Skip to content
Question of 112

Q.Explain the hybridization involved in PCl5 molecule.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2026Subjective· 4mImportance★★★★★
0% · 0/112 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Phosphorus in PCl5 is sp3d hybridised, giving a trigonal bipyramidal shape with three equatorial and two (longer) axial P-Cl bonds.

Ground and excited state of P: The electronic configuration of phosphorus (Z = 15) is [Ne] 3s2 3p3. To form five bonds it must provide five unpaired (half-filled) orbitals. On excitation, one electron from the filled 3s orbital is promoted to a vacant 3d orbital:

Excited state: 3s1 3px1 3py1 3pz1 3d1 (five half-filled orbitals).

Hybridization: These one s, three p and one d orbitals mix to form five equivalent sp3d hybrid orbitals.

Bond formation and shape: Each sp3d hybrid orbital overlaps with a singly occupied 3p orbital of a chlorine atom, forming five P-Cl sigma bonds. The five hybrid orbitals point to the corners of a trigonal bipyramid:

  • Three orbitals lie in a plane at 120 degrees to one another (equatorial bonds).
  • Two orbitals are perpendicular to this plane, one above and one below, at 90 degrees (axial bonds). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.