For electron in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are :

### Core Logic The element with atomic number 9 is **Fluorine (F)**. Let's write out its ground state electronic configuration: F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$$F (Z=9) = 1s^2 \, 2s^2 \, 2p^5$$ Let's systematically audit each statement: * **Statement A**: Within the total population of 9 electrons, the pairing distribution across shells shows: 1s^2$1s^2$ (one up, one down), 2s^2$2s^2$ (one up, one down), 2p^5$2p^5$ (three up, two down). Summing up-spins (m_s = +frac12$m_s = +\frac{1}{2}$) yields 1 + 1 + 3 = 5$1 + 1 + 3 = 5$ electrons. Down-spins (m_s = -frac12$m_s = -\frac{1}{2}$) yield 1 + 1 + 2 = 4$1 + 1 + 2 = 4$ electrons. **Statement A is fully correct.**

* **Statement B**: By Hund's Rule, the 5 electrons in the 2p subshell occupy the degenerate p_x, p_y, p_z$p_x, p_y, p_z$ states. This produces two fully-filled sub-orbitals and one half-filled sub-orbital. The unpaired slot can reside arbitrarily in *any* of the three orbitals (p_x, p_y$p_x, p_y$, or p_z$p_z$) due to spatial symmetry. It is not constrained to p_z$p_z$. **Statement B is incorrect.** * **Statement C**: The highest energy valence electron enters the 2p subshell, which is defined by principal number n = 2$n = 2$ and azimuthal index l = 1$l = 1$. **Statement C is fully correct.** * **Statement D**: Angular nodes are given directly by the quantum number l$l$. For s-orbitals (1s, 2s$1s, 2s$), angular nodes = 0$= 0$. For each of the three populated p-orbitals (2p$2p$), angular nodes = 1$= 1$. The sum total of angular nodes across all orbitals is 0 + 0 + 3 = 3$0 + 0 + 3 = 3$. **Statement D is incorrect.** ### Pattern Recognition Total angular nodes equals the total number of p-electrons' spatial orientation count, not simply the subshell boundary value. Recognizing that degenerate p-orbitals share uniform probability status exposes the restriction in Statement B instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Bohr energy level formula for hydrogenic species: E_n = -13.6 times fracZ^2n^2 quad texteV$$E_n = -13.6 \times \frac{Z^2}{n^2} \quad \text{eV}$$ where: Z$Z$ = atomic number of the species n$n$ = principal quantum number of the orbit ### Execution Step 1: Identify the parameters for the first excited state of textBe^3+$\text{Be}^{3+}$: * For Beryllium (textBe$\text{Be}$), the atomic number is Z = 4$Z = 4$. * The term 'first excited state' refers to the second energy level, so n = 2$n = 2$. Step 2: Substitute these values into the Bohr energy equation: E_textBe^3+ = -13.6 times frac4^22^2 = -13.6 times frac164$$E_{\text{Be}^{3+}} = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4}$$ E_textBe^3+ = -13.6 times 4 = -54.4 text eV$$E_{\text{Be}^{3+}} = -13.6 \times 4 = -54.4 \text{ eV}$$ Step 3: Extract the magnitude and round to the nearest integer value: |E_textBe^3+| = 54.4 approx 54$$|E_{\text{Be}^{3+}}| = 54.4 \approx 54$$ ### Pattern Recognition For the first excited state of Beryllium (Z=4, n=2$Z=4, n=2$), the term fracZ^2n^2 = frac4^22^2 = frac164 = 4$\frac{Z^2}{n^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$. Thus, the energy value is exactly 4 times that of the ground-state hydrogen atom (13.6 times 4 = 54.4$13.6 \times 4 = 54.4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula E_n = -13.6 fracZ^2n^2text eV$$E_n = -13.6 \frac{Z^2}{n^2}\text{ eV}$$ ### Core Logic For a constant principal quantum number n$n$, the energy E$E$ is directly proportional to -Z^2$-Z^2$. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. ### Pattern Recognition Since energy values are inherently negative for bound states, as Z$Z$ increases, E$E$ becomes rapidly more negative following a parabolic curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Related Formula Delta x cdot Delta p ge frach4pi$$Delta x cdot Delta p ge \frac{h}{4pi}$$ ### Core Logic Statement I is a verbatim definition of Heisenberg's Uncertainty Principle, hence it is completely true. For Statement II, we are given that Delta x = Delta p$Delta x = Delta p$: Delta p cdot Delta p ge frach4pi implies (Delta p)^2 ge frach4pi $$Delta p cdot Delta p ge \frac{h}{4pi} implies (Delta p)^2 ge \frac{h}{4pi} $$ Delta p ge sqrtfrach4pi = frac12sqrtfrachpi$$Delta p ge sqrt{\frac{h}{4pi}} = \frac{1}{2}sqrt{\frac{h}{pi}}$$ Since Delta p = m cdot Delta v$Delta p = m cdot Delta v$: m cdot Delta v ge frac12sqrtfrachpi implies Delta v ge frac12msqrtfrachpi$$m cdot Delta v ge \frac{1}{2}sqrt{\frac{h}{pi}} implies Delta v ge \frac{1}{2m}sqrt{\frac{h}{pi}}$$ This perfectly matches Statement II, so it is also true. ### Pattern Recognition When solving inequality bounds for identical uncertainties, always substitute directly to obtain a clean quadratic form before taking the square root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

### Core Logic In multi-electron systems, the energy profile depends on both the primary shell (n$n$) and azimuthal subshell (l$l$) quantum values via the (n+l)$(n+l)$ rule. Orbitals sharing the exact same n$n$ and l$l$ values are energy-degenerate as long as external field metrics are zero. Let us map each set: - **A:** 1s$1s$ - **B:** 2s$2s$ - **C:** 2p$2p$ - **D:** 3d$3d$ (m_l = 1$m_l = 1$) - **E:** 3d$3d$ (m_l = 0$m_l = 0$) Since both D and E describe subcomponents of the same 3d$3d$ subshell (n=3, l=2$n=3, l=2$), they share identical energies, establishing perfect degeneracy. ### Pattern Recognition Sees: Orbitals matching energy level in multi-electron space. Trap: Confusing single-electron hydrogen atoms (where energy depends solely on n$n$) with multielectron atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom