Chapter 11:  Molecular Geometry, Polarity of Molecules, and Advanced Bonding Theory

 

 

Section 11-1: Molecular Geometry: Using VSEPR Theory to Determine Three-Dimensional Shapes and Bond Angles

Section 11-2: Polarity of Molecules

Section 11-3: Hybridization

Section 11-4: Sigma and Pi Bonds

Section 11-5: Molecular Orbital (MO) Theory

Chapter 11 Practice Exercises and Review Quizzes

 

 

 

 

 

Section 11-1:  Molecular Geometry:  Using VSEPR Theory to Determine Three-Dimensional Shapes and Bond Angles

 

Diatomic molecules and ions are linear, regardless of the number of covalent bonds between the atoms.  For molecules or ions with three or more bonded atoms, the Lewis structure indicates which specific atoms are bonded together and the number of bonds between the atoms.  However, the Lewis structure does not necessarily provide an accurate depiction of the three-dimensional shape and bond angles in the molecule or ion.

 

Valence-Shell Electron-Pair Repulsion (VSEPR) Theory suggests that the valence electron domains comprised of covalent bonds and lone pairs adjacent to the central atom in a molecule or ion will be arranged in a three-dimensional shape that minimizes electron repulsion between the bonds and lone pairs.  We will use the VSEPR notation ABxEy to categorize molecules and ions, where A represents the center atom, x represents the number of outer atoms (B) bonded to the center atom A, and y represents the number of lone pairs (E) on the center atom A.  Once we know the VSEPR notation for an atom or ion, we can determine the three-dimensional shape and bond angles.  When we draw three-dimensional sketches of molecules or ions on paper, we will use solid straight lines to represent bonds oriented in the plane of the paper, dashed wedges to represent bonds oriented back behind the plane of the paper, and solid wedges to represent bonds oriented forward out of the paper.  We will also only include lone pairs on the center atom in sketches, but not lone pairs on the outer atoms as these generally do not affect the shape according to VSEPR Theory:

 

I. AB2E0 = AB2 = two outer atoms bonded to center atom + no lone pairs on center atom = molecule is linear:

 

 

II. AB3E0 = AB3 = three outer atoms bonded to center atom + no lone pairs on center atom = molecule is trigonal planar:

 

 

III. AB2E1 = two outer atoms bonded to center atom + one lone pair on center atom = molecule is bent:

 

 

The bond angle is decreased below 120 because the lone pair has a slightly greater repulsive effect than the electrons in the covalent bonds between A and B.

 

IV. AB4E0 = AB4 = four outer atoms bonded to center atom + no lone pairs on center atom = tetrahedral:

 

 

V. AB3E1 = three outer atoms bonded to center atom + one lone pair on center atom = trigonal pyramidal:

 

 

The bond angles are decreased below 109.5 because the lone pair has a slightly greater repulsive effect than the electrons in the covalent bonds between A and B. 

 

VI. AB2E2 = two outer atoms bonded to center atom + two lone pairs on center atom = bent:

 

 

In the AB2E2 = bent case, the extra electron repulsion caused by the second lone pair on the center atom may cause the B-A-B bond angle to decrease even further below 109.5 than the decrease expected in the AB3E1 = trigonal pyramidal case with only one lone pair on the center atom.

 

In contrast to the Lewis structures, our three-dimensional sketches will neglect the differences between and, thus, not distinguish among single, double, and triple bonds, as demonstrated in the following problem:

 

Sample Exercise 11A:

 

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the bond angle for each of the following molecules and ions:

 

(a) CO2

(b) NH3

(c) CO32-

(d) CH4

(e) NO2-

(f) H2O

 

Solution:

 

(a)

Lewis structure     

      

AB2 = linear

 

3-D sketch:     

      

(no need to distinguish between single v. double v. triple bonds in sketch)

 

(b)

Lewis structure:    

      

 AB3E1 = trigonal pyramidal 

 

3-D sketch:

 

 

(c)

Lewis structure:

 

 

Note that we do not actually need to draw all the resonance structures to determine the shape as we can see from any one of the three resonance structures that the carbonate ion = AB3 = trigonal planar.

