A completely filled d-orbital (d^10) is of — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Spatial Shapes of Individual $d$-Orbitals An atomic $d$-subshell corresponds to the azimuthal quantum number $l = 2$. It consists of exactly five degenerate spatial orbitals corresponding to the magnetic quantum numbers $m_l \in \{-2, -1, 0, +1, +2\}$: * $d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, and $d_{z^2}$ Individually, none of these orbitals are spherically symmetrical — their electron probability distributions depend heavily on spatial angles. ### Step 2 - Total Electron Cloud of a Filled $d^{10}$ Subshell When a $d$-subshell is completely filled ($d^{10}$), each of the five degenerate orbitals contains 2 electrons. The total probability density is: $$\rho(r, \theta, \phi) = 2 \sum_{m_l=-2}^{+2} \left|\psi_{n,2,m_l}(r, \theta, \phi)\right|^2$$ Factoring the wave function into radial and angular parts: $$\rho(r, \theta, \phi) = 2 \left|R_{n,2}(r)\right|^2 \sum_{m_l=-2}^{+2} \left|Y_{2,m_l}(\theta, \phi)\right|^2$$ ### Step 3 - Unsöld's Theorem Proves Spherical Symmetry According to **Unsöld's Theorem**, the sum of squares of spherical harmonics for a complete subshell is a constant: $$\sum_{m_l=-l}^{+l} \left|Y_{l,m_l}(\theta, \phi)\right|^2 = \frac{2l+1}{4\pi}$$ For $l = 2$: $$\sum_{m_l=-2}^{+2} \left|Y_{2,m_l}(\theta, \phi)\right|^2 = \frac{5}{4\pi}$$ Therefore: $$\rho(r) = \frac{5}{2\pi} \left|R_{n,2}(r)\right|^2$$ This is independent of $\theta$ and $\phi$ — the $d^{10}$ configuration has **spherical symmetry**. ### Step 4 - Evaluation of Options * **Option (A) Spherical symmetry:** Correct — Unsöld's theorem proves uniform angular distribution for filled subshells. * **Option (B) Octahedral symmetry:** Incorrect — describes molecular complex geometry, not isolated atomic subshells. * **Option (C) Tetrahedral symmetry:** Incorrect — discrete molecular geometry, not applicable here. * **Option (D) Unsymmetry:** Incorrect — completely filled subshells are highly symmetrical. $$\text{Correct Option: } \boxed{\text{A}}$$