Question

The \(2s\) and the \(2p\) orbital energies of hydrogen atom are \(E_{2s}(\mathrm{H})\) and \(E_{2p}(\mathrm{H})\), respectively. The \(2s\) and the \(2p\) orbital energies of lithium atom are \(E_{2s}(\mathrm{Li})\) and \(E_{2p}(\mathrm{Li})\), respectively. The correct option(s) about the orbital energies is(are)

• A. \(E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li})\)
• B. \(E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H})\)
• C. \(E_{2p}(\mathrm{H}) < E_{2s}(\mathrm{Li})\)
• D. \(E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li})\)

Hint:

For hydrogen-like species: \[ \text{Energy depends only on } n \] Hence, \[ 2s = 2p \] For multi-electron atoms: \[ \text{Energy depends on both } n \text{ and } l \] Due to better penetration: \[ s < p < d < f \] in energy for the same principal quantum number.

Answer 1

The Correct Option is A

Concept: Orbital energies depend strongly on:
• Number of electrons present
• Nuclear charge
• Shielding effect
• Penetration effect For hydrogen-like atoms containing only one electron, the orbital energy depends only on the principal quantum number \(n\). Thus, all orbitals having the same value of \(n\) possess identical energy. However, in multi-electron atoms, orbital energies also depend on the azimuthal quantum number \(l\) because shielding and penetration become important.

Step 1: Comparing \(E_{2s}(\mathrm{H})\) and \(E_{2p}(\mathrm{H})\). Hydrogen atom contains only one electron. Therefore, electron-electron repulsion does not exist. Hence the energy depends only on \(n\). For \(n=2\), \[ E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H}) \] Thus option (B) is correct.

Step 2: Understanding orbital energies in lithium atom. Lithium has electronic configuration: \[ 1s^2\,2s^1 \] Lithium is a multi-electron atom. Hence shielding and penetration effects become important. Now compare \(2s\) and \(2p\) orbitals. The \(2s\) orbital penetrates closer to the nucleus than the \(2p\) orbital. Therefore:
• \(2s\) electrons experience greater effective nuclear charge
• \(2s\) orbital becomes more stabilized
• its energy becomes lower than \(2p\) Hence, \[ E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li}) \] Thus option (A) is correct.

Step 3: Comparing hydrogen and lithium orbital energies. For hydrogen atom: \[ E_n = -\frac{13.6}{n^2}\ \text{eV} \] Thus for \(n=2\), \[ E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H}) = -3.4\ \text{eV} \] In lithium atom, because of larger nuclear charge, the \(2s\) orbital experiences stronger attraction toward the nucleus. Hence the \(2s\) orbital of lithium becomes more stable than the \(2s\) orbital of hydrogen. Therefore, \[ E_{2s}(\mathrm{Li}) < E_{2s}(\mathrm{H}) \] which implies, \[ E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li}) \] Thus option (D) is correct.

Step 4: Checking option (C). Option (C) states: \[ E_{2p}(\mathrm{H}) < E_{2s}(\mathrm{Li}) \] But we know: \[ E_{2s}(\mathrm{Li}) \] is more stabilized and lower in energy than hydrogen \(2p\). Therefore actually, \[ E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{H}) \] Hence option (C) is incorrect.

Step 5: Final conclusion. The correct relations are: \[ E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li}) \] \[ E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H}) \] \[ E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li}) \] Therefore, the correct options are: \[ \boxed{(A),\ (B)\ \text{and}\ (D)} \] Final Answer: \[ \boxed{(A),\ (B)\ \text{and}\ (D)} \]