 

3-D sketch:

 

 

(d)

Lewis structure:

           

AB4 = tetrahedral

 

3-D sketch:

 

 

(e)

Lewis structure:     

         

AB2E1 = bent (no need to show all resonance structures to determine shape)

 

3-D sketch:

 

 

(f)

Lewis structure:  

         

AB2E2 = bent

 

3-D sketch:

 

 

 

For molecules or ions with more than one center atom, we can describe the shape in the region of each center atom as demonstrated in the following problem:

 

Sample Exercise 11B:

 

Draw the Lewis structure for acetic acid.  Name the molecular geometry and indicate the bond angles in the region of each center atom.

 

Solution:

 

Acetic acid = CH3COOH.  All organic acids with the ending COOH have a group of atoms (in this case CH3) single-bonded to the carbon in the COOH.  The carbon in the COOH is double-bonded to one oxygen and single-bonded to the second oxygen, with the hydrogen in the COOH single-bonded to the second oxygen.  Therefore, the Lewis structure is:

 

 

carbon on the left = AB4 = tetrahedral, bond angles = 109.5

 

carbon in center = AB3 = trigonal planar, bond angles = 120

 

oxygen on right = AB2E2 = bent, bond angle = <109.5

 

 

For molecules or ions with an “expanded octet” on the center atom, we will ignore any effects of lone pair repulsion on bond angles and instead only indicate the ideal bond angle(s) in each case:

 

VII. AB5E0 = AB5 = five outer atoms bonded to center atom + no lone pairs on center atom = trigonal bipyramidal.  The two outer atoms occupying an axial position (Ba) are not equivalent to the three outer atoms occupying an equatorial position (Be):

 

 

VIII. AB4E1 = four outer atoms bonded to center atom + one lone pair on center atom = seesaw.  The lone pair must occupy an equatorial position to minimize electron repulsion:

 

 

IX. AB3E2 = three outer atoms bonded to center atom + two lone pairs on center atom = T-shaped.  Both lone pairs must occupy equatorial positions to minimize electron repulsion:

 

 

X. AB2E3 = two outer atoms bonded to center + three lone pairs on center = linear.  All three lone pairs must occupy equatorial positions to minimize electron repulsion:

 

 

XI. AB6E0 = AB6 = six outer atoms bonded to center + no lone pairs on center = octahedral:

 

 

XII. AB5E1 = five outer atoms bonded to center + one lone pair on center = square pyramidal:

 

 

XIII. AB4E2 = four outer atoms bonded to center + two lone pairs on center = square planar:

 

 

Sample Exercise 11C:

 

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the ideal bond angle(s) for each of the following molecules and ions:

 

(a) BrCl4-

(b) BrF6+

(c) I3-

(d) IBr4+

(e) PCl5

(f) SF3-

(g) SeCl5-

 

Solution:

 

(a)

Lewis structure:

AB4E2 = square planar, 3-D sketch:

 

(b)

Lewis structure:

AB6 = octahedral, 3-D sketch:

 

(c)

Lewis structure:

AB2E3 = linear, 3-D sketch:

 

(d)

Lewis structure:

AB4E1 = seesaw, 3-D sketch:

 

(e)

Lewis structure:

AB5 = trigonal bipyramidal, 3-D sketch:

 

(f)

Lewis structure:

AB3E2 = T-shaped, 3-D sketch:

 

(g)

Lewis structure:

AB5E1 = square pyramidal, 3-D sketch:

 

 

 

Section 11-2:  Polarity of Molecules

 

In a polar covalent bond, the electrons will be more attracted toward the more electronegative atom.  We can indicate the direction in which the electrons are shifted in a polar covalent bond by placing a bond dipole arrow parallel to the bond in a sketch of the molecule, with the head of the arrow closer to the more electronegative atom and the + end of the arrow closer to the less electronegative atom.  For the molecule BrF, the head of the bond dipole arrow will be closer to the more electronegative F atom and the + end of the bond dipole arrow will be closer to the less electronegative Br atom:

 

 

A dipole moment (μ) is essentially a measurement of the overall net shift of electrons toward a particular direction in a neutral molecule.  A molecule with no identifiable direction toward which the electrons are shifted is said to be a nonpolar molecule with zero dipole moment (μ = 0).  A molecule with an identifiable direction toward which the electrons are shifted is said to be a polar molecule with a dipole moment greater than zero (μ > 0). 

 

In a sketch of the linear carbon dioxide molecule, the heads of the two bond dipole arrows will be closer to the more electronegative outer oxygen atoms:

 

However, since the two bond dipole arrows essentially cancel each other out because they are equal in magnitude but are oriented in opposite directions, carbon dioxide is a nonpolar molecule (μ = 0).

