The orbital having two nodal surfaces is — Atomic Structure Chemistry Question

💡 Solution & Explanation

**Step 1 - Define Nodal Surfaces and the Radial Node Formula** In quantum mechanics, a nodal surface (also referred to as a radial node) is a spherical shell-like boundary centered around the nucleus of an atom where the probability density of finding an electron is exactly zero: $$\psi^2 = 0$$ The number of radial nodes or nodal surfaces ($N_{\text{radial}}$) for any given atomic orbital is determined by its principal quantum number ($n$) and azimuthal quantum number ($l$) according to the formula: $$N_{\text{radial}} = n - l - 1$$ Where: * $n$ is the principal quantum number, representing the main energy level or shell. * $l$ is the azimuthal quantum number, representing the subshell. The values of $l$ for various subshells are: * For $s$-orbitals: $l = 0$ * For $p$-orbitals: $l = 1$ * For $d$-orbitals: $l = 2$ **Step 2 - Apply the Formula to Each Option** We systematically evaluate the number of nodal surfaces for each of the given choices: * **Option (A) 1s orbital:** * Principal quantum number, $n = 1$ * Azimuthal quantum number, $l = 0$ * Substituting the values into the formula: $$N_{\text{radial}} = 1 - 0 - 1 = 0$$ The 1s orbital has $0$ nodal surfaces. Thus, this option is incorrect. * **Option (B) 2s orbital:** * Principal quantum number, $n = 2$ * Azimuthal quantum number, $l = 0$ * Substituting the values into the formula: $$N_{\text{radial}} = 2 - 0 - 1 = 1$$ The 2s orbital has exactly $1$ nodal surface. Thus, this option is incorrect. * **Option (C) 3s orbital:** * Principal quantum number, $n = 3$ * Azimuthal quantum number, $l = 0$ * Substituting the values into the formula: $$N_{\text{radial}} = 3 - 0 - 1 = 2$$ The 3s orbital has exactly $2$ nodal surfaces. Thus, this option is correct. * **Option (D) 2p orbital:** * Principal quantum number, $n = 2$ * Azimuthal quantum number, $l = 1$ * Substituting the values into the formula: $$N_{\text{radial}} = 2 - 1 - 1 = 0$$ The 2p orbital has $0$ radial nodal surfaces (though it contains $1$ angular node or nodal plane, it has $0$ spherical nodal surfaces). Thus, this option is incorrect. **Step 3 - Final Selection** Comparing the calculated values, only the 3s orbital possesses exactly two nodal surfaces. $$\boxed{\text{Correct Option: (C) } 3s}$$