DISSOCIATION ENTHALPIES, BOND ANGLES AND VSEPR THEORY
1.1.5 VESPR Theory and Molecular Geometry
We can predict the arrangement of atom in molecules and ions on the basis of a relatively simple idea called the valence shell electron pair repulsion (VSEPR) theory.
We can apply this theory on the basis of the following considerations:
1. In a molecule or ion, the central atom is covalently bonded to two or more atoms or groups.
2. Covalent bonds contain shared pair of electrons which are often called bond pairs or bonding pairs. The unshared electrons of the central atom are called nonbonding pairs or unshared pairs or lone pairs.
3. Since electron pairs repel each other, the electron pairs of the valence shell tend to stay as far apart as possible to avoid electronic repulsion.
4. If the central atom is surrounded by bond pairs as well as lone pairs of electrons, the repulsions among themselves are different. As a result, the molecule possesses an irregular or distorted geometry. The repulsive interactions of various electron pairs decrease in the order: lone pair–lone pair (lp – lp)> lone pair–bond pair (lp – bp) > bond pair–bond pair (bp – bp).
5. The geometry of a molecule is to be settled by considering all of the electron pairs, bonding and nonbonding. However, the shape of the molecule is to be described by referring to the positions of the atoms and not by the position of the electron pairs.
Let us consider the following examples.
Methane (CH4) molecule: In CH4 molecule, the total number of electrons surrounding the central carbon atom = 4 valence electrons of C atom + 4 electrons of four singly-bonded H atom = 8 electron or 4 electrons pairs = 4 s bond pairs. The four bond pairs experience minimum repulsion if they are tetrahedrally oriented, i.e., if all the H—C—H bond angles are of 109.5°. Hence, the shape of CH4 molecule is tetrahedral.
C 109.5°
H
H H
H
Tetrahedral methane molecule
Ammonia (NH3) molecule: In NH3 molecule, the total number of electrons in the valence shell of the central N atom = 5 valence electrons of N atom + 3 electrons of three singly-bonded H atom = 8 electron or 4 electron pairs = 3 s bond pairs + 1 lone pair. All four electron pairs experience minimum repulsion if they occupy the four corners of a tetrahedron. As lone pair–bond pair repulsion is greater than the bond pair–bond pair repulsion, the H—N—H bond angle is slightly deviated from the normal tetrahedral angle
(109.5°) and is reduced to 107°, i.e., the tetrahedron is somewhat distorted. Excluding the lone pair, the shape of the molecule is trigonal pyramidal.
Water (H2O) molecule: In water molecule, the total number of electrons surrounding the central O atom = 6 valence electron of oxygen atom + 2 electrons of two singly-bonded H atoms = 8 electrons or 4 electron pairs = 2 s bond pairs + 2 lone pairs.
In order to minimise the extent of mutual repulsion, these four electron pairs are oriented towards the four corners of a tetrahedron. However, the tetrahedron is somewhat distorted due to the strong repulsive forces exerted by the lone pairs on each bond pair of electrons.
In fact, the H—O—H bond angle is reduced to 104.5° from the normal tetrahedral angle of 109.5°. Excluding the lone pairs, the shape of the molecule is angular or V-shaped.
Angular or V-shaped water molecule O
104.5°
H H
or O
104.5°
H H
Boron tifl uoride (BF3) molecule: In boron trifl uoride molecule, the total number of electrons in the valence shell of the central boron atom = 3 valence electrons of B atom + 3 electrons of three singly-bonded F atom = 6 electrons or 3 electrons pairs = 3 s bond pairs. The three bond pairs experience minimum propulsion if they remain at 120° angle with respect to each other. Therefore, the geometrical shape of BF3 molecule is trigonal planar.
B F 120° F 120° 120°
F
Trigonal planar boron trifluoride molecule
Acetylene (HCCH) molecule: The number of electrons surrounding each carbon atom of acetylene molecule = 4 valence electron of carbon + 3 electrons of one triply-bonded C atom + 1 electron of one singly-bonded H atom = 8 electrons = 4 electron pairs = 2 s bond
pairs + 2p bond pairs. In order to minimise the repulsive forces between the bond pairs, the shape of acetylene molecule is linear. The effect of electrons involved in the formation of a p bond is not generally considered in determining the geometrical shape of a molecule.