 

In a sketch of the bent water molecule, the head of the two bond dipole arrows will be closer to the more electronegative center oxygen atom:

 

 

In the case of water, the two bond dipoles arrows are equal in magnitude but do not cancel each other out.  We see that there is an overall net shift of electrons and, therefore, a dipole moment oriented toward the oxygen end of the molecule.  Thus, water is a polar molecule (μ > 0). 

 

The following general guideline indicates which categories of molecules will be nonpolar and which categories of molecules will be polar:

 

Molecules with no lone pairs on the center atom will generally be nonpolar if all the outer atoms are the same element.  Molecules with one or more lone pairs on the center atom and molecules with outer atoms that are different elements will generally be polar.

 

For each different category, the table below summarizes the molecular geometry (shape), bond angle, and whether the molecule is polar or nonpolar:

 

VSEPR Notation

Name of Molecular Geometry (Shape)

Bond Angle

Polar or Nonpolar Molecule?

AB2

linear

180

nonpolar*

AB3

trigonal planar

120

nonpolar*

AB2E1

bent

<120

polar

AB4

tetrahedral

109.5

nonpolar*

AB3E1

trigonal pyramidal

<109.5

polar

AB2E2

bent

<109.5

polar

 

*unless outer atoms are different elements

 

Sample Exercise 11C:

 

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, indicate the bond angle, and state whether each of the following molecules is polar or nonpolar:

 

(a) BCl3

(b) CH2F2

(c) HCN

(d) SCl2

(e) PF3

(f) HNO

 

Solution:

 

(a)

Lewis structure:     

     

AB3 = trigonal planar

 

3-D sketch:

 

nonpolar molecule

 

(b)

Lewis structure:

           

AB4 = tetrahedral

 

3-D sketch:

 

polar molecule (different outer elements)

 

(c)

Lewis structure:   

         

AB2 = linear

 

3-D sketch:  

   

polar molecule (different outer elements)

 

(d)

Lewis structure:    

       

AB2E2 = bent

 

3-D sketch:

 

polar molecule

 

(e)

Lewis structure:    

       

AB3E1 = trigonal pyramidal 

 

3-D sketch:

 

polar molecule

 

(f)

Lewis structure:     

    

 AB2E1 = bent

 

3-D sketch:

 

polar molecule

 

 

For molecules with an “expanded octet” on the center atom, we will focus only on those where all the outer atoms are the same element.  Although the general guideline above stating that no lone pairs on center atom = nonpolar molecule and one or more lone pairs on center atom = polar molecule still applies in many cases, note that the orientation of the bonds in AB2E3 = linear and AB4E2 = square planar molecules results in those molecules being nonpolar.   

 

For each different category of molecules with “expanded octets”, the table below summarizes the molecular geometry (shape), ideal bond angle(s), and whether the molecule is polar or nonpolar:

 

VSEPR Notation

Name of Molecular Geometry (Shape)

Ideal Bond Angle(s)

Polar or Nonpolar Molecule?

AB5

trigonal bipyramidal

90°, 120°

nonpolar

AB4E1

seesaw

90°, 120°

polar

AB3E2

T-shaped

90°

polar

AB2E3

linear

180°

nonpolar

AB6

octahedral

90°

nonpolar

AB5E1

square pyramidal

90°

polar

AB4E2

square planar

90°

nonpolar

 

 

Sample Exercise 11E:

 

Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, indicate the ideal bond angle(s), and state whether each of the following molecules is polar or nonpolar:

 

(a) AsF5

(b) BrF5

(c) ICl3

(d) KrF4

(e) SF4

(f) SeCl6

(g) XeCl2

 

Solution:

 

(a)

Lewis structure:

AB5 = trigonal bipyramidal, 3-D sketch:

nonpolar molecule

 

(b)

Lewis structure:

AB5E1 = square pyramidal, 3-D sketch:

polar molecule

 

(c)

Lewis structure:

AB3E2 = T-shaped, 3-D sketch:

polar molecule

 

(d)

Lewis structure:

AB4E2 = square planar, 3-D sketch:

nonpolar molecule

 

(e)

Lewis structure:

AB4E1 = seesaw, 3-D sketch:

polar molecule

 

(f)

Lewis structure:

AB6 = octahedral, 3-D sketch:

nonpolar molecule

 

(g)

Lewis structure:

AB2E3 = linear, 3-D sketch:

nonpolar molecule

 

 

Section 11-3:  Hybridization


Valence bond theory generally suggests that a covalent bond is formed when an orbital on one atom occupied by an unpaired valence electron overlaps partially with an orbital on a second atom occupied by an unpaired valence electron.  Increased electron density in the region of overlap between the atoms attracts the positively-charged nuclei toward each other. 

