Molecular Geometry Bonding Theories

Molecular Geometry Bonding Theories
Chapter 9 Molecular Geometry Bonding Theories

Molecular Shape A bond angle is the angle defined by lines joining the centers of two atoms to a third atom to which they are covalently bonded The molecular geometry or shape is defined by the lowest energy arrangement of its atoms in three-dimensional space.

VSEPR Valence-Shell Electron-Pair Repulsion Theory The geometric arrangement of atoms bonded to a given atom is determined principally by minimizing electron pair repulsions of bonding and non-bonding electrons.

Central Atoms without Lone Pairs
Steric number (SN) is the number of volumes of space occupied by electrons surrounding a central atom

Geometric Forms

Predicting a VSEPR Structure
Draw Lewis structure. Determine the steric number of the central atom. Use the SN to determine the geometry around the central atom. The name for molecular structure is determined by the number of volumes of space occupied by bonding electrons.

Examples What is the molecular geometry of BF3?
What is the molecular geometry of CH4

Examples F B F F What is the molecular geometry of BF3?
What is the molecular geometry of CH4 Lewis Structure (exception to Law of Octaves) F B F F

Examples F B F F What is the molecular geometry of BF3?
What is the molecular geometry of CH4 Lewis Structure (exception to Law of Octaves) F Bond Angles = 120° Trigonal Planar B F F

Examples F B F F C What is the molecular geometry of BF3?
What is the molecular geometry of CH4 Lewis Structure (exception to Law of Octaves) F Bond Angles = 120° Trigonal Planar B F F H C H H H

Examples F B F F C What is the molecular geometry of BF3?
What is the molecular geometry of CH4 Lewis Structure (exception to Law of Octaves) F Bond Angles = 120° Trigonal Planar B F F H Bond Angles = 120° Tetrahedral C H H H

Central Atoms with Lone Pairs
Electron-pair geometry describes the arrangement of atoms and lone pairs of electrons about a central atom. The electron-pair geometry will always be one of the five geometries presented previously. The molecular geometry in these molecules describes the shape of the atoms present (it excludes the lone pairs).

Lone Pairs Lone pairs of electrons occupy more space around a central atom than do bonding electrons. Lone pair-lone pair repulsion is the largest. Lone pair-bonding pair repulsion is the next largest. Bonding pair-bonding pair repulsion is the smallest. In structures with lone pairs on the central atom, the bond angles are a little smaller than predicted based on the electron-pair geometry.

SN = 3, Electron-pair Geometry = Trigonal Planar
No. of Bonded Atoms No. of Lone Pairs Molecular Geometry Bond Angles 3 Trigonal Planar 120o 2 1 Bent <120o Like 119.6o

Non-bonding Electrons & Shape
Angles = 120 Trigonal Planar Angles < 120 Bent

SN = 4, Electron-pair Geometry = Tetrahedral
No. of Bonded Atoms No. of Lone Pairs Molecular Geometry Bond Angles 4 Tetrahedral 109.5o 3 1 Trigonal Pyramidal <109.5o 2 Bent

Tetrahedral Trigonal Pyramid v-shape

SN = 5, Electron-pair Geometry = Trigonal Bipyramidal
No. of Bonded Atoms No. of Lone Pairs Molecular Geometry Bond Angles 5 Trigonal Bipyramidal 120o & 90o 4 1 Seesaw <120o & 90o 3 2 T-shaped Linear 180o The lone pairs of electrons are always found in the trigonal planar part of the structure to minimize repulsion.

SN = 6, Electron-pair Geometry = Octahedral
No. of Bonded Atoms No. of Lone Pairs Molecular Geometry Bond Angles 6 Octahedral 90o 5 1 Square Pyramidal <90o 4 2 Square Planar 3 Although these arrangements are possible, we will not encounter any molecules with these arrangements.