H——C∫∫C——H 180° 180°
Linear acetylene molecule
1. Give the state of hybridization of the central atom of each of the following species and predict their shapes.
(a) H2O (b) BF3 (c) CH ! 3 (d) @
CH 3 (e) @ :N H2
(f)
H S:2 (g)
NH 3 (h) BF4* (i) ≈
H O 3 (j) HCN (k) CCl4 (l) CO2 (m) BeH2 (n) NH4≈
Solution
(a) (b)
F
sp2; trigonal planar
F
F B
(c)
H
sp2; trigonal planar
H
H C
+
(d)
C H
H
sp3; trigonal pyramidal –
H
(e)
N H
H
sp3; bent or angular –
(f)
S H
H
sp3; bent or angular
(g)
N H H
sp3; trigonal pyramidal
H
(h)
B F
F
sp3; tetrahedral –
F
F (i)
O+ H H H
sp3; trigonal pyramidal
(j)
H—C∫∫N
sp; linear
(k)
C Cl Cl
sp3; tetrahedral
Cl
Cl (l)
O==C==O
sp; linear
(m)
H—Be—H
sp; linear
(n)
N H
H
sp3; tetrahedral
H H
+
2. Draw the structure of a hydrocarbon which contains:
(a) Three sp3-hybridized carbon atoms;
(b) One sp3 and two sp2-hybridized carbon atoms;
(c) One sp and two sp2-hybridized carbon atoms;
(d) Two sp3 and two sp-hybridized carbon atoms.
Solution
(a) (b) (c)
(d)
3. Which orbitals are used to form each bond in methylamine, CH NH3 2? Solution
Each C—H bond is formed by the overlap of an sp3 orbital of carbon with the s orbital of hydrogen (C 3
sp —H1s). The C—N bond is formed by the overlap of an sp3 orbital of carbon with sp3 orbital of nitrogen (C 3
sp —N 3
sp ) and each N—H bond is formed by the overlap of an sp3 orbital of nitrogen with the s orbital of hydrogen (N 3
sp —H1s).
4. Answer the following question for the acetaldehyde ||
— —
: : 3
O
(CH C H)
molecule:
(a) Determine the hybridization of oxygen and the two carbon atoms.
(b) Which orbitals are involved in forming the carbon–oxygen double bond?
(c) Mention the type of orbital in which the lone pairs reside.
Solution (a)
(b) The s bond is formed by the end-on overlap of an sp2 orbital of carbon with an sp2 orbital of oxygen and the p bond is formed by the side-by-side overlap of the 2p orbital of carbon with the 2p orbital of oxygen.
(c) The two sp2 hybrid orbitals are occupied by the two lone pairs of oxygen.
5. Mention the state of hybridization of the starred (*) carbon atoms in each of the following compounds:
(a) (b) CH CN 3 * (c) HC∫∫C—CHO* (d) CH == C == CH 2 * 2 (e) * (f)
* –
Solution
(a) Three groups around the starred carbon: sp2-hybridized; (b) two groups around the starred carbon: sp-hybridized; (c) three groups around the starred carbon: sp2-hybridized;
(d) two groups around the starred carbon: sp-hybridized; (e) three groups around the starred carbon: sp2-hybridized; (f) four groups around the starred carbon; sp3-hybridized.
6. How many s and p bonds are present in each of the following molecules?
(a) CH3—C ∫∫ C—CH == CH—CH3 (b) –CH CH2 3
(c) CH3 CH == C == CH CH2 CH3
Solution
(a) s bond = 13; p bond = 3;
(b) s bond = 18; p bond = 3;
(c) s bond 15; p bond 2
7. Designate the state of hybridization of all the atoms of the following molecule:
Solution The hybridized atoms within the molecule is designated as a, b and c according to their hybridization status: a = sp3; b = sp2 and c = sp.
