Category Archives: Organic chemistry

center of symmetry
Molecular Symmetry

 

Molecular Symmetry

Any object is called as symmetrical if it has mirror symmetry, or ‘left-right’ symmetry i.e. it would look the same in a mirror.

For example : a cube , a matchbox, a circle 

Further it can be said ; a sphere is more symmetrical compare to a cube. Cube looks the same after rotation through any angle about the diameter while during the rotation of a cube, it looks similar only with certain angles like 90°, 180°, or 270° about an axis passing from the centers of any of its opposite faces, or by 120° or 240° about an axis passing from any of the opposite corners.

Similarly Molecules can also be classified as Symmetric or Asymmetric.

Symmetry Operations

An action on an object which leaves the object at same position after the action carried out. Such type of action is called as Symmetry operations.

All known molecules can classify in groups possess the same set of symmetry elements.

Such type of classification is helpful to assign the molecular properties without calculation and in the determination of polarity and degeneracy of molecular states.

It provides the systematic treatment of symmetry in chemical systems in a mathematical framework which is called as group theory.

Group theory is also helpful is some other investigations

  • Prediction of polarity and chirality of molecule.
  • In examination of bonding and visualizing molecular orbitals of molecules.
  • In prediction of polarization a molecule.
  • In investigation of vibrational motions of the molecule.

 

The simplest example of symmetry operations is water molecule. If we rotate the molecule by 180° about an axis which is passing through the central Oxygen atom it will look the same as before. Similarly reflection of molecule through both axis of molecule show same molecule.

There are five types of symmetry operations and five types of symmetry elements.

1. The identity (E)
This symmetry operation is consists of doing nothing. In other words; any object undergo this symmetry operation and every molecule consists of at least this symmetry operation. For example; bio molecules like DNA and bromo fluoro chloro methane consist of only this symmetry operation. ‘E’ notation used to represent identity operation which is coming from a German word ‘Einheit’stands for unity.

 

2. An n-fold axis of symmetry (Cn)

  • This symmetry operation involves the clockwise rotation of molecule through an angle of 2 π /n radian  or  360º/n where n is an integer. The notation used for n-fold of axis is Cn
  • For principal axis, the value of n will be highest. The rotation through 360°/n angle is equivalent to identity (E). 
  • For example, one twofold axis rotation of water (H2O) towards oxygen axis leaves molecule at same position, hence has C2axis of symmetry. Similarly ammonia (NH3) has one threefold axis, C3 and benzene (C6H6) molecule has one sixfold axis C6and six twofold axis (C2) of symmetry.

Linear diatomic molecules like hydrogen, hydrogen chloride have C∞ axis as the rotation on any angle remains the molecule the same.

 

3. Improper rotation (Sn)

Improper clockwise rotation through the angle of 2π/n radians is represented by notation ‘Sn’ and called as n-fold axis of symmetry which is a combination of two successive transformations. During improper rotation, the first rotation is through 360°/n and the second transformation is a reflection through a plane perpendicular to the axis of the rotation. Improper rotation is also known as alternating axis of symmetry or rotation-reflection axis. For example; methane (CH4) molecule has three S4axis of symmetry.

4. A plane of symmetry (σ)

There are some plane in molecule through which reflection leaves the molecule same. The vertical mirror plane is labelled as σv and one perpendicular to the axis is called a horizontal mirror plane is labelled as σh , while the vertical mirror plane which bisects the angle between two C2axes is known as a dihedral mirror plane, σd. For example, H2O molecule contains two mirror planes (a YZ Reflection (σyz) and a XZ Reflection (σxz)) which are mirror planes contain the principle axis and called as vertical mirror planes (σv).


5. Center of symmetry (i)
It is a symmetry operation through which the inversion leaves the molecule unchanged. For example, a sphere or a cube has a centre of inversion. Similarly molecules like benzene, ethane and SF6have a center of symmetry while water and ammonia molecule do not have any center of symmetry.

Overall the symmetry operations can be summarized as given below.

Inversion Center

The inversion operation is a symmetry operation which is carried out through a single point, this point is known as inversion center and notated by ‘ i’. This point is located at the center of the molecule and may or may not coincide with an atom in the molecule. 

When we are moving each atom in a molecule along a straight line through the inversion center to a point an equal distance from the inversion center and get same configuration, we say there is an inversion center in the molecule. It can be in such molecules which do not have any atom at center like benzene, ethane. 