Polar Bonds and Polar Molecules
Two covalently bonded atoms with different electronegativities have partial electric charges of opposite sign creating a bond dipole. A molecule is called a polar molecule when it has polar bonds and a shape where the bond dipoles don’t offset each other. 21

Examples 22

Measuring Polarity The permanent dipole moment () is a measured value that defines the extent of separation of positive and negative charge centers in a covalently bonded molecule. 23

Valence Bond Theory Hybridization is the mixing of atomic orbitals to generate new sets of orbitals that are then available to overlap and form covalent bonds with other atoms. A hybrid atomic orbital is one of a set of equivalent orbitals about an atom created when specific atomic orbitals are mixed. 24

Valence-Bond Theory Valence-bond theory assumes that covalent bonds form when orbitals on different atoms overlap or occupy the same region of space. A sigma () bond is a covalent bond in which the highest electron density lies between the two atoms along the bond axis connecting them. 25

Atomic Orbitals and Bonds
A tetrahedral molecule requires that four orbitals of the central atom must overlap with an orbital of an outer atom to form a bond. The central atom would use its s orbital and its three p orbitals, but these orbitals would not yield the 109° bond angles observed in the tetrahedral molecule. 26

Hybrid Orbitals You may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules?

Hybrid Orbitals You may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules? This brings us to the concept of hybrid orbitals, combinations of atomic orbitals, or molecular orbitals (from wave equations of electrons in molecules)

Hybrid Orbitals You may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules?

Hybrid Orbitals You may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules? This brings us to the concept of hybrid orbitals, combinations of atomic orbitals, or molecular orbitals (from wave equations of electrons in molecules)

Hybrid Orbitals Hybridization is a concept you might be familiar with. For example a grapefruit is a hybrid of what two fruits?

Hybrid Orbitals Hybridization is a concept you might be familiar with. For example a grapefruit is a hybrid of what two fruits? Lemon and orange

Hybrid Orbitals How about a nectarine?

Hybrid Orbitals How about a nectarine? Plumb and a peach.

Hybrid Orbitals How about a nectarine? Plumb and a peach.
Broccoaflower? Broccoli and cauliflower

Broccoaflower? Broccoli and cauliflower And a Cocapoo?

Broccoaflower? Broccoli and cauliflower And a Cocapoo? Cocker spaniel and poodle

Hybrid Orbitals On to Chemistry! How about an s-orbital and a p-orbital? Yes, sp orbital.

Hybrid Orbitals On to Chemistry! How about an s-orbital and a p-orbital? Yes, sp orbital. How about one s-orbital and two p-orbitals?

Hybrid Orbitals On to Chemistry! How about an s-orbital and a p-orbital? Yes, sp orbital. How about one s-orbital and two p-orbitals? Yes an sp2 orbital.

Examples 42

Tetrahedral Geometry: sp3 Hybrid Orbitals
A tetrahedral orientation of valence electrons is achieved by forming four sp3 hybrid orbitals form one s and three p atomic orbitals.

Other sp3 Hybrid Examples

sp2 Hybridization In a covalent pi () bond, electron density is greatest above and below the bonding axis.

sp Hybridization Pi bonds will not exist between two atoms unless a sigma bond forms first.

The Bonding in Carbon Dioxide
The carbon atom is sp hybridized and these orbitals form the two sigma bonds. The  bonds are rotated 90° from one another.

d2sp3 Hybridization

dsp3 Hybridization

Delocalization of Electrons
The electrons in the  system with alternating single and double bonds can be delocalized over several atoms or even an entire molecule. Localized Delocalized

Hybrid Orbital Notation
In order to construct hybrid orbital notation, we need to separate the central atom from the surrounding electrons, usually the central atom is the largest, the most electronegative, or the one that there is one of.

Hybrid Orbital Notation
In order to construct hybrid orbital notation, we need to separate the central atom from the surrounding electrons, usually the central atom is the largest, the most electronegative, or the one that there is one of. When constructing a hybrid orbital diagram, all of the valence electrons of the central atom are used and only the single electrons of the atoms attached to the central atom are use.

Hybrid Orbital Example
Suppose we want to make a diagram of SF6 First we separate the central atom from the other atoms. The central atom is A and the other atoms are called X’s A X’s SF6

Suppose we want to make a diagram of SF6 First we separate the central atom from the other atoms. The central atom is A and the other atoms are called X’s A X’s SF6 Then we generate a set of degenerate hybrid orbitals to house the valence electrons

Suppose we want to make a diagram of SF6 First we separate the central atom from the other atoms. The central atom is A and the other atoms are called X’s A X’s SF6 Then we generate a set of degenerate hybrid orbitals to house the valence electrons F F F F F F F Insert single electrons into the degenerate hybrid orbitals