8. Draw the orbital picture for each of the following molecules: (a) ethylene (b) acetylene (c) ketene (CH2 == C == O) (d) acraldehyde (e) acrylonitrile (f) but-1, 2, 3-triene.
Solution (a) Ethylene:
(b) Acetylene:
(c) Ketene:
(d) Acraldehyde:
(e) Acrylonitrile:
(f) But-1, 2, 3-triene:
9. Which atoms in each of the following molecules always remain in the same plane and why?
(a) CH3CH==CH CH3 (b) C6H5C ≡≡ CCH3 (c) CH3CH==C==C==CH2 (d) CH2==CH—C ≡≡ CH
Solution The two sp2-hybridized carbon atoms and the atom directly attached to them always remain in the same plane. This is applicable also when the two sp2-hybridzed carbon atoms are linked through an even number of sp-hybridized carbon atoms. Therefore, in (a), all atoms excluding the six methyl hydrogens remain in the same plane. In (b), all atoms excluding the three methyl hydrogens remain in the same plane. In (c), all atoms excluding the three methyl hydrogens remain in the same plane and in (d), all atoms remain in the same plane. This is also because the two sp-hybridized carbon atoms and the atoms directly attached to them remain in the same line.
10. Explain why a p bond is weaker than a s bond.
Solution A p bond is weaker than a s bond because the end-on overlap that forms s bonds is better than the side-to-side overlap that forms p bonds. This is also because the electron density in a p bond is farther from the two nuclei as compared to that in a s bond.
11. In CH3CH3, the C—H bond is shorter and stronger than the C—C bond – explain.
Solution The s orbital of hydrogen is closer to the nucleus than the sp3 orbital of carbon having less s character. So, the carbon and hydrogen nuclei are closer together in sp3-s overlap than the carbon nuclei in sp3-sp3 overlap. Because of this, the C—H bond in ethane (CH3—CH3) is shorter and stronger than the C—C bond. Again, as the percentage of s character of the overlapping orbitals increases, the electron density in the region of overlap increases and as a result, the strength of the bond increases. Since there is greater electron density in the region of sp3-s overlap than in the region of sp3-sp3 overlap, therefore, the C—H bond is stronger than the C—C bond.
12. Arrange the indicated bonds in each of the following compounds in order of increasing bond strength and increasing bond length:
(a) ≠
HC∫∫C—CH==CH—CH —CH2 3
≠ ≠
bond 1 bond 2 bond 3
(b) H C3
CH —C2 ∫∫N NH—CH CH2 3
N
≠ bond 2
bond 3 bond 1
Solution Greater the bond multiplicity, shorter the bond and greater the bond strength.
Therefore, in compound (a), bond length increases in the order: bond 1 < bond 2 < bond 3 and bond strength increase in the order: bond 3 < bond 2 < bond 1. In compound (b), the bond length increases in the order: bond 3 < bond 2 < bond 1 and the bond strength increases in the order: bond 1 < bond 2 < bond 3.
13. Which of the indicated bonds in each pair of compounds is shorter and why?
(a)
(b)
(c) CH —C—H and H—CH OH3 2
==
O
≠ ≠
Solution As the s-character of the overlapping orbitals increases, the nuclei involved in bond formation becomes closer and the bond becomes shorter. Therefore, (a) the Csp—H bond in the second compound is shorter than C 2
sp —H bond in the fi rst compound;
(b) the N 2
sp —H bond in the second compound is shorter than the N 3
sp —H bond in the fi rst compound, and (c) the C 2
sp —H bond in the fi rst compound is shorter than the C 3
sp —H bond in the second compound.
14. Determine the state of hybridization of the indicated atom in each of the following species:
(a)
= =
–
O
(b)
O N2
CH2
–
≠
(c)
O N2
NH2 –
≠
(d)
C H —C6 5
O= = O–ỉ
(e)
O= =
= = ỉ–
O
(f)
H—C—NH2
≠
==
O
Solution In any system of the type X = Y – Z:, Z is sp2-hybridized and the unshared electron pair on it occupies a p orbital to delocalize the electron pair and make the system conjugated. Therefore, the indicated atoms in (a), (b), (c), (d) and (f) are sp2-hybridized.
