Match List I with List II and select the correct answer using the code given below the Lists : List I (Element) A. Boron B. Nitrogen C. Oxygen D. Neon List II (Unpaired electron) 1. Zero 2. Two 3. One 4. Three Code :

Correct — D, A-3, B-4, C-2, D-1. Write out the configurations and apply Hund's rule, which says electrons occupy the orbitals of a subshell singly before any orbital is doubled. Boron, Z = 5, is 1s2 2s2 2p1, so one unpaired electron, and A goes with 3. Nitrogen, Z = 7, is 1s2 2s2 2p3, and the three 2p electrons occupy the three 2p orbitals singly, giving three unpaired, so B goes with 4. Oxygen, Z = 8, is 1s2 2s2 2p4, so one 2p orbital holds a pair and the remaining two hold one electron each, giving two unpaired, and C goes with 2. Neon, Z = 10, is 1s2 2s2 2p6, a filled subshell with every electron paired, so D goes with 1.

• (a)A-1, B-4, C-2, D-3 — This gives boron zero unpaired electrons and neon one, which is impossible. Boron's odd electron count of five means at least one electron must be unpaired, and neon's closed shell means none can be.
• (b)A-1, B-2, C-4, D-3 — This assigns nitrogen two unpaired electrons and oxygen four. Nitrogen has only three 2p electrons in total, so four is out of reach, and by Hund's rule all three sit unpaired.
• (c)A-3, B-2, C-4, D-1 — The first two pairings here are right, but it then gives oxygen four unpaired electrons and neon three. Oxygen has four 2p electrons of which two must pair, and neon is a noble gas with a completely filled subshell.

Electrons fill orbitals by three rules. The Aufbau principle fills the lowest energy orbitals first; the Pauli exclusion principle allows at most two electrons per orbital and only with opposite spins; Hund's rule of maximum multiplicity keeps electrons unpaired across the orbitals of a subshell as long as empty orbitals remain. A p subshell has three orbitals, so the unpaired count for p1 to p6 runs 1, 2, 3, 2, 1, 0. Unpaired electrons make a substance paramagnetic, that is weakly attracted into a magnetic field; a species with none is diamagnetic.

The p1 to p6 sequence 1, 2, 3, 2, 1, 0 is the whole item once it is memorised, and it rises to a peak at the half-filled subshell and falls symmetrically after it. Two of the four elements can be placed without any configuration work: neon is a noble gas, so it must take zero, and nitrogen sits at the half-filled p3, so it must take the maximum of three. Fixing those two alone eliminates every option but the correct one. Oxygen's two unpaired electrons are also why molecular oxygen is paramagnetic and is drawn towards the poles of a magnet.

• Hund's rule fills each orbital of a subshell singly before any is doubled.
• For a p subshell the unpaired count from p1 to p6 runs 1, 2, 3, 2, 1, 0.
• Boron 1s2 2s2 2p1 has one unpaired electron; nitrogen 1s2 2s2 2p3 has three.
• Oxygen 1s2 2s2 2p4 has two unpaired electrons, which makes molecular oxygen paramagnetic.
• Neon 1s2 2s2 2p6 has none, which is the closed-shell configuration of a noble gas.

• Pairing electrons within a subshell before all its orbitals are singly occupied.
• Counting total valence electrons instead of unpaired ones.
• Forgetting that a noble gas must have zero unpaired electrons.

A matching item that is really a configuration drill. Two of the four pairings can be fixed by inspection, which collapses the code list at once.

• practice — not a real PYQ

  How many unpaired electrons are there in a carbon atom in its ground state?

  • (a)Zero
  • (b)One
  • (c)Two
  • (d)Four

  Answer(c) Two — carbon is 1s2 2s2 2p2, and the two p electrons occupy separate orbitals.

• practice — not a real PYQ

  A species with no unpaired electrons is described as

  • (a)paramagnetic
  • (b)ferromagnetic
  • (c)diamagnetic
  • (d)antiferromagnetic

  Answer(c) diamagnetic — it is weakly repelled by a magnetic field.