Geometries like tetrahedral, triangles, pentagons don’t contain an inversion center. Hence a cube, a sphere contains a center of inversion but tetrahedron does not contain this symmetry operation. The molecule must be achiral for the presence of inversion center.

Some of the common examples of molecules contain center of inversion are as follow.

(a) Benzene molecule: Inversion center located at the center of molecule.

 

(b) 1,2-Dichloroethane: The staggered form of 1,2-Dichloroethane contains one inversion center at the center of molecule.

 

 

(c) trans-diaminedichlorodinitroplatinum complex: trans- form of some Coordination compounds like trans diaminedichlorodinitroplatinum complex contains inversion center.

Another example of coordination compound is hexacarbonylchromium complex [Cr(CO)6], where the inversion center located at the position of metal atom in complex.

(d) Ethane molecule: The staggered form of ethane contains inversion center while eclipsed form does not.

 

(e) Meso-tartaric acid: The anti-periplanar conformer of meso-tartaric acid has an inversion center.

(f) Dimer of D and L-Alanine: The dimer of two configurations of Alanine; D-alanine and L-alanine contains one inversion center.

(g) 18-Crown-6: An organic compound with the formula [C2H4O]6 named as 18-crown-6 (IUPAC name: 1,4,7,10,13,16-hexaoxacyclooctadecane) also contains inversion center located at center of molecule.

(h) Cyclohexane: The chair conformation of cyclohexane contains an inversion center while boat form does not.

Molecular Symmetry Examples

A molecule or an object may contains one or more than one symmetry elements, therefore molecules can be grouped together having same symmetry elements and classify according to their symmetry. Such type of groups of symmetry elements are known as point groups because there is at least one point in space which remains unchanged no matter which symmetry operation from the group is applied. 

For the labelling of symmetry groups, two systems of notation are given, known as the Schoenflies and Hermann-Mauguin (or International) systems. The Schoenflies notations are used to describe the symmetry of individual molecule. The molecular point groups with their example are listed below.

Point group  Explanation  Example 
C1 Contains only identity operation(E) as the C1 rotation is a rotation by 360o Bromochlorofloromethane (CFClBrH)
Ci  Contains the identity (E) and a center of inversion center (i). Anti-conformation of 1, 2-dichloro-1, 2-dibromoethane.
Cs Contains the identity E and plane of reflection σ. Hypochlorus acid (HOCl), Thionyl chloride (SOCl2).
Cn Have the identity and an n-fold axis of rotation. Hydrogen Peroxide (C2)
Cnv Have the identity, an n-fold axis of rotation, and n vertical mirror planes (σv). Water (C2v), Ammonia (C3v)
Cnh Have the identity, an n-fold axis of rotation, and σh (a horizontal reflection plane). Boric acid H3BO3 (C3h), trans-1,2-dichloroethane (C2h)
Dn Have the identity, an n-fold axis of rotation with n2-fold rotations about the axis which is perpendicular to the principal axis. Cyclohexane twist form (D2)
Dnh Contains the same symmetry elements as Dn with the addition of a horizontal mirror plane. Ethene (D2h), boron trifluoride (D3h), Xenon tetrafluoride (D4h).
Dnd Contains the same symmetry elements as Dwith the addition of n dihedral mirror planes. Ethane (D3d), Allene(D2d)
Sn Contains the identity and one Sn axis. CClBr=CClBr
Td Contains all the symmetry elements of a regular tetrahedron, including the identity, four C3 axis, three-C2 axis, six dihedral mirror planes, and three S4 axis. Methane (CH4)
T
Th
Same as Td but no planes of reflection.
Same as for T but contains a center of inversion .
 
Oh
O
The group of the regular octahedron.
Same as Oh but with no planes of reflection.
Sulphur hexafluoride (SF6)

 

Different point groups correspond to certain VSEPR geometry of molecule. Out of them some are as follow.

VSEPR Geometry of molecule  Point group 
Linear D∞h
Bent or V-shape  C2v 
Trigonal planar  D3h 
Trigonal pyramidal  C3v 
Trigonal bipyramidal  D5h 
Tetrahedral  Td 
Sawhorse or see-saw  C2v 
T-shape  C2v
Octahedral  Oh 
Square pyramidal C4v
Square planar  D4h 
Pentagonal bipyramidal  D5h

 

A molecule may contain more than one symmetry operation and show symmetrical nature. Some of the examples of symmetry operation on molecule with their point group are as given below.

