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Key concepts and summary

We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp 2 hybridization; four, sp 3 hybridization; five, sp 3 d hybridization; and six, sp 3 d 2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals ( p or d orbitals).

Chemistry end of chapter exercises

Why is the concept of hybridization required in valence bond theory?

Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

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Give the shape that describes each hybrid orbital set:

(a) sp 2

(b) sp 3 d

(c) sp

(d) sp 3 d 2

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Explain why a carbon atom cannot form five bonds using sp 3 d hybrid orbitals.

There are no d orbitals in the valence shell of carbon.

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What is the hybridization of the central atom in each of the following?

(a) BeH 2

(b) SF 6

(c) PO 4 3−

(d) PCl 5

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A molecule with the formula AB 3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.

trigonal planar, sp 2 ; trigonal pyramidal (one lone pair on A) sp 3 ; T-shaped (two lone pairs on A sp 3 d , or (three lone pairs on A) sp 3 d 2

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Methionine, CH 3 SCH 2 CH 2 CH(NH 2 )CO 2 H, is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?

A Lewis structure is shown in which a carbon atom is single bonded to three hydrogen atoms and single bonded to a sulfur atom with two lone pairs of electrons. The sulfur atom is attached to a chain of four singly bonded carbon atoms, the first two of which are single bonded to two hydrogen atoms each, and the third of which is single bonded to a hydrogen atom and single bonded to a nitrogen atom which has one lone electron pair. The nitrogen atom is also single bonded to two hydrogen atoms. The fourth andfinal carbon in the chain is double bonded to an oxygen with two lone pairs of electrons and single bonded to an oxygen atom with two lone pairs of electrons. The second oxygen atom is single bonded to a hydrogen atom.
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Sulfuric acid is manufactured by a series of reactions represented by the following equations:
S 8 ( s ) + 8 O 2 ( g ) 8 SO 2 ( g )
2 S O 2 ( g ) + O 2 ( g ) 2 SO 3 ( g )
SO 3 ( g ) + H 2 O ( l ) H 2 SO 4 ( l )

Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:

(a) circular S 8 molecule

(b) SO 2 molecule

(c) SO 3 molecule

(d) H 2 SO 4 molecule (the hydrogen atoms are bonded to oxygen atoms)

(a) Each S has a bent (109°) geometry, sp 3
A Lewis structure is shown in which eight sulfur atoms, each with two lone pairs of eletrons, are single bonded together into an eight-sided ring.
(b) Bent (120°), sp 2
Two Lewis structure are shown, connected by a double-ended arrow. The left structure shows a sulfur atom with one lone pair of electrons double bonded to an oxygen atom with two lone pairs of electrons on the left and single bonded to an oxygen atom with three lone pairs of electrons on the right. The right structure shows the same molecule, except that the double bonded oxygen is on the right side of the sulfur and the single bonded oxygen is to the left of the sulfur.
(c) Trigonal planar, sp 2
A Lewis structure of a sulfur atom singly bonded to two oxygen atoms, each with three lone pairs of electrons, and double bonded to a third oxygen atom with two lone pairs of electrons is shown.
(d) Tetrahedral, sp 3
A Lewis structure is shown in which a sulfur atom is single bonded to four oxygen atoms. Two of the oxygen atoms have three lone pairs of electrons while the other two each have two lone pairs of electrons and are each singly bonded to a hydrogen atom.

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Two important industrial chemicals, ethene, C 2 H 4 , and propene, C 3 H 6 , are produced by the steam (or thermal) cracking process:

2 C 3 H 8 ( g ) C 2 H 4 ( g ) + C 3 H 6 ( g ) + CH 4 ( g ) + H 2 ( g )

For each of the four carbon compounds, do the following:

(a) Draw a Lewis structure.

(b) Predict the geometry about the carbon atom.

(c) Determine the hybridization of each type of carbon atom.

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For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.

(a) What is the formula of the compound?

(b) Write a Lewis structure for the compound.

(c) Predict the shape of the molecules of the compound.

(d) What hybridization is consistent with the shape you predicted?

(a) XeF 2
(b)
A Lewis structure is shown in which a xenon atom that has three lone pairs of electrons is single bonded to two fluorine atoms, each of which has three lone pairs of electrons.
(c) linear (d) sp 3 d

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Consider nitrous acid, HNO 2 (HONO).

(a) Write a Lewis structure.

(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO 2 molecule?

(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO 2 ?

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Strike-anywhere matches contain a layer of KClO 3 and a layer of P 4 S 3 . The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO 3 contains the ClO 3 ion. P 4 S 3 is an unusual molecule with the skeletal structure.

A Lewis structure is shown in which three phosphorus atoms are single bonded together to form a triangle. Each phosphorus is bonded to a sulfur atom by a vertical single bond and each of those sulfur atoms is then bonded to a single phosphorus atom so that a six-sided ring is created with a sulfur in the middle.

(a) Write Lewis structures for P 4 S 3 and the ClO 3 ion.

(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.

(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.

(d) Determine the oxidation states and formal charge of the atoms in P 4 S 3 and the ClO 3 ion.

(a)
Two Lewis structure are shown, the left of which depicts three phosphorus atoms single bonded together to form a triangle. Each phosphorus is bonded to a sulfur atom by a vertical single bond and each of those sulfur atoms is then bonded to a single phosphorus atom so that a six-sided ring is created with a sulfur in the middle. Each sulfur atom in this structure has two lone pairs of electrons while each phosphorus has one lone pair. The second Lewis structure shows a chlorine atom with one lone pair of electrons single bonded to three oxygen atoms, each of which has three lone pairs of electrons.
(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp 3 ; (d) Oxidation states P +1, S 1 1 3 , Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1

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Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)

A Lewis structure is shown that is missing all of its bonds. Six carbon atoms form a chain. There are three hydrogen atoms located around the first carbon, two located around the second, one located near the fifth, and two located around the sixth carbon.
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Write Lewis structures for NF 3 and PF 5 . On the basis of hybrid orbitals, explain the fact that NF 3 , PF 3 , and PF 5 are stable molecules, but NF 5 does not exist.

Two Lewis structures are shown. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to three fluorine atoms, each of which has three lone pairs of electrons. The right structure shows a phosphorus atoms single bonded to five fluorine atoms, each of which has three lone pairs of electrons.

Phosphorus and nitrogen can form sp 3 hybrids to form three bonds and hold one lone pair in PF 3 and NF 3 , respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp 3 d hybrid orbitals to bind five fluorine atoms in NF 5 . Phosphorus has d orbitals and can bind five fluorine atoms with sp 3 d hybrid orbitals in PF 5 .

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In addition to NF 3 , two other fluoro derivatives of nitrogen are known: N 2 F 4 and N 2 F 2 . What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?

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Questions & Answers

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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