 

In some cases, valence bond models depicting overlap between unmodified s and p atomic orbitals would lead to predictions of bond angles and other properties that do not match experimental data.  Hybridization theory suggests that the valence s orbital and one or more valence p orbitals from the center atom undergo a change during the bonding process where the original atomic orbitals involved are mixed to create an equivalent number of new hybrid orbitals as follows:

 

A. If the ideal bond angle in the molecule or ion being formed is 109.5°, one s orbital and three p orbitals from the center atom will mix to create four identical sp3 hybrid orbitals arranged tetrahedrally 109.5° away from each other around the center atom.

 

B. If the ideal bond angle in the molecule or ion being formed is 120°, one s orbital and two p orbitals from the center atom will mix to create three identical sp2 hybrid orbitals arranged in a plane 120° away from each other around the center atom.

 

C. If the ideal bond angle in the molecule or ion being formed is 180°, one s orbital and one p orbital from the center atom will mix to create two identical sp hybrid orbitals arranged linearly 180° away from each other on opposite sides of the center atom. 

 

The table below summarizes the center atom hybridization based on VSEPR notation:

 

VSEPR Notation

Center Atom Hybridization

AB2

sp

AB3

sp2

AB2E1

sp2

AB4

sp3

AB3E1

sp3

AB2E2

sp3

 

 

Sample Exercise 11F:

 

Draw the Lewis structure and indicate the center atom hybridization for each of the following molecules and ions:

 

(a) BrNO

(b) ClCN

(c) SF3+

 

Solution:

 

(a)

AB2E1 = sp2

 

(b)

AB2 = sp

 

(c)

AB3E1 = sp3

 

 

Section 11-4:  Sigma and Pi Bonds


A bond formed as a result of end-to-end overlap of orbitals from two neighboring atoms is known as a sigma (σ) bond.  Consider the Lewis structure of C2H4:

 

 

Both carbon atoms in the molecule have three sp2 hybrid orbitals (AB3), each of which is occupied by an unpaired valence electron, that are 120° away from each other in the same plane.  A sigma bond will be formed between the carbon atoms as a result of end-to-end overlap between two sp2 orbitals, one from each carbon.  In addition, sigma bonds will be formed between each hydrogen and a carbon as a result of end-to-end overlap between the 1s orbital occupied by an unpaired valence electron from each hydrogen and an sp2 orbital from a carbon.  A single bond is always a sigma bond:

 

single bond = one σ bond

 

The third valence p orbital on each carbon that was not involved in hybridization will be perpendicular to the plane of the three sp2 orbitals.  Side-to-side overlap between the parallel p orbitals, each of which is occupied by carbon’s fourth unpaired valence electron that was not used in sigma bonding, results in a second type of bond between the carbons known as a pi (π) bond:

 

 

Therefore, a double bond is comprised of one sigma and one pi bond:

 

double bond = one σ bond + one π bond 

 

Consider the Lewis structure of C2H2:

 

 

Each carbon atom in the molecule has two sp hybrid orbitals (AB2), each of which is occupied by an unpaired valence electron, that are 180° away from each other on opposite sides of the carbon.  A sigma bond will be formed between the carbon atoms as a result of end-to-end overlap between two sp orbitals, one from each carbon.  In addition, sigma bonds will be formed between each hydrogen and a carbon as a result of end-to-end overlap between the 1s orbital occupied by an unpaired valence electron from each hydrogen and an sp orbital from a carbon. 

 

The other two valence p orbitals on each carbon that were not involved in hybridization will be perpendicular to each other and also to the line of the two sp orbitals.  Side-to-side overlap between two pairs of parallel p orbitals, each of which is occupied by one of carbon’s other unpaired valence electrons that was not used in sigma bonding, results in two pi bonds between the carbons:

 

 

Therefore, a triple bond is comprised of one sigma and two pi bonds:

 

triple bond = one σ bond + two π bonds 

 

Sample Exercise 11G:

 

Draw the Lewis structure for (NH2)2CO and determine the number of sigma and pi bonds in the molecule.