(a) Benzene: The point group of benzene molecule is D6h with given symmetry operations.

  • Inversion center: i
  • The Proper Rotations: seven C2axis and one C3 and one C6 axis
  • The Improper Rotations: Sand S3axis
  • The Reflection Planes: one σh , three σvand three σd

(b) Ammonia: The point group of ammonia molecule is C3v with following symmetry operations.

  • The Proper Rotations: one C3axis
  • The Reflection Planes: three σplane

(c) Cyclohexane: The chair conformation of Cyclohexane has D3d point group with given symmetry operations;

  • Inversion center: i
  • The Proper Rotations: Three C2axis and one C3 axis
  • The Improper Rotations: S6axis
  • The Reflection Planes: Three σdplane

(d) Methane: The point group of methane is Td (tetrahedral) with C3 as principal axis and other symmetry operations are as follows;

  • The Proper Rotations: Three Caxis and Four C3axis
  • The Improper Rotations: Three S4axis
  • The Reflection Planes: Five σdplane

(e) 12-Crown-4: This has S6 point group with C3 and S6 axis with inversion center (i).

(f) Allene: The point group of methane is D2d with given symmetry operations.

  • The Proper Rotations: Three C2axis 
  • The Improper Rotations: One S4axis
  • The Reflection Planes: Two σplane

Molecular Symmetry and Group Theory

Group theory deals with symmetry groups which consists of elements and obey certain mathematical laws. Each point group is a set of symmetry operation or symmetry elements which are present in molecule and belongs to this point group. To obtain the complete group of a molecule, we have to include all the symmetry operation including identity ‘E’. A character table represents all the symmetry elements correspond to each point group. Hence we can make separate character table for each point group like C2v, C3v, D2h… etc.

For example, in the character table of C2v point group; all the symmetry elements has to written in first row and the symmetry species or Mulliken labels are listed in first column. These symmetry species specify different symmetries within one point group. For C2v, there are four symmetry species or Mulliken labels; A1, A2, B1, B2.
Remember

  • The symmetry species for one-dimensional representations: A or B
  • The symmetry species for two-dimensional representations: E
  • The symmetry species for three-dimensional representations: T

The best example of C2v point group is water which has oxygen as center atom. The px orbital of oxygen atom is perpendicular to the plane of water molecule, hence it is not symmetric with respect to the plane σv(yz). So this orbital is anti-symmetric with respect to the mirror plane and its sign get change when symmetry operations applied. On the other hand, the s orbital is symmetric with respect to mirror plane. The symmetric and anti-symmetric nature can be represents by using mathematical sign; +1 and -1; here +1 stands for symmetric and –1 stands for anti-symmetric which are the characters in character table.

Hence the symmetry operations for the px orbitals are as follow.

1. E: Symmetric hence character will be 1

2. C2:Anti-symmetric, hence character will be 1

3. σv(xz):Symmetric; character :1

4. σv(yz):Anti-symmetric, character: -1

Hence the character table for C2v point group.

C2v  E C2  σv(XZ)  σv(YZ)     
A1 1 1 1 z x2, y2,z2 
A2  -1  -1  Rz  xy 
B1  -1  -1  x, Ry  xz 
B2  -1  -1  y, Rx  yz 

 

Similarly character can be assigned for other symmetry species. The last two columns of character table make it easier to understand the symmetric nature. For example; x in second last column of Bsymmetry indicates that the x-axis has Bsymmetry in C2v point group and the Rx notation indicates the rotation around the x-axis. Similarly the character table for C3v point group will be

C3v  2C3  3σv    
A1 1 1 1 x2+y2, z2 
A2  -1  Iz  
-1  (x, y), (Ixy, Iz (x2-y2, xy), (xz, yz)

 

For doubly degenerate, the character for E will be 2 and for triply degenerate it will be 3, because in this case we have two and three orbitals respectively which are symmetric with respect to E. Some of the character tables with their point groups are as follow

a. Character table for Oh point group, for example Sulfur fluorine (SF6)

 

b. Character table Td point group, methane (CH4)

 

 

c. Character table for D3d point group, for example, staggered ethane

 

d. Character table for D6h point group, example Benzene (C6H6)

 