However, due to violation of Bredt’s rule (introduction of a double bond is not possible at the bridgehead position in bridged bicyclic compounds with small rings) delocalization of the unshared electron pair is not possible in the compound (e) and therefore, the indicated carbon atom is sp3-hybridized.
15. “Bromination of methane is less exothermic than that of chlorination” — explain with DH° calculation.
[Bond dissociation energies for C—H = 104 kcal/mol; Br—Br = 46 kcal/mol; H—Br = 87.5 kcal/mol; Cl—Cl = 58 kcal/mol; H—Cl = 103 kcal/mol; C—Cl = 83.5 kcal/mol and C—Br = 70 kcal/mol]
Solution CH4+Br – Br ổổặhn CH3 -Br HBr+
DH° (bromination) = [(–70) + (–87.5)] – [(–104) + (–46)]
= (–157.5) – (–150) = –7.5 kcal/mol CH4 +Cl – Clổổặhn CH3-Cl HCl+
DH° (chlorination) = [(–83.5) + (103)] – [(–104) + (–58)]
= (–186.5) – (–162) = – 24.5 kcal/mol
Thus, bromination of methane is less exothermic than chlorination of methane.
16. The H—X bond in hydrogen halides becomes longer and weaker as the atomic mass of the halogen increases—explain.
Solution A p orbital of halogen overlaps with the s orbital of hydrogen to form the hydrogen–halogen (H—X) bond in hydrogen halides. For bond formation, fl uorine uses the p orbital that belongs to the second shell of electrons while chlorine uses the p orbital that belongs to the third shell of electrons. Since the average distance from the nucleus is greater for an electron in the third shell than for an electron in the second shell, therefore, the average electron density is less in a 3p orbital than in a 2p orbital. Consequently, the electron density in the region where the s and p orbitals overlap decreases as the size of the halogen increases. Thus, the hydrogen–halogen bond becomes longer and weaker as the atomic mass of the halogen increases.
1. A carbon-carbon bond formed by sp2–sp2 overlap is stronger than the one formed by sp3–sp3 overlap. Explain.
2. Which atom in the ammonium ion (NH )≈4 has the least electron density and why?
3. What is the state of hybridization of each of the C, O and N atom in the following compound?
H C3 C O
C∫∫C—CH—CH —O—CH2 3
NH2
==
4. Predict the geometry around each of the indicated atoms:
(a) BH≠ 3 (b) H N3 ặBH ≠ 3 (c)
3 ≠ 3
O
CH —C—OCH
||
(d) H O 3≠! (e)
≠
!
C(CH )3 3 (f)
≠
!
N H4
(g) HCN ≠ (h) (CH ) N H 3 2 ≠ (i) Et N2 ≠@
(j) CH CH CH3≠ 2 3 (k) CH2 CH– CH2
= ≠@
5. Predict the mentioned bond angles:
(a) C—O—C bond angle in CH3COCH3 (b) F—B—F bond angle in BF4@ (c) C—C—N bond angle in C2H5CN (d) H—N—H bond angle in NH≈4 (e) C—O—H bond angle in CH3CH2OH (f) C—N—C bond angle in (CH )3 2 N H! 2 6. Which of the indicated bonds is shorter and why?
(a) (b)
≠ ≠
= ∫ -
3 3
CH CH CH – C C CH
(c) 9 ≠ ≠Cl (d) ≠
≠ NH2
OH
(e) CH CH CH – CH CH CH3 3
≠ ≠
= = -
(f) ¨
≠
||
3 2 3
O
CH — O — CH — C — CH (g) N—CH CH N==CHCH CH2 2 2 3
≠ ≠
7. Give the state of hybridization of the indicated atoms:
(a) ≈
3 2 ≠ 2
(CH ) N H (b) ≈
3 2 ≠
(CH ) O H (c)
≠ 4@
BF (d) CH NO 3 ≠ 2 (e)
ặ≠
3 3
H N BF (f)
== ≠
2 2
CH CH — CH@ (g) CH CH3 ==CH — NH (h) ≠ 2 CH2== ==C≠ CH2