spin Multiplicity} = 2(0) + 1 = \boxed{1}$$ ### Step 4 - Evaluation of Options * **Option (A) 1, 8:** Incorrect order — gives minimum first, then maximum. * **Option (B) 8, 1:** Correct — maximum spin multiplicity is 8, minimum is 1. * **Option (C) 6, 1:** Incorrect — underestimates maximum (would require only 5 unpaired electrons). * **Option (D) 7, 0:** Incorrect — spin multiplicity can never be 0 since $2S+1 \geq 1$. $$\text{Correct Option: } \boxed{\text{B}}$$