 

Solution:

 

6 single + 1 double = 6 sigma + 1 (1 sigma + 1 pi) = 7 sigma bonds + 1 pi bond

 

 

 

Section 11-5:  Molecular Orbital (MO) Theory


The Lewis structure shown below suggests that O2 is diamagnetic because all electrons are paired:

 

 

However, we know from experiment that O2 is, in fact, paramagnetic due to unpaired electrons.  Molecular orbital (MO) theory provides an explanation for the paramagnetism of O2.  MO theory suggests that the entire pool of electrons in a molecule occupy a series of molecular orbitals rather than being assigned to orbitals on individual atoms. 

 

For homonuclear (same element) diatomic molecules and ions from the second row of the periodic table, the valence electrons will be assigned to molecular orbitals as follows:

 

A. When Li, Be, B, C, or N is bonded, the order is:     σ2s      σ2s*     π2p     σ2p     π2p*     σ2p*   

 

B. When O, F, or Ne is bonded, the order is:     σ2s      σ2s*     σ2p     π2p     π2p*     σ2p*   

 

Electrons that occupy bonding molecular orbitals (no asterisk) will strenthen the bond between the atoms, whereas electrons that occupy antibonding molecular orbitals (*) will weaken the bond between the atoms.  Sigma (σ) molecular orbitals occur singly and can be occupied by a maximum of two electrons.  Pi (π) molecular orbitals occur in pairs, with each orbital occupied by a maximum of two electrons.  Therefore, each set of two pi orbitals can be occupied by a maximum of four electrons. 

 

When molecular orbital diagrams are written, two electrons that occupy the same orbital must have opposite spins.  Two electrons in a set of pi orbitals must occupy separate orbitals and also must have parallel spins.  We can see from the molecular orbital diagram below that O2 (6 + 6 = 12 valence electrons) will have two unpaired electrons and, therefore, be paramagnetic:

 

 

σ2s

σ2s*

σ2p

π2p

π2p

π2p*

π2p*

σ2p*

 

Bond order in MO theory is analagous to the number of covalent bonds in a Lewis structure.  We can determine the bond order for O2 as follows:

 

bond order = 0.5(number of bonding electrons – number of antibonding (*) electrons) = 0.5(8–4) = 2

 

A bond order can be a fraction, and a higher bond order will result in a shorter bond length.  For example, O2+ has one less antibonding π2p* electron than O2, which gives O2+ a bond order of 0.5(8-3) = 2.5.  Since O2+ has a higher bond order than O2, the bond length of O2+ will be shorter than the bond length of O2.

 

Sample Exercise 11H:

 

(a) Write the molecular orbital diagram for N2 and determine the bond order.  Also state whether N2 is diamagnetic or paramagnetic.

(b) Is the bond length of N2- shorter or longer than the bond length of N2?  Explain.

 

Solution:

 

(a) Recall that the π2p orbitals are filled before the σ2p orbital when Li, Be, B, C, or N is bonded.  An N2 molecule has 5 + 5 = 10 valence electrons:

 

 

 

 

σ2s

σ2s*

π2p

π2p

σ2p

π2p*

π2p*

σ2p*

 

bond order = 0.5(8-2) = 3

no unpaired electrons = diamagnetic

 

(b) N2- has one more antibonding π2p* electron than N2, which gives N2- a bond order of 0.5(8-3) = 2.5.  Since N2- has a lower bond order than N2, N2- has a longer bond length than N2.

 

 

If the bond order is determined to be zero, a molecule will not be stable.  For example, Ne2 (8 + 8 = 16 valence electrons) will not be stable:

 

 

σ2s

σ2s*

σ2p

π2p

π2p

π2p*

π2p*

σ2p*

 

bond order = 0.5(8-8) = 0

 

However, if the molecule is ionized, the resulting ion may be stable.  For example, Ne2+ has one less antibonding σ2p* electron than Ne2, which gives Ne2+ a bond order of 0.5(8-7) = 0.5.  Since the bond order is greater than zero, Ne2+ will be stable.

