Question

For hydrogen atom, the orbitals with the lowest energy among the given orbitals are: (i) $4s$ (ii) $2p_x$ (iii) $3d_{z^2}$ (iv) $2p_y$

• A. (i) & (iii)
• B. (ii) & (iv)
• C. (ii) & (iii)
• D. (ii) only
• E. (i) only

Hint:

For H-atoms, energy levels are $1s < 2s = 2p < 3s = 3p = 3d$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question is about the energy of atomic orbitals in a hydrogen atom. A key feature of the hydrogen atom (and other single-electron species like He\(^+\), Li\(^{2+}\)) is that the energy of its orbitals depends only on the principal quantum number (n).
Step 2: Key Formula or Approach:
For a hydrogen atom, the energy of an orbital is determined solely by its principal quantum number, n. \[ E_n = -\frac{R_H}{n^2} \] where \( R_H \) is the Rydberg constant. Orbitals with the same value of n are degenerate (have the same energy). We need to find the principal quantum number for each of the given orbitals and identify the smallest value of n. Step 3: Detailed Explanation:
Let's identify the principal quantum number (the number in front of the orbital letter) for each given orbital: - (i) 4s: \( n = 4 \) - (ii) 2p\(_{x}\): \( n = 2 \) - (iii) 3d\(_{z^2}\): \( n = 3 \) - (iv) 2p\(_{y}\): \( n = 2 \) The energies of these orbitals are related by their n values. A lower value of n corresponds to a lower (more negative, i.e., more stable) energy. Comparing the n values: \( n=2 \) is the smallest principal quantum number among the options. The orbitals with \( n=2 \) are 2p\(_{x}\) and 2p\(_{y}\). Since they have the same principal quantum number, they are degenerate and have the same, lowest energy among the given choices. - Energy order: \( E_{2p_x} = E_{2p_y}<E_{3d_{z^2}}<E_{4s} \) Step 4: Final Answer:
The orbitals with the lowest energy are (ii) 2p\(_{x}\) and (iv) 2p\(_{y}\).