Let x be the number of electrons present in the ^ * (pi-antibonding) molecular orbitals of the anion in Na ₂ O ₂ , and y be the number of electrons present in the ^ * molecular orbitals of the anion in KO ₂ . What is the value of (x + y) and the magnetic nature of the anion present in Na ₂ O ₂ ?

Step-by-step solution

The anion in Na ₂ O ₂ is the peroxide ion ( O ₂²⁻ ). Total number of electrons in O ₂²⁻ = 18. The molecular orbital configuration is: 1s^2, ^ * 1s^2, 2s^2, ^ * 2s^2, 2p_z^2, 2p_x^2 = 2p_y^2, ^ * 2p_x^2 = ^ * 2p_y^2 . Number of electrons in ^ * orbitals, x = 4 . Since all electrons are paired, the peroxide ion is diamagnetic. The anion in KO ₂ is the superoxide ion ( O ₂⁻ ). Total number of electrons in O ₂⁻ = 17. The molecular orbital configuration is: 1s^2, ^ * 1s^2, 2s^2, ^ * 2s^2, 2p_z^2, 2p_x^2 = 2p_y^2, ^ * 2p