Question

Which of the following hybridizations is associated with a bond angle on the central atoms of \[{{109.5}^{o}}\] degrees?
A) sp
B) $ s{{p}^{2}} $
C) $ s{{p}^{3}} $
D) $ ds{{p}^{3}} $

Answer 1

Hint: In valence bond theory, orbital hybridisation is the notion of mixing atomic orbitals to produce new hybrid orbitals (with different energies, shapes, and other properties than the component atomic orbitals) appropriate for electron pairing to form chemical bonds.

Complete answer:
The valence-shell s orbital mixes with three valence-shell p orbitals to produce four equivalent $ s{{p}^{3}} $ mixtures that are placed in a tetrahedral configuration around the carbon to connect to four distinct atoms in a carbon atom that makes four single bonds. Hybrid orbitals are symmetrically arranged in space and are important in explaining molecular geometry and atomic bonding characteristics. Hybrid orbitals are usually created by combining atomic orbitals with similar energies.
Hybrid orbitals are thought to be a combination of atomic orbitals superimposed in varying proportions on each other. The C hybrid orbital that generates each carbon–hydrogen bond in methane, for example, is characterised as $ s{{p}^{3}} $ hybridised because it has 25% s character and 75% p character. When $ s{{p}^{3}} $ orbitals develop, they want to spread out as much as possible. This is a tetrahedral configuration with a \[{{109.5}^{o}}\] degree angle.
The energy released by the creation of two extra bonds more than compensates for the necessary excitation energy, favouring the formation of four C-H bonds energetically.
The lowest energy is attained quantum mechanically if the four bonds are equal, which means they are generated from equivalent orbitals on the carbon. The four $ s{{p}^{3}} $ hybrids are a collection of four analogous orbitals that are linear combinations of the valence-shell (core orbitals are practically never involved in bonding) s and p wave functions.
Hence option c is correct.

Note:
In the presence of four hydrogen atoms, Pauling theorised that the s and p orbitals create four equivalent combinations, which he dubbed hybrid orbitals. To identify its makeup, each hybrid is labelled $ s{{p}^{3}} $ and is oriented along one of the four C-H bonds.