Electronic Configurations of Elements — NEET Chemistry MCQs with Solutions

Solution: Half-filled subshells gain stability from symmetrical electron distribution (A), smaller coulombic repulsion (B), larger exchange energy from parallel spins (D) and relatively smaller mutual shielding (E). Statement C is wrong: the parallel-spin electrons occupy degenerate orbitals of the same subshell, not non-degenerate ones.

Solution: F⁻, Ne and Na⁺ all have 10 electrons, so n and electron-electron repulsion are essentially the same. The size difference comes from the nuclear charge: the higher Z (Na⁺, Z = 11) pulls the same 10 electrons in more strongly, so radius order is F⁻ > Ne > Na⁺.

Solution: Rb (Z = 37) has configuration [Kr]5s¹. The valence electron is in 5s, so n = 5, l = 0, and for l = 0 the only value of m is 0; spin +½.

Solution: This is chromium (Z = 24). Cr adopts 3d⁵4s¹ instead of 3d⁴4s² in its ground state because the exactly half-filled 3d subshell gives extra exchange-energy stability. It is a neutral atom (24 electrons), so it is the ground state, not an excited state or ion.

Solution: Cr (Z = 24) is an exception to the aufbau order: one 4s electron shifts to 3d so that both subshells are half-filled ([Ar]3d⁵4s¹), which maximises exchange energy and gives a symmetrical, extra-stable arrangement.

Solution: Cl (Z = 17) is [Ne]3s²3p⁵, so the unpaired electron sits in a 3p orbital: n = 3, l = 1, and m can be −1, 0 or +1. Among the given sets only n = 3, l = 1, m = 1 is possible.

Solution: The most electronegative element is fluorine, a halogen with outer configuration 2s²2p⁵, i.e. the general halogen pattern ns²np⁵ (one electron short of the noble-gas octet ns²np⁶).

Solution: Cr is [Ar]3d⁵4s¹ (half-filled stability) — correct. m ranges from −l to +l, so negative values are allowed — correct. Ag is [Kr]4d¹⁰5s¹: 46 electrons are paired (23 of each spin) and the lone 5s electron makes it 23 vs 24 — correct. In hydrazoic acid HN₃, H is +1 so the three N atoms together are −1, giving an average oxidation state of −1/3, not −3 — incorrect.

Solution: Completely filled subshells with n = 4 are 4s, 4p, 4d and 4f, holding 2 + 6 + 10 + 14 = 32 electrons in total. In a filled subshell exactly half the electrons have s = +½, so the count is 32/2 = 16.

Solution: Sr²⁺ has 36 electrons ([Kr]) and Cs⁺ has 54 ([Xe]) — noble gas configurations. La²⁺ (55 e⁻) is [Xe]5d¹, Pb²⁺ (80 e⁻) is [Xe]4f¹⁴5d¹⁰6s², Yb²⁺ (68 e⁻) is [Xe]4f¹⁴ and Fe²⁺ (24 e⁻) is [Ar]3d⁶ — none are noble gas cores. So only 2 qualify.

Solution: Ge: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p². Filled orbitals with mₗ = 0: the s orbitals 1s, 2s, 3s, 4s (four), plus the mₗ = 0 orbital of each filled subshell 2p, 3p and 3d (three more). The 4p² electrons occupy separate orbitals singly (Hund), so no 4p orbital is full. Total x = 7.

Solution: Ga is [Ar]3d¹⁰4s²4p¹; removing one electron takes away the 4p electron, so Ga⁺ is [Ar]3d¹⁰4s². Its valence electrons are the 4s pair, for which l = 0.