Structure of Sulfurhexafluoride
Shape- Octahedral Hybrid Orbitals- sp3d2 Bond angles- 90° Polarity- Nonpolar F F s F F F

Noble Gas Compounds Critics of the hybrid orbital theory argued that the hybrid orbital theory suggests that compounds of Noble gases should exist or be made. In 1962 Neil Bartlett created a compound of xenon, platinum and fluorine. Today there are now several hundred Noble gas compounds known. University of British Columbia

Noble Gas Compounds Practice XeO2 KrF4 XeO2F22+

Problems with Bonding Theories
Lewis structure and valence bond theory help us understand the bonding capacities of elements. VSEPR and valence bond theories account for the observed molecular geometries. None of these models enables us to explain why O2 is attracted to a magnetic field while N2 is repelled slightly.

Molecular Orbital (MO) Theory
The wave functions of atomic orbitals of atoms are combined to create molecular orbitals (MOs) in molecules. Each MO is associated with an entire molecule, not just a single atom. MOs are spread out, or delocalized over all the atoms in a molecule.

Types of MOs Electrons in bonding orbitals serve to hold atoms together in molecules by increasing the electron density between nuclear centers. Electrons in antibonding orbitals in a molecule destabilize the molecule because they do not increase the electron density between nuclear centers.

MO Guidelines The total number of MO formed equals the number of atomic orbitals used in the mixing process. Orbitals with similar energy and shape mix more effectively than do those that are different. Orbitals of different principal quantum numbers have different sizes and energies resulting in less effective mixing. A MO can accommodate two electrons with opposite spin. Electrons are placed in MO diagrams according to Hund’s rule.

MOs for H2 The two 1s orbitals may be added or subtracted to yield two MOs.

Bond Order The bond order is zero in He2 and the molecule is not
Bond Order = 1/2 (# bonding electrons - # antibonding electrons) The bond order is zero in He2 and the molecule is not stable.

BOND ORDER (s1s)2(s1s*)2 Useful concept:
The net number of bonds existing after the cancellation of bonds by antibonds. the electronic configuration is…. In He2 (s1s)2(s1s*)2 the two bonding electrons were cancelled out by the two antibonding electrons. There is no BOND! BOND ORDER = 0!!!!! So……..

Bond Types A sigma, , bond is a covalent bond in which the highest electron density lies along the bond axis. A pi, , bond is formed by the mixing of atomic orbitals that are not oriented along the bonding axis in a molecule.

= BOND ORDER A measure of bond strength and molecular stability.
If # of bonding electrons > # of antibonding electrons the molecule is predicted to be stable Bond order =

BOND ORDER A measure of bond strength and molecular stability. If # of bonding electrons > # of antibonding electrons the molecule is predicted to be stable Bond order { } = # of bonding electrons(nb) # of antibonding electrons (na) 1/2 – = 1/2 (nb - na) A high bond order indicates high bond energy and short bond length. Consider H2+,H2,He2+,He2……….

PARAMAGNETIC SUBATANCES
Substances containing unpaired electrons are attracted to a magnetic field and are called paramagnetic. Hund’s rule states that electrons placed in degenerate orbitals are placed with parallel spins (same direction). Spinning charges create a magnetic field. The number of single electrons the stronger the magnetic field a compound will exert. In the laboratory the number of unpaired electrons can be measured using a Gouy Balance. The strength of the magnetic field of a substance can be determined by adding weights to the balance pan to counter the force of the magnetic field. The mass of the added weights is directly proportional to the number of single electrons.

Gouy Balance

First row diatomic molecules and ions
s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 H2+ He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- 225 106 He2+ He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- 225 106 He2+ Para- 251 108 He2 E

s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- 225 106 He2+ Para- 251 108 He2 — E

HOMONUCLEAR DIATOMICS
Now look at second period….. First is Li2 Li : 1s22s1 Both the 1s and 2s overlap to produce s bonding and anti-bonding orbitals. This is the energy level diagram…..