 

 

 

Chapter 11 Practice Exercises and Review Quizzes:

 

11-1) Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the bond angle for each of the following molecules and ions:

 

(a) PH3

(b) NH2-

(c) NO3-

(d) O3

(e) NH4+

(f) NO2+

Click for Solution

 

11-1)

(a)

Lewis structure:  

         

AB3E1 = trigonal pyramidal 

 

3-D sketch:

 

 

(b)

Lewis structure:     

     

AB2E2 = bent

 

3-D sketch:

 

 

(c)

Lewis structure:

 

AB3 = trigonal planar

 

3-D sketch:

 

 

(d)

Lewis structure:  

         

AB2E1 = bent

 

3-D sketch:

 

 

(e)

Lewis structure:

           

AB4 = tetrahedral

 

3-D sketch:

 

 

(f)

Lewis structure:     

     

AB2 = linear

 

3-D sketch:

 

 

 

 

11-2) Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, indicate the bond angle, and state whether each of the following molecules is polar or nonpolar:

 

(a) SiH4

(b) HOBr

(c) CS2

(d) NI3

(e) ClNO

(f) BF3

Click for Solution

 

11-2)

(a)

Lewis structure:   

        

AB4 = tetrahedral

 

3-D sketch:

 

 

nonpolar molecule

 

(b)

Lewis structure:    

      

AB2E2 = bent

 

3-D sketch:

 

 polar molecule

 

(c)

Lewis structure:     

     

AB2 = linear

 

3-D sketch: 

 

 

nonpolar molecule

 

(d)

Lewis structure:    

      

AB3E1 = trigonal pyramidal 

 

3-D sketch:

 

polar molecule

 

(e)

Lewis structure:    

      

AB2E1 = bent

 

3-D sketch:

 

 

polar molecule

 

(f)

Lewis structure:    

       

AB3 = trigonal planar

 

3-D sketch:

 

 

 

nonpolar molecule

 

 

11-3) Draw the Lewis structure, name the molecular geometry (shape), draw a three-dimensional sketch, and indicate the ideal bond angle(s) for each of the following molecules and ions.  Also state whether the neutral molecules are polar or nonpolar.

 

(a) AsCl5

(b) IF3

(c) ICl6+

(d) RnF2

(e) SeCl4

(f) SF5-

(g) XeF4

Click for Solution

11-3)

(a)

Lewis structure:

AB5 = trigonal bipyramidal, 3-D sketch:

nonpolar molecule

 

(b)

Lewis structure:

AB3E2 = T-shaped, 3-D sketch:

polar molecule

 

(c)

Lewis structure:

AB6 = octahedral, 3-D sketch:

 

(d)

Lewis structure:

AB2E3 = linear, 3-D sketch:

nonpolar molecule

 

(e)

Lewis structure:

AB4E1 = seesaw, 3-D sketch:

polar molecule

 

(f)

Lewis structure:

AB5E1 = square pyramidal, 3-D sketch:

 

(g)

Lewis structure:

AB4E2 = square planar, 3-D sketch:

nonpolar molecule

 

 

 

11-4) Draw the Lewis structure and indicate the center atom hybridization for each of the following molecules and ions:

 

(a) H2CO

(b) PH2-

(c) OCS

Click for Solution

11-4)

(a)

AB3 = sp2

 

(b)

AB2E2 = sp3

 

(c)

AB2 = sp

 

 

 

11-5) Draw the Lewis structure for CH3CN and determine the number of sigma and pi bonds in the molecule.

Click for Solution

11-5)

4 single + 1 triple = 4 sigma + 1(1 sigma + 2 pi) = 5 sigma bonds + 2 pi bonds

 

 

 

11-6)

(a) Write the molecular orbital diagram for B2 and determine the bond order.  Also state whether B2 is diamagnetic or paramagnetic.

(b) Is the bond length of B2+ shorter or longer than the bond length of B2?  Explain.

Click for Solution

11-6) (a) A B2 molecule has 3 + 3 = 6 valence electrons:

 

 

 

 

σ2s

σ2s*

π2p

π2p

σ2p

π2p*

π2p*

σ2p*

 

bond order = 0.5(4-2) = 1

has unpaired electrons = paramagnetic

 

(b) B2+ has one less bonding π2p electron than B2, which gives B2+ a bond order of 0.5(3-2) = 0.5.  Since B2+ has a lower bond order than B2, B2+ has a longer bond length than B2.

 

Click for Review Quiz 1

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