Symmetry Adapted Linear Combinations

In some molecules like water, ammonia, methane which have more than one symmetry equivalent atom, the combinations of the symmetry equivalent orbitals can transform according to a irreducible representations of the molecules point group which are refer as Symmetry Adapted Linear Combinations. For the formation of an n-dimensional representation a set of equivalent functions -f1, f2, …, fn- can be used. The representation can be expressed as a sum of irreducible representations with the use of calculation of characters for this representation and by the use of the great orthogonality theorem. The n-linear combinations of f1, …, fn which transform the irreducible representations is given by the projection operator which denoted as p^p^ Gi;

Here 

  • p^p^ = The operator which projects out of a set of equivalent functions the Gi Irreducible representation of the point group.
  • In n/g factor; n = dimension of the irreducible representation
  • g = the order of the group

The function fj can be chosen by any one of n which belongs to the equivalent set. For example, in the C3v character table; the 2C3^C3^ represents the class composed by C13^C31^ and C13^C31^operations. With Cclass, there are three different C3^C3^ operations would also perform separately on fj which produce different results. Let’s take an example of the O-H stretches along the ‘yz’ plane as molecular plane in water molecule; the formula can be apply to tabulate the characters of the irreducible representations and list the effect of O^O^R on one of the functions at the bottom of the table.

C2v  E C2(Z)  sv(XZ)  sv(YZ) 
A1 1 1 1
B1  -1  +1  -1
B2  -1  -1  +1 
OR(OH2 OHa  OHb OHb OHa 

 

After applying the projection operator for A1
p^p^ A1 (O-Ha) = ¼ (O-Ha + O-Hb + O-Hb + O-Ha) = 1/2 (O-Ha + O-Hb)

According to the orthogonality theorem; it shouldn’t be possible to obtain a B1 linear combination, and indeed the projection operator will be zero.
p^p^ B(O-Ha) = ¼ (O-Ha + O-Hb + O-Hb + O-Ha)=0

Application of the B2 projection operator gives
p^p^ B2 (O-Ha) = ¼ (O-Ha + O-Hb + O-Hb + O-Ha)

=1/2 (O-Ha + O-Hb)

 

These linear combinations relate to the symmetric (A1) and anti-symmetric (B2) stretches of water as given below.

When two equivalent real functions are involved, the correct linear combinations will be equals to the sum and difference functions. In case of degenerate representations like in case of N-H stretching vibrations in ammonia, it is more difficult to construct the symmetry adapted linear combinations. 

 

We can create symmetry-adapted linear combinations of atomic orbitals in exactly the same way. The point group is D2h with four carbon-hydrogen sigma bonds which are symmetry-equivalent and can make four carbon-hydrogen bonding symmetry-Adapted Linear Combinations ‘s. The character table will be as follow.

Result of symmetry operations on s1
  E C2(Z) C2(Y)  C2(X)  i s(XY)  s(XZ) s(YZ)
s1 s1  s3 s4  s2 s3 s1  s2 s4 

 

There are four non-zero symmetry-adapted linear combinations can be possible. 

 

Hybridization

 

HYBRIDIZATION

In the Valence Bond (VB) theory an atom may rearrange its atomic orbitals prior to the bond formation. Instead of using the atomic orbitals directly, mixture of them (hybrids) are formed. For carbon (and other elements of the second row) the hybridization is limited to mixing one 2s and three 2p orbitals, as appropriate.

We recognize three basis types of hybridization: sp3, sp2 and sp. These terms specifically refer to the hybridization of the atom and indicate the number of p orbitals used to form hybrids.

  • In sp3 hybridization all three p orbitals are mixed with the s orbital to generate four new hybrids (all will form σ type bonds or hold lone electron pairs).
  • If two p orbitals are utilized in making hybrids with the s orbital, we get three new hybrid orbitals that will form σ type bonds (or hold lone electron pairs), and  the “unused” p may participate in π type bonding.  We call such an arrangement sp2 hybridization.
  • If only one p orbital is mixed with the s orbital, in sp hybridization, we produce two hybrids that will participate in σ type bonding (or hold a lone electron pair). In this case, the remaining two p orbitals may be a part of two perpendicular π systems.

The most important rule is that the number of orbitals must be preserved

in the mixing process. The mixing principles can be illustrated on a simple

example of one s and one p orbital making two equivalent sp hybrids

(only one is shown here for clarity). The constructive interference of

http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/making-sp-hybrids.gif

the wavefunctions at the “right” half gives a large lobe of the hybrid

orbital, while the destructive interference on the left (opposite signs

of the s and p wavefunctions) yields a small “tail”.  The second hybrid

formed in this case is a 180°-rotated version of the one shown.  In

this case each of the two hybrids is constructed from ½ of s and ½ of p.