DI-LITHIUM s2s* Li2 2s 2s E s2s Put the electrons in……. s1s* 1s 1s s1s

ELECTRONS FOR DILITHIUM Li2
s2s* Li2 2s 2s E s2s Put the electrons in the MO’s s1s* 1s 1s s1s

Electron configuration for DILITHIUM
s2s* Li2 (s1s)2(s1s*)2(s2s)2 2s 2s E s2s Bond Order ?????? What do we need? 1s 1s s1s

Electron configuration for DILITHIUM
s2s* Li2 (s1s)2(s1s*)2(s2s)2 2s 2s nb = 4 na = 2 E s2s (nb - na) Bond Order = 1/2 = 1/2(4 - 2) =1 A single bond. 1s 1s s1s

Li2 Electron configuration for DILITHIUM s2s* (s1s)2(s1s*)2(s2s)2 2s
nb = 4 na = 2 E s2s Note: The s1s and s1s* orbitals cancel! 1s 1s So……. s1s

Li2 Electron configuration for DILITHIUM s2s* (s1s)2(s1s*)2(s2s)2 2s
The s1s and s1s* orbitals can be ignored when both are FILLED! E s2s 1s 1s We often omit the inner shell! s1s

Only valence orbitals contribute to molecular bonding
Li2 (s2s)2 Li Li2 Li s2s* E 2s 2s s2s (s1s)2(s1s*)2 assumed The Li2 configuration…. But can be included.

Li2 (s2s)2 Li Li2 Li s2s* E 2s 2s s2s The complete configuration is: (s1s)2(s1s*)2 (s2s)2 What is the bond order????

Li2 (s2s)2 Li Li2 Li s2s* Ignoring the filled (s1s)2(s1s*)2 E 2s 2s nb = 2 na = 0 s2s Bond Order = 1/2(nb - na) = 1/2(2 - 0) =1 A single bond. Is the molecule stable or unstable???

Li2 (s2s)2 Li Li2 Li s2s* Ignoring the filled (s1s)2(s1s*)2 E 2s 2s nb = 2 na = 0 s2s Bond Order = 1/2(nb - na) = 1/2(2 - 0) =1 A single bond. STABLE! Now Be2…….

Be2 VALENCE ELECTRONS FOR DIBERYLLIUM Be Be2 Be s2s* E 2s 2s s2s
Put the electrons in the MO’s...

Be2 Electron configuration for DIBERYLLIUM Be Be2 Be s2s* E 2s 2s s2s
Bond order? Configuration: (s2s)2(s2s*)2

Electron configuration for DIBERYLLIUM
(s2s)2(s2s*)2 s2s* nb = 2 E 2s 2s na = 2 s2s Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0 No bond!!! We conclude???????

Electron configuration for DIBERYLLIUM
(s2s)2(s2s*)2 s2s* nb = 2 E 2s 2s na = 2 s2s Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0 Now B2... No bond!!! The molecule is not stable!

B2 The Boron atomic configuration is 1s22s22p1 So we expect B to use 2p orbitals to form molecular orbitals. How do we do that??? Combine them by ??? Addition and subtraction…. This is what they look like…….

- - + + s molecular orbitals s2p* antibonding + - - s2p bonding
SUBTRACT s2p* antibonding - - + + ADD + - - s2p bonding PHH picture

s-molecular orbitals - + PHH picture p- MO’s

The p molecular orbitals.
SUBTRACT - + - + p2p* antibonding + ADD PHH... - p2p bonding

The p molecular orbitals.
+ - ENERGY LEVELS?

ENERGY LEVEL DIAGRAM When we form MO’s we get orbitals of different energies Example the s2s s2s* From 2s E 2s s2s* s2s The p- MO’s……..

The M.O.’s formed by p orbitals
s2p* p2p* p2p s2p The p do not split as much because of weaker overlap. Combine this with the s-orbitals…..

Modified Molecular Orbital Diagram
It should be noted that both sigma 1s and sigma 2p orbitals have similar shape and energy, thus mixing occurs lowering the sigma 1s orbital and elevating the sigma 2p orbital above the bonding pi molecular orbitals. This is illustrated on the next slide.

Expected orbital splitting:
p2p p2p* The p do not split as much because of weaker overlap. 2p But the s and p along the internuclear axis interact……. E 2s s2s* s2s This pushes the s2p up..

MODIFIED ENERGY LEVEL DIAGRAM
2s s2s* s2s 2p s2p* s2p p2p p2p* Shows additional s interaction. E Notice that the s2p and p2p have changed places!!!! Now look at B2...

