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Chapter 1: Problem 56

State the orbital hybridization of each highlighted atom. (a) CC (b) C=C (c) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) (d) C=O (e) O=CO (f) CCO (g) CN (h) \(\mathrm{H}-\stackrel{*}{\mathrm{O}}-\mathrm{N}=\stackrel{*}{\mathrm{O}}\) (i) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}\)

Short Answer

Expert verified
Question: Determine the orbital hybridization of each highlighted atom in the following structures: (a) CC (b) C=C (c) H-C鈮-H (d) C=O (e) O=CO (f) CCO (g) CN (h) H-O-N=O (i) CH2=C=CH2 Answer: (a) Both carbon atoms: sp3 (b) Both carbon atoms: sp2 (c) Both carbon atoms: sp (d) Carbon: sp2, Oxygen: sp2 (e) Carbon: sp2, Both oxygen atoms: sp2 (f) Both carbon atoms: sp3, Oxygen: sp3 (g) Carbon: sp, Nitrogen: sp2 (h) First oxygen: sp2, Nitrogen: sp3, Second oxygen: sp2 (i) All carbon atoms: sp2

Step by step solution

01

Carbon atom hybridization

For both carbon atoms in the structure, there are only single covalent bonds. Each carbon atom is surrounded by four electron domains. This leads to a sp3 hybridization for each carbon atom. (b) C=C:
02

Carbon atom hybridization

In this structure, the carbon atoms are double-bonded to each other. Each carbon atom has three electron domains in the structure. This corresponds to a sp2 hybridization for both carbon atoms. (c) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\):
03

Carbon atom hybridization

Here, the carbon atoms are connected with a triple bond. Each carbon atom has two electron domains in the structure. This leads to a sp hybridization for both carbon atoms. (d) C=O:
04

Carbon and Oxygen atom hybridization

In this structure, the carbon atom is double-bonded to the oxygen atom. The carbon atom has three electron domains, resulting in sp2 hybridization. The oxygen atom, with a double bond and two lone pairs (four electron domains), is also sp2 hybridized. (e) O=CO:
05

Carbon and Oxygen atom hybridization

In this structure, the carbon atom is double-bonded to two oxygen atoms. The carbon atom has three electron domains, resulting in sp2 hybridization. Both oxygen atoms, with a double bond and two lone pairs each (four electron domains), are also sp2 hybridized. (f) CCO:
06

Carbon and Oxygen atom hybridization

In this structure, both carbon atoms are single-bonded to other atoms. Each carbon atom has four electron domains and exhibits sp3 hybridization. The oxygen atom has two single bonds and two lone pairs (four electron domains) and is also sp3 hybridized. (g) CN:
07

Carbon and Nitrogen atom hybridization

The carbon atom is triple-bonded to the nitrogen atom. The carbon atom has two electron domains, resulting in sp hybridization. The nitrogen atom, with a triple bond and one lone pair (three electron domains), is sp2 hybridized. (h) \(\mathrm{H}-\stackrel{*}{\mathrm{O}}-\mathrm{N}=\stackrel{*}{\mathrm{O}}\):
08

Oxygen and Nitrogen atom hybridization

The first oxygen atom is single-bonded to the hydrogen and nitrogen. With two lone pairs and a single bond, it has three electron domains and is thus sp2 hybridized. The nitrogen atom is double-bonded to the second oxygen. With two single bonds and a double bond, it has four electron domains, resulting in sp3 hybridization. The second oxygen atom has two lone pairs and a double bond, which equates to three electron domains, and is therefore sp2 hybridized. (i) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}\):
09

Carbon atom hybridization

Each outer carbon atom has two single bonds and a double bond (three electron domains), which corresponds to sp2 hybridization. The central carbon atom is double-bonded to both outer carbon atoms, also resulting in three electron domains, thus it is also sp2 hybridized.

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Most popular questions from this chapter

Which statements are true about electronegativity? (a) Electronegativity increases from left to right in a period of the Periodic Table. (b) Electronegativity increases from top to bottom in a column of the Periodic Table. (c) Hydrogen, the element with the lowest atomic number, has the smallest electronegativity. (d) The higher the atomic number of an element, the greater its electronegativity.

Draw a Lewis structure for the azide ion, \(\mathrm{N}_{3}{ }^{-}\). (The order of atom attachment is \(\mathrm{N}-\mathrm{N}-\mathrm{N}\), and they do not form a ring.) How does the resonance model account for the fact that the lengths of the \(\mathrm{N}-\mathrm{N}\) bonds in this ion are identical?

Draw condensed structural formulas for the three ketones with the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\).

Use VSEPR to predict the geometry of these ions. (a) \(\mathrm{NH}_{2}{ }^{-}\) (b) \(\mathrm{NO}_{2}{ }^{-}\) (c) \(\mathrm{NO}_{2}^{+}\) (d) \(\mathrm{NO}_{3}^{-}\)

Draw structural formulas for (a) The four primary (1 \(\left.1^{\circ}\right)\) amines with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\). (b) The three secondary \(\left(2^{\circ}\right)\) amines with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\). (c) The one tertiary \(\left(3^{\circ}\right)\) amine with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\).
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