Within each type of hybridization, one can produce infinite number of different hybrids (mixtures). The hybrids are defined by the p to s ratio of the contributing orbitals.  Thus, an spm hybrid is composed of m+1 parts: one part of s and m parts of p, and the p/s ratio is equal m, called the hybridization index.  For example, an sp3 hybrid has ¼ (25%) of s and ¾ (75%) of p. This fraction is called an s (or p) character of the orbital. Thus, an sp3 hybrid has 25% s character.

http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hybrid-orbitals.gif

The hybrids with larger s character have bigger front lobes (and smaller “tails”) than the hybrids with smaller s character, as illustrated above for an sp (left) and sp3 (right) hybrids.  The s/p ratio is, thus, responsible for the bonding properties of the hybrid.  Increased s contribution brings electrons closer to the nuclei, increasing stabilizing Coulomb interactions. The more s character the hybrid orbital has the lower its energy, the better its overlap (with bonding partners), and the stronger (and shorter) bonds it can form.

It is important not to confuse the hybridization type that applies to the atom (see above) with the individual hybrid character that is described by the hybridization index m. The first (indirectly) indicates the number of p orbitals set aside (for participation in π systems), the second precisely describes the specific mixture of s and p used to construct the given hybrid.

The mixing of s and p orbitals in different ratios also results in changes of the angle between the resulting hybrids.  Since the atomic p orbitals are 90° from each other, their various degree of participation in the mixing will yield hybrids separated by different angles, depending on their p character.  For two identical hybrids, in general, the more p character in the hybrids the smaller the angle between them. Thus, two pure p orbitals (100% p) are 90° apart, two sp3 hybrids are 109.5° apart,  two sp2 hybrids are 120° apart, and two sp hybrids are 180° apart.  More generally, the angle (α) between any two hybrids (spm and spn) is given by cosα = –1/(m·n)0.5.

An atom will adjust its hybridization in such a way as to form the strongest possible bonds and keep all its bonding and lone-pair electrons in as low-energy hybrids as possible, and as far from each other as possible (to minimize electron-electron repulsions).  This adjustment  is accomplished by varying s and p characters of individual mixtures, but “moving” s character between hybrids (to lower energy of some) also changes the angles between them (potentially increasing electron-electron repulsion).  Thus, it is all a compromise game.

Let us look at some examples. Something simple to start: methane. The carbon atom forms four identical bonds using four identical hybrid orbitals. These orbitals are the result of sp3 hybridization (here we talk about the hybridization type), i.e. one s and three p orbitals are mixed to form four sp3 hybrids (here we talk about the composition, or character, of each hybrid).  Each of these hybrids is composed of ¼ of s and ¾ of p (the p/s ratio is 3, i.e. m = 3). The angle between any two such hybrid orbitals is (yes…  that’s the cosine formula above) 109.5o.  The table below gives more examples of different hybrid orbitals involved in making C-H bonds. Note that for the same hybridization type (sp3 in our example) one can have quite different hybrids involved in making C-H bonds.

Table 1. Hybrids Involved in Making C–H Bonds in Simple Hydrocarbons

Hydrocarbon Hybridization

Type

Bond angle Hybrid  involveda s characterb Bond length (A) BDEc
http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hyb-m.gif sp3 109.5o sp3 25% 1.100 105  
http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hyb-ea.gif sp3 107.3o sp3.36 23% 1.100 100  
http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hyb-cp.gif sp3 115o sp2.37 30% 1.089 106 46
http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hyb-ee.gif sp2 116.6o sp2.23 31% 1.076 106 45
http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/hyb-a.gif sp 180o sp(e) 50% 1.060 132 25

a. Hybrid orbitals on carbon involved in the formation of the indicated (in green) bonds with hydrogen atoms calculated from the formula: m = -1/cosα

b. Percent s character in the hybrid orbitals [(1/(m+1)) ×100]

c. Homolytic Bond Dissociation Energies (BDE). BDE’s depend to a much larger degree on the stability of the radicals formed than on the hybridization type of the bond broken (see below).

d. Acidity of the C–H bonds (the pKa values for methane and ethane are very approximate).

e. Strictly speaking, in this case the hybrid cannot be determined just from the bond angle. An angle between any two sp-type hybrids is always 180°, regardless of their specific s characters. In this case, the two hybrids made by mixing s and p orbitals are not equivalent, one makes bond to carbon, one to hydrogen.

