Question

For a multielectron atom, the CORRECT order of the orbital energy is

• A. \( 6s>5p>4s>3d \)
• B. \( 5p>6s>4f>4s \)
• C. \( 6s>5p>4f>4s \)
• D. \( 4f>6s>5p>4s \)

Hint:

Remember the \(n + l\) rule when determining orbital energies for multielectron atoms: Lower \(n + l\) values correspond to lower energy orbitals.

Answer 1

The Correct Option is D

The correct order of orbital energy for a multielectron atom is determined by the general rule of orbital filling in the \(n + l\) scheme. This rule states that the lower the sum of \(n + l\), the lower the energy of the orbital. In cases where two orbitals have the same value of \(n + l\), the orbital with the lower \(n\) value will have lower energy.
- \(6s\) has \(n = 6, l = 0\), so its energy is relatively high.
- \(5p\) has \(n = 5, l = 1\), and it is higher in energy compared to \(6s\).
- \(4f\) has \(n = 4, l = 3\), and the \(f\)-orbitals are lower in energy than both \(6s\) and \(5p\), which makes \(4f>6s>5p>4s\) the correct order.
Thus, the correct order of the orbital energies is \(4f>6s>5p>4s\).