Electron configuration for B2
B is [He] 2s22p1 s2p* p2p* s2p 2p 2p p2p E Place electrons from 2s into s2s and s2s* s2s* 2s 2s s2s

Electron configuration for B2:
s2p* p2p* s2p 2p 2p p2p E Place electrons from 2p into p2p and p2p s2s* 2s 2s Remember HUND’s RULE s2s

ELECTRONS ARE UNPAIRED 2p s2p* s2p p2p p2p* Abbreviated configuration (s2s)2(s2s*)2(p2p)2 E Complete configuration 2s s2s* s2s (s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)2 Bond order????

s2p* (s2s)2(s2s*)2(p2p)2 p2p* na = 2 nb = 4 s2p 2p 2p p2p E Bond order 1/2(nb - na) s2s* = 1/2(4 - 2) =1 2s 2s Molecule is predicted to be stable and paramagnetic. s2s

HOMONUCLEAR DIATOMICS
Li2 B2 C2 N2 O2 F2 Which energy level diagram??? The one with s and p interaction or the one without? We find……….

SECOND ROW DIATOMICS Li2 B2 C2 N2 O2 F2 2s s2s* s2s 2p s2p* s2p p2p
USE USE E 2s s2s* s2s s2p* 2p s2p p2p p2p*

Second row diatomic molecules
s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 C2 N2 O2 F2 E

s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 N2 O2 F2 E

s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 O2 F2 E

s2p* p2p* s2p p2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 F2 E

NOTE SWITCH OF LABELS s2p* p2p* p2p s2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 Para- 2 495 121 F2 E

NOTE SWITCH OF LABELS s2p* p2p* p2p s2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 Para- 2 495 121 F2 Dia- 1 154 143 E

QUESTION 1 B2+ 2 C2+ 3 N2+ 4 O2+ 5 F2+ ANSWER…...
The molecule X+2 that has the molecular orbital configuration (s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)4 (s2p)2 (p2p*)1 is 1 B2+ 2 C2+ 3 N2+ 4 O2+ 5 F2+ ANSWER…...

QUESTION How? 1 B2+ The molecule X2 has 16 electrons 2 C2+
The molecule X+2 that has the molecular orbital configuration (s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)4 (s2p)2 (p2p*)1 is How? 1 B2+ The molecule X2 has 16 electrons 2 C2+ Therefore Z = 8 3 N2+ Therefore O2+ 4 O2+ PARAMAGNETIC? YES! 5 F2+ 2.5 BOND ORDER ANSWER…... EXAMPLE…..

Question Example: Give the electron configuration and bond order for O2, O2+ , O2- & O22-. Place them in order of bond strength and describe their magnetic properties. Step 1:Determine the number of valence electrons in each: O2 : = 12 O2+ : = 11 O2– : = 13 O22- : = 14

Step 2: Determine the valence electrons configurations:
s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : O2+ : O2– : O22-:

O2 : (s2s)2(s2s*)2 (s2p)2(p2p)4 (p2p*)2 O2+ : O2– : s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

O2 : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)2 O2+ : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)1 O2– : s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

O2 : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)2 O2+ : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)1 O2– : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)3 s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

Step 2: What about O22- ???: s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)2 O2+ : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)1 O2– : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)3 O22- : (s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)4

Step 3: Determine the bond orders of each species:
p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : B.O. = (8 - 4)/2 = 2 O2+ : B.O. = (8 - 3)/2 = 2.5 O2– : B.O. = (8 - 5)/2 = 1.5 O22- : B.O. = (8 - 6)/2 = 1

O2+ >O2 >O2– > O22- BOND ENERGY ORDER
Step 4: Use the bond orders to place the species in order of bond strength: s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E O2 : B.O. = 2 BOND ENERGY ORDER O2+ : B.O. = 2.5 O2+ >O2 >O2– > O22- O2– : B.O. = 1.5 HETERONUCLEAR DIATOMICS…... O22- : B.O. = 1

SECOND ROW DIATOMICS Can be used for s2p* E s2p* p2p* p2p* s2p p2p 2p
heteronuclear diatomics 2s s2s* s2s 2s s2s* s2s Which????

