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In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom ( [link] ). We find this situation in acetylene, H−C≡C−H , which is a linear molecule. The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ bond between the carbon atoms ( [link] ). The remaining sp orbitals form σ bonds with hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.

A diagram of a carbon atom with two balloon-like purple orbitals labeled, “sp” arranged in a linear fashion around it is shown. Four red balloon-like orbitals are aligned in pairs in the y and z axes around the carbon and are labeled, “unhybridized p orbital,” and, “Second unhybridized p orbital.”
Diagram of the two linear sp hybrid orbitals of a carbon atom, which lie in a straight line, and the two unhybridized p orbitals at perpendicular angles.
Two diagrams are shown and labeled, “a” and “b.” Diagram a shows two carbon atoms with two purple balloon-like orbitals arranged in a plane around each of them, and four red balloon-like orbitals arranged along the y and z axes perpendicular to the plane of the molecule. There is an overlap of two of the purple orbitals in between the two carbon atoms. The other two purple orbitals that face the outside of the molecule are shown interacting with spherical blue orbitals from two hydrogen atoms. Diagram b depicts a similar image to diagram a, but the red, vertical orbitals are interacting above and below and to the front and back of the plane of the molecule to form two areas labeled, “One pi bond,” and, “Second pi bond,” each respectively.
(a) In the acetylene molecule, C 2 H 2, there are two C–H σ bonds and a C C triple bond involving one C–C σ bond and two C–C π bonds. The dashed lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized p orbitals. (b) This shows the overall outline of the bonds in C 2 H 2 . The two lobes of each of the π bonds are positioned across from each other around the line of the C–C σ bond.

Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.

For example, molecule benzene has two resonance forms ( [link] ). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp 2 . The electrons in the unhybridized p orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)

A diagram is shown that is made up of two Lewis structures connected by a double ended arrow. The left image shows six carbon atoms bonded together with alternating double and single bonds to form a six-sided ring. Each carbon is also bonded to a hydrogen atom by a single bond. The right image shows the same structure, but the double and single bonds in between the carbon atoms have changed positions.
Each carbon atom in benzene, C 6 H 6 , is sp 2 hybridized, independently of which resonance form is considered. The electrons in the π bonds are not located in one set of p orbitals or the other, but rather delocalized throughout the molecule.

Assignment of hybridization involving resonance

Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO 2 , is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO 2 ?

Solution

The resonance structures of SO 2 are

Two Lewis structures connected by a double-ended arrow are shown. The left structure shows a sulfur atom with one lone pair of electrons and a positive sign which is single bonded on one side to an oxygen atom with three lone pairs of electrons and a negative sign. The sulfur atom is double bonded on the other side to another oxygen atom with two lone pairs of electrons. The right-hand structure is the same as the left except that the position of the double bonded oxygen atom is switched. In both structures the attached oxygen atoms form an acute angle in terms of the sulfur atom.

The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp 2 .

Check your learning

Another acid in acid rain is nitric acid, HNO 3 , which is produced by the reaction of nitrogen dioxide, NO 2 , with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO 2 ? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)

Answer:

sp 2

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Key concepts and summary

Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary.

Chemistry end of chapter exercises

The bond energy of a C–C single bond averages 347 kJ mol −1 ; that of a C C triple bond averages 839 kJ mol −1 . Explain why the triple bond is not three times as strong as a single bond.

A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.

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For the carbonate ion, CO 3 2− , draw all of the resonance structures. Identify which orbitals overlap to create each bond.

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A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H 3 CCN. It is present in paint strippers.

(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.

(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.

(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.

(a)
A Lewis structure is shown in which a carbon atom is attached by single bonds to three hydrogen atoms. It is also attached by a single bond to a carbon atom that is triple bonded to a nitrogen atom with one lone electron pair. Below the structure is a right facing arrow with its head near the nitrogen and its tail, which looks like a plus sign, located near the carbon atoms. The arrow is labeled, “dipole moment.”
(b) The terminal carbon atom uses sp 3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2 p orbital on carbon and a nitrogen 2 p orbital.

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For the molecule allene, H 2 C = C = CH 2 , give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?

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Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:

(a) ClNO (N is the central atom)

(b) CS 2

(c) Cl 2 CO (C is the central atom)

(d) Cl 2 SO (S is the central atom)

(e) SO 2 F 2 (S is the central atom)

(f) XeO 2 F 2 (Xe is the central atom)

(g) ClOF 2 + (Cl is the central atom)

(a) sp 2 ; (b) sp ; (c) sp 2 ; (d) sp 3 ; (e) sp 3 ; (f) sp 3 d ; (g) sp 3

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Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.

(a) H 3 PO 4 , phosphoric acid, used in cola soft drinks

(b) NH 4 NO 3 , ammonium nitrate, a fertilizer and explosive

(c) S 2 Cl 2 , disulfur dichloride, used in vulcanizing rubber

(d) K 4 [O 3 POPO 3 ], potassium pyrophosphate, an ingredient in some toothpastes

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For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:

(a) ozone (O 3 ) central O hybridization

(b) carbon dioxide (CO 2 ) central C hybridization

(c) nitrogen dioxide (NO 2 ) central N hybridization

(d) phosphate ion ( PO 4 3− ) central P hybridization

(a) sp 2 , delocalized; (b) sp , localized; (c) sp 2 , delocalized; (d) sp 3 , delocalized

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For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:

(a) Hybridization of each carbon

A Lewis structure is shown in which a carbon atom is single bonded to three hydrogen atoms and a second carbon atom. This second carbon atom is, in turn, double bonded to an oxygen atom with two lone pairs of electrons. The second carbon atom is also single bonded to another carbon atom that is single bonded to three hydrogen atoms.

(b) Hybridization of sulfur

A Lewis structure is shown in which a sulfur atom with two lone pairs of electrons and a positive sign is double bonded to an oxygen with two lone pairs of electrons. The sulfur atom is also single bonded to an oxygen with three lone pairs of electrons with a negative sign. It is drawn in an angular shape.

(c) All atoms

A Lewis structure is shown in which a hexagonal ring structure is made up of five carbon atoms and one nitrogen atom with a lone pair of electrons. There are alternating double and single bonds in between each carbon atom. Each carbon atom is also single bonded to one hydrogen atom.
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Draw the orbital diagram for carbon in CO 2 showing how many carbon atom electrons are in each orbital.

A diagram is shown in two parts, connected by a right facing arrow labeled, “Hybridization.” The left diagram shows an up-facing arrow labeled, “E.” To the lower right of the arrow is a short, horizontal line labeled, “2 s,” that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, “2 p.” Above both sets of lines is the phrase, “Orbitals in an isolated C atom.” There are two upward facing arrows on two of these lines. The right side of the diagram shows two short, horizontal lines placed halfway up the space and each labeled, “s p.” An upward-facing half arrow is drawn vertically on each line. Above these lines are two other short, horizontal lines, each labeled, “2 p,” and which have two upward facing arrows on them. Above both sets of lines is the phrase, “Orbitals in the s p hybridized C in C O subscript 2.”

Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.

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Questions & Answers

what is hydrocarbon
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this is an organic compound that consist of hydrogen and ox
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Hydrocarbon are organic compound consisting of hydrogen and carbon atoms
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relationship between orbit and a shell
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INETIABOR
they are mixture of same particles or substance
INETIABOR
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it is a branch of science that deals with the mixture of chemicals and properties and uses of matter
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the third ionization energy of aluminium
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Metalloids are substances having the characteristics of both metals and non-metals
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Acid is an electron acceptor
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The number of chemical bonds the atoms of a certain element can form.
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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