Question

The order of energy of orbitals in the same subshell is

• A. \( E_{2s}(\text{Na}) > E_{2s}(\text{Li}) > E_{2s}(\text{K}) \)
• B. \( E_{2s}(\text{Li}) > E_{2s}(\text{Na}) > E_{2s}(\text{K}) \)
• C. \( E_{2s}(\text{K}) > E_{2s}(\text{Na}) > E_{2s}(\text{Li}) \)
• D. \( E_{2s}(\text{Na}) > E_{2s}(\text{K}) > E_{2s}(\text{Li}) \)
• E. \[ E_{2s}(\text{Na}) > E_{2s}(\text{K}) > E_{2s}(\text{Li}) \]

Hint:

The energy of orbitals in the same subshell decreases with increasing atomic number due to the increased effective nuclear charge, which pulls the electrons closer.

Answer 1

The Correct Option is B

The order of energy of orbitals in the same subshell is influenced by the effective nuclear charge (\( Z_{\text{eff}} \)) and shielding effects.
- For \( E_{2s} \) orbitals, the energy depends on the atomic number (Z). As the number of protons in the nucleus increases, the effective nuclear charge increases, pulling the electrons closer to the nucleus and lowering the energy of the orbital.
- Lithium (Li) has the smallest atomic number, so its 2s orbital will be at a higher energy level than the other elements in the same subshell due to less shielding.
- Sodium (Na) has more protons, and the 2s orbital experiences more shielding than in lithium, lowering the energy further. - Potassium (K) has the largest atomic number and experiences the most shielding, so its 2s orbital has the least energy.
Thus, the correct order of energy is \( E_{2s}(\text{Li}) > E_{2s}(\text{Na}) > E_{2s}(\text{K}) \).