Question

consider the lewis structure below and answer the following five (5) questions.
what is the hybridization of ( c_{b} )?
( s p ) ( s p^{2} ) ( s p^{3} )
what is the hybridization of ( o_{y} )?
( s p ) ( s p^{2} ) ( s p^{3} )
what is the ideal ( o - c_{a} - o_{y} ) bond angle?
( 90^{circ} ) ( 109^{circ} ) ( 120^{circ} ) ( 180^{circ} )
what is the ideal ( c_{a} - o_{y} - c_{b} ) bond angle?
( 90^{circ} ) ( 109^{circ} ) ( 120^{circ} ) ( 180^{circ} )
which of the following describes the ( c_{a} - o_{y} ) bond?
( sigma(2 p - 2 p) ) ( sigma(s p^{3} - s p^{3}) ) ( sigma(s p^{2} - s p^{3}) )

Explanation:

Hybridization of \(C_b\)

• Step1: Count the electron - group around \(C_b\)

\(C_b\) is bonded to four atoms (three \(H\) atoms and one \(O\) atom). According to the VSEPR (Valence - Shell Electron - Pair Repulsion) theory, the number of electron - groups \(n = 4\).

• Step2: Determine the hybridization

The formula for hybridization is \(hybridization= s + p\) orbitals, and if \(n = 4\), the hybridization is \(sp^{3}\) (because \(s + 3p\) orbitals are used for bonding).

Hybridization of \(O_y\)

• Step1: Count the electron - group around \(O_y\)

\(O_y\) has two bonding pairs (one with \(C_a\) and one with \(C_b\)) and two lone pairs. So the number of electron - groups \(n=4\) ( \(2\) bonding + \(2\) non - bonding).

• Step2: Determine the hybridization

Using the formula \(hybridization = s + p\) orbitals, when \(n = 4\), the hybridization is \(sp^{3}\) ( \(s+3p\) orbitals, where two \(p\) orbitals are used for lone pairs and two for bonding).

Ideal \(O - C_a - O_y\) bond angle

• Step1: Determine the hybridization of \(C_a\)

\(C_a\) has three electron - groups (two single bonds and one double bond). The number of electron - groups \(n = 3\), so the hybridization is \(sp^{2}\).

• Step2: Determine the bond angle

For \(sp^{2}\) hybridization, the ideal bond angle is \(120^{\circ}\) (trigonal planar geometry).

Ideal \(C_a - O_y - C_b\) bond angle

• Step1: Recall the hybridization of \(O_y\)

We found that \(O_y\) has \(sp^{3}\) hybridization.

• Step2: Determine the bond angle

For \(sp^{3}\) hybridization (tetrahedral electron - pair geometry, but with two lone pairs), the ideal bond angle (based on the parent tetrahedral geometry with \(n = 4\) electron - groups) is \(109^{\circ}\) (bent geometry due to lone pairs, but the ideal angle from the electron - pair geometry is \(109^{\circ}\)).

\(C_a - O_y\) bond description

• Step1: Recall the hybridization of \(C_a\) and \(O_y\)

\(C_a\) is \(sp^{2}\) hybridized and \(O_y\) is \(sp^{3}\) hybridized.

• Step2: Determine the bond type

A single bond is a \(\sigma\) bond. The \(\sigma\) bond is formed between the \(sp^{2}\) hybrid orbital of \(C_a\) and \(sp^{3}\) hybrid orbital of \(O_y\), so it is \(\sigma(sp^{2}-sp^{3})\).

Answer:

• Hybridization of \(C_b\): \(sp^{3}\)
• Hybridization of \(O_y\): \(sp^{3}\)
• Ideal \(O - C_a - O_y\) bond angle: \(120^{\circ}\)
• Ideal \(C_a - O_y - C_b\) bond angle: \(109^{\circ}\)
• \(C_a - O_y\) bond: \(\sigma(sp^{2}-sp^{3})\)