NITRIC OXIDE (NO) Number of valence electrons: 5 + 6 = 11 s2p*
p2p* s2p p2p s2s* s2s E Number of valence electrons: = 11 use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation Put the electrons in…..

s2p* NITRIC OXIDE (NO) Number of valence electrons: 5 + 6 = 11
p2p* s2p p2p s2s* s2s E NITRIC OXIDE (NO) Number of valence electrons: = 11 use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation

Number of valence electrons: 5 + 6 = 11
s2p* p2p* s2p p2p s2s* s2s E NITRIC OXIDE (NO) Number of valence electrons: = 11 use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation Bond order

NITRIC OXIDE (NO) s2p Number of valence electrons: 5 + 6 = 11 p2p s2s*
use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation s2p* Bond order p2p* Molecule is stable and paramagnetic. Experimental data agrees. NO+ and CN-

Localized/MO Combination
Recalling the molecular orbital diagram for carbon we notice that the sigma and sigma antibonding fill before the pi bonds. Since this does not contribute to the bond order it makes sense that we can use localized bonding (line) to describe a sigma bond and use molecular orbitals on just the pi bonding electrons.

Consider ozone, O3, we can use lines to connect the three oxygen atoms together and do a molecular orbital diagram on just the unhybridized p-orbitals. Then we arrange the three combinations of p-orbitals in increasing potential energy order by counting the nodes. Potential energy is inversely proportional to the number of nodes.

A review of adding orbitals Constructive Interference Destructive Interference + + + + - - + + + - - - Makes anti-bonding orbitals Makes bonging orbitals

P=2 , which means two pi bonding electrons Ozone Molecular Orbital Diagram two nodes Pi antibonding one node non bonding zero Node Pi bonding Note: Odd electron systems will always have a nonbonding molecular orbital, which is equal in potential energy of the p-orbital they were derived from.

P=2 , which means two pi bonding electrons Ozone Molecular Orbital Diagram two node Pi antibonding one node non bonding zero Node Pi bonding Now fill Pi MO with electrons

P=2 , which means two pi bonding electrons Ozone Molecular Orbital Diagram one node Pi antibonding one node non bonding zero Node Pi bonding Now fill Pi MO with electrons

P=2 , which means two pi bonding electrons Ozone Molecular Orbital Diagram one node Pi antibonding ( π*) one node non bonding zero Node Pi bonding ( π) Now fill Pi MO with electrons

P=2 , which means two pi bonding electrons Ozone Molecular Orbital Diagram one node Pi antibonding ( π*) one node non bonding zero Node Pi bonding ( π) The MO diagram shows one pi bond. And the diagram above shows two sigma bonds. Now fill Pi MO with electrons

Each oxygen atom is sp2 hybridized, thus one electron in each unhybridized p-orbital and two electrons in the central oxygen unhybridized p-orbital. We will add 4 electrons to the MO below. O O O Ozone Molecular Orbital Diagram one node Pi antibonding ( π*) one node non bonding zero Node Now fill Pi MO with 4 electrons Pi bonding ( π) The MO diagram shows one pi bond over 3 atoms. And the diagram above shows two sigma bonds. Therefore the B.O. = 1+1/2 =1.5 Stable and diamagnetic.

Comparison of Theories
MO theory may provide the most complete picture of covalent bonding, but it is also the most difficult to apply to large molecules and it does not account for molecular shape.

The End

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Click to launch animation PC | Mac This animated tutorial offers a patient explanation of molecular orbital theory, an alternative to the bonding theory depicted by Lewis dot structures. Includes Practice Exercises.

A) H2C=C=CH2 B) H2C=C=C=CH2 C) Neither
Ethylene, which has the molecular formula C2H4, is a rigid molecule in which all 6 atoms lie in a plane. Which of the following molecules also has a rigid planar structure? A) H2C=C=CH2 B) H2C=C=C=CH2 C) Neither © 2008 W. W. Norton & Company Inc. All rights reserved. Planar Hydrocarbons

Consider the following arguments for each answer and vote again:
A combination of 3 carbons and 4 hydrogens can form the rigid planar molecule H2C=C=CH2. The orientations of the π bonds in H2C=C=C=CH2 alternate in such a way as to create a planar structure. The hybridization of the atomic orbitals on the carbons prevents the retention of a planar structure in molecules longer than C2H4. Answer: B Planar Hydrocarbons

The majority of the bonds in NO3- are single bonds, so the bond order is 1. The N-O bond is twice as likely to be a single bond as it is to be a double bond, so the bond order should be 11/3. The bond order is dictated by the strongest bond, which in NO3- is a double bond. Answer: B Bond Order of Nitrate

