Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Describe each highlighted bond in terms of the overlap of atomic orbitals. (a) C=C (b) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) (c) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}\) (d) C=O (e) O=CO (f) CCO (g) CN (h) COC=O (i) \(\mathrm{H}-\mathrm{O}-\mathrm{N}=\mathrm{O}_{*}^{*+}\)

Short Answer

Expert verified
Question: Describe the type of chemical bond and orbital overlap in the following molecules: (a) \(C=C\), (b) \(H-C \equiv C-H\), (c) \(CH_2=C=CH_2\), (d) \(C=O\), (e) \(O=CO\), (f) \(CO\), (g) \(CN\), (h) \(COC=O\), and (i) \(O=NO\).

Step by step solution

01

(a) \(C=C\)

For molecule (a), the bond is a double bond between two carbon atoms. This double bond consists of one sigma bond formed by the overlap of two sp2 hybridized orbitals and one pi bond formed by the overlap of two parallel p orbitals.
02

(b) \(H-C \equiv C-H\)

In molecule (b), we have a triple bond between two carbon atoms, and each carbon is bonded to a hydrogen atom. The triple bond consists of one sigma bond formed by the overlap of two sp hybridized orbitals and two pi bonds formed by the overlap of parallel p orbitals. The bond between each carbon and hydrogen atom is a single sigma bond formed by the overlap of sp hybridized orbitals in carbon and the 1s atomic orbital of the hydrogen.
03

(c) \(CH_2=C=CH_2\)

In molecule (c), there is a double bond between two carbon atoms, and each carbon is bonded to two hydrogen atoms. The double bond consists of one sigma bond formed by the overlap of two sp2 hybridized orbitals and one pi bond formed by the overlap of two parallel p orbitals. The bond between each carbon and hydrogen atom is a single sigma bond formed by the overlap of sp2 hybridized orbitals in carbon and the 1s atomic orbital of the hydrogen.
04

(d) \(C=O\)

In molecule (d), there is a double bond between a carbon and an oxygen atom. The double bond consists of one sigma bond formed by the overlap of one sp2 hybridized orbital from carbon and one sp2 hybridized orbital from oxygen. The pi bond formed by the overlap of two parallel p orbitals.
05

(e) \(O=CO\)

In molecule (e), we have a double bond between an oxygen and a carbon atom, and a double bond between the same carbon atom and another oxygen atom. Each double bond is made up of one sigma bond formed by the overlap of sp2 hybridized orbitals and one pi bond formed by the overlap of two parallel p orbitals.
06

(f) \(CO\)

In molecule (f), there is a single sigma bond between one carbon and one oxygen atom, formed by the overlap of sp3 hybridized orbital in carbon and sp3 hybridized orbital in oxygen.
07

(g) \(CN\)

In molecule (g), there is triple bond between one carbon and one nitrogen atom. The triple bond consists of one sigma bond formed by the overlap of sp hybridized orbitals and two pi bonds formed by the overlap of parallel p orbitals.
08

(h) \(COC=O\)

In molecule (h), we have a single sigma bond between two carbon atoms, formed by the overlap of sp3 hybridized orbitals. The carbon atom bonded to an oxygen atom has a double bond formed by one sigma bond (overlap of sp2 hybridized orbitals) and one pi bond formed by the overlap of parallel p orbitals.
09

(i) \(O=NO\)

In molecule (i), there is double bond between one oxygen atom and one nitrogen atom. The double bond consists of one sigma bond formed by the overlap of sp2 hybridized orbitals and one pi bond formed by the overlap of parallel p orbitals.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using cartoon representations, draw a molecular orbital mixing diagram for a \(\mathrm{C}-\mathrm{O}\) \(\sigma\) bond. In your picture, consider the relative energies of \(\mathrm{C}\) and \(\mathrm{O}\) and how this changes the resulting bonding and antibonding molecular orbitals relative to a \(\mathrm{C}-\mathrm{C} \sigma\) bond.

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) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}\)

Which compounds have nonpolar covalent bonds, which have polar covalent bonds, and which have ions? (a) \(\mathrm{LiF}\) (b) \(\mathrm{CH}_{3} \mathrm{~F}\) (c) \(\mathrm{MgCl}_{2}\) (d) \(\mathrm{HCl}\)

(a) Draw a Lewis structure for the ozone molecule, \(\mathrm{O}_{3}\). (The order of atom attachment is \(\mathrm{O}-\mathrm{O}-\mathrm{O}\), and they do not form a ring.) Chemists use ozone to cleave carbon-carbon double bonds (Section 6.5C). (b) Draw four contributing resonance structures; include formal charges. (c) How does the resonance model account for the fact that the length of each \(\mathrm{O}-\mathrm{O}\) bond in ozone \((128 \mathrm{pm})\) is shorter than the \(\mathrm{O}-\mathrm{O}\) single bond in hydrogen peroxide \((\mathrm{HOOH}\), \(147 \mathrm{pm}\) ) but longer than the \(\mathrm{O}-\mathrm{O}\) double bond in the oxygen molecule ( \(123 \mathrm{pm})\) ?

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?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free