Chemical Bonding - Result Question 65
####40. On hybridisation of one $s$ and one $p$-orbital we get
(a) two mutually perpendicular orbitals
$(1984,1 M)$
(b) two orbitals at $180^{\circ}$
(c) four orbitals directed tetrahedrally
(d) three orbitals in a plane
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Solution:
- Hybridisation of one ’ $s$ ’ and one ’ $p$ ’ orbitals gives two $s p$ hybrid orbitals oriented linearly at $180^{\circ}$.
$$ s+p \longrightarrow 2 s p \text { hybrid orbitals } $$
