JEE Main 2022 · 28 June · morning · Numerical answer · +4 / −0
The hybridization of P exhibited in PF₅ is spˣdʸ. The value of y is __________.
Answer and solution
Answer 1
Solution
Chemistry ma’am · 3 steps
Count the electron pairs around P: five bond pairs need five hybrid orbitals, so one d orbital joins in.
- Electron pairs on P. Phosphorus has 5 valence electrons and uses all of them to form 5 P-F bonds. That leaves 5 bond pairs and 0 lone pairs on P.
steric number = 5 + 0 = 5
- Pick the orbitals. Five hybrid orbitals need five atomic orbitals: one 3s, three 3p and one 3d. This is sp³d hybridisation, which gives the trigonal bipyramidal shape of PF₅.
- So. Matching spˣdʸ with sp³d¹ gives x = 3 and y = 1.
The common slip: 3. Reading off the p superscript instead of the d superscript gives 3, which is x, not y.
The idea to keep. The number of hybrid orbitals equals bond pairs plus lone pairs on the central atom. Fill s first, then the three p, then add d orbitals for anything beyond 4.
Answer from the NTA final answer key. Solution by PrepWiser.