When we say that bonds made out of hybrids containing more s character are stronger, we have to be very precise in our meaning. Bond strength are commonly measured by bond dissociation energies (BDE’s) that reflect enthalpies of homolysis (i.e. the energy required to break a bond, forming two radicals).  What counts in such considerations is the actual strength of the bond broken (that is related to the s character of the hybrid) and the stability of the radicals formed which is influenced strongly by other effects (such as hyperconjugation or resonance) that may not even be present in the molecule before homolysis.  In general, the stability of the radicals is a more important contribution to BDE’s than the hybridization effect (i.e. s character).

Similarly the s character of the orbital containing the lone electron pair will influence the stability of the anion, and therefore, the pKa value of the hydrocarbon precursors of this anion.  If the geometry of the hydrocarbon is similar to that of the anion, the more s character in the hybrid involved in making the “acidic” C-H bond, the lower the pKa value of the hydrocarbon (and the more stable the anion).  But, again caution is required in making comparisons:  the stability of the anion also strongly depends on other effects (such as inductive and resonance stabilization, ion pairing and solvation, i.e. interactions with solvent molecules). In particular, if the resonance stabilization is involved, the hybridization of the hydrocarbon and the anion derived from it are going to be quite different.

In general, very rarely all hybrids formed by an atom are exactly equal.  We may have an infinite number of combinations of mixtures.  For example, let us assume that carbon forms two sp2 hybrids (with an angle of 120°) to two identical substituents and additional two (equal to each other) spx hybrid orbitals that are going to be used to form two additional bonds, as shown in A below. How can we find x? It is very simple, just a little fraction arithmetic! Each of the two sp2 hybrids contains 1/3 of s (total of 2/3 s). The remaining 1/3 s must be divided between the two identical spx hybrids; i.e. 1/6 s per orbital. To make the “full” hybrid the “missing” 5/6 of the hybrid must be composed of p. So, the ratio of p/s = 5 (or x = 5) and we have our answer (sp5). If you do not believe it, you may check the math doing the balance for the p orbitals.   Here, how it goes: the two sp2 hybrids contain 2/3 of p each (total 4/3 p).  Since all three p orbitals participate in hybridization, we have 3 – 4/3 = 5/3 of p left to be used in the two spx orbitals. That leaves (5/3)/2, or 5/6 of p per hybrid; exactly the same answer we got doing the balance for the s orbital.

http://courses.chem.psu.edu/chem210/mol-gallery/hybridization/mixed-hybrids.gif

Now we can look at the hybridization of nitrogen in ammonia (B) or oxygen in water (C) with more precision. Since each of these central atoms has four electron pairs around, and no π bonds (that information is available from a simple Lewis structure) we may say that oxygen and nitrogen are sp3 hybridized. We mean that each atom uses three p’s and one s atomic orbitals to make four hybrids. But none of these hybrids is an sp3 hybrid! The angle between hydrogens in : NH3 is 107°  Since all the hydrogens are identical, we can calculate that the hybrids used to make N-H bonds are sp3.42 (cos(107°) = –1/m or m = 3.42); i.e they have 1/(1+3.42) = 22.6% s character each). That leaves 32.2% s character for the hybrid containing the lone pair, or the lone pair is an sp2.10 hybrid (the p character in the lone-pair hybrid must be 100 –-32.2 = 67.8%, or m/(m+1) = 0.678 from where m = 2.10). For a moment, forget the arithmetic! The point is that the lone pair hybrid has increased its s character in comparison to what it would be in the pure sp3 hybrid (32.2% vs 25% s) stabilizing the lone pair (more s character more stabilization). And the lone pair needs more stabilization: these are, by definition, unshared electrons! The price to be paid is the decrease in the H-N-H angle from the ideal tetrahedral 109.5o to 107o and the increased repulsion between the bonding pairs. The observed situation is the compromise between these two trends. You may recognize this concept as being equivalent to saying that the lone pair needs “more space” than the bonding pairs. Similar situation is found in H2O. Here, the H-O-H bond is 104.5°. Now, you do the math! The O-H bonds use sp4 hybrids and the lone pairs (two here) are sp2.3, or have about 30% s character each. Still, that is better than the 25% s of the pure sp3 hybrids.