NO+ is isoelectronic with N2, which has no unpaired electrons and hence is not paramagnetic. NO has no electrical charge and thus cannot be paramagnetic. By pairing an additional electron with the one unpaired electron in NO, a diamagnetic anion, NO-, is formed. Answer: A Bond Order of Nitrate

Molecular Geometry of SF2, SF3-, and SF4
According to Valence Shell Electron Pair Repulsion (VSEPR) theory, 4 objects around a central atom will have the tetrahedral arrangement shown to the left with bond angles of ~109.5º. Which of the following compounds has a bond angle of ~109.5º? © 2008 W. W. Norton & Company Inc. All rights reserved. A) SF2 B) SF3- C) SF4 Molecular Geometry of SF2, SF3-, and SF4

Molecular Geometry of SF2, SF3-, and SF4
Please consider the following arguments for each answer and vote again: SF2 consists of a sulfur atom surrounded by 2 lone electron pairs and bonded to 2 fluorine atoms, therefore, it has an approximately tetrahedral bond angle. The tetrahedral VSEPR arrangement of SF3- is formed by a sulfur atom surrounded by 3 fluorine atoms and by the additional electron (from the negative charge). Sulfur tetrafluoride is the only molecule with a central atom (sulfur) surrounded by 4 additional atoms (4 fluorines) and so is the only molecule with a bond angle of ~109.5º. Answer: A Molecular Geometry of SF2, SF3-, and SF4

Bond Angles of BrF2 and ICl2
Which of the following is true of the bond angle (θ1) in BrF2+ compared to the bond angle (θ2) in ICl2-? A) θ1 = θ2 B) θ1 > θ2 C) θ1 < θ2 © 2008 W. W. Norton & Company Inc. All rights reserved. Bond Angles of BrF2 and ICl2

Bond Angles of BrF2 and ICl2
Please consider the following arguments for each answer and vote again: Both BrF2+ and ICl2- consist of a central halogen atom bonded to two halogen atoms, and therefore should have the same arrangement of atoms. ICl2- has 1 more lone pair of electrons than BrF2+, which forces the chlorine atoms closer together. ICl2-, with 3 lone pairs, is linear whereas BrF2+, with 2 lone pairs, is bent. Answer: C Bond Angles of BrF2 and ICl2

Reaction of Boron Trifluoride
Boron trifluoride (BF3), which has the structure shown to the left, is capable of reacting with an unknown compound to form a new compound without breaking any bonds. Which of the following could be the unknown compound? © 2008 W. W. Norton & Company Inc. All rights reserved. A) BF3 B) CH4 C) NH3 Reaction of Boron Trifluoride

Reaction of Boron Trifluoride
Please consider the following arguments for each answer and vote again: BF3 can dimerize to BF3-BF3 by forming a boron-boron single bond. By forming a boron-carbon bond, the carbon atom in CH4 will increase its steric number to 5, thus expanding its octet to compensate for boron's incomplete octet. The nitrogen lone electron pair can form a nitrogen-boron bond yielding BF3-NH3, isoelectronic with CH3-CH3. Answer: C Reaction of Boron Trifluoride

Dipole Moments of Dichloroethylene
Pictured to the left is the planar molecule ethylene, C2H4, which does not have a permanent electric dipole moment. If chlorine atoms were substituted for two hydrogen atoms, how many of the possible structures would also not possess a dipole moment? © 2008 W. W. Norton & Company Inc. All rights reserved. A) B) C) 2 Dipole Moments of Dichloroethylene

Dipole Moments of Dichloroethylene
Consider the following arguments for each answer and vote again: Chlorine atoms always draw electron density away from carbon atoms, so all possible structures will possess a dipole moment. Only if the chlorine atoms are diagonally opposite will the two carbon-chlorine dipole moments cancel each other. So long as the two chlorine atoms are on different carbon atoms, no permanent dipole moment will form. Answer: B Dipole Moments of Dichloroethylene

Molecular Geometry of XF42-
Please consider the following arguments for each answer and vote again: CF42- forms a structure in which the 4 fluorine atoms form a square plane with one negative charge on either side of the plane. With 2 lone electron pairs on the sulfur in SF42-, its steric number is 6. To maximize fluorine-fluorine distances, the 4 fluorine atoms in XeF42- will lie in a plane. Answer: B Molecular Geometry of XF42-

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Molecular Geometry Bonding Theories

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