If the nitrogen atom has electronic configuration 1s^7, it would have energy lower than that of the — Atomic Structure Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Proposed Hypothetical Electronic Configuration A neutral nitrogen atom has atomic number $Z = 7$, so it has $7$ electrons. * Normal ground state: $\ce{1s^2 2s^2 2p^3}$ * Hypothetical: $\ce{1s^7}$ In $\ce{1s^7}$, all $7$ electrons are in the $1s$ subshell — closest to the nucleus, theoretically lower energy. Step 2 - Determine Maximum Occupancy of the $\ce{1s}$ Subshell For the $\ce{1s}$ orbital: $n=1$, $l=0$, $m_l=0$ — there is exactly **one** spatial orbital. Step 3 - Apply Pauli's Exclusion Principle Pauli's Exclusion Principle states: no two electrons in an atom can have the same set of all four quantum numbers ($n, l, m_l, m_s$). Since $m_s$ can only be $+1/2$ or $-1/2$, any single orbital holds at most **2 electrons**. * Maximum capacity of $\ce{1s}$: $2$ electrons * $\ce{1s^7}$ requires 7 electrons in one orbital — at least 5 pairs would share identical quantum numbers * Therefore $\ce{1s^7}$ violates Pauli's Exclusion Principle Step 4 - Evaluation of Options * **Option (A) is incorrect:** Heisenberg's uncertainty principle ($\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$) governs measurement precision, not orbital occupancy. * **Option (B) is incorrect:** Hund's rule governs filling of degenerate orbitals, not single-orbital capacity. * **Option (C) is correct:** Pauli's exclusion principle limits each orbital to maximum 2 electrons. $\ce{1s^7}$ violates this. * **Option (D) is incorrect:** Bohr's postulate governs allowed orbits and quantized angular momentum, not spin. $$\text{Correct Option: } \boxed{\text{C}}$$