For hydrogen atom, the orbital/s with lowest energy is/are : (A) 4 s (B) $3 \mathrm{p}_x$ (C) $3 \mathrm{~d}_{x^2-y^2}$ (D) $3 \mathrm{~d}_{z^2}$ (E) $4 \mathrm{p}_z$ Choose the correct answer from the options given below :

• Correct answer
  A

  (B), (C) and (D) only
• B

  (A) and (E) only
• C

  (A) only
• D

  (B) only

Solution

For hydrogen atom and one electron species, the energy of orbitals is decided by the value of principal quantum number. Higher the value of principal quantum number, higher will be the energy of orbital.
$\therefore \quad(B),(C)$ and (D) have orbitals with the lowest energy.

Hence, the correct answer is option: 'A'.