8.2 Tetravalence of Carbon: Shapes of Organic Compounds

The Shapes of Carbon Compounds

The knowledge of fundamental concepts of molecular structure helps in understanding and predicting the properties of organic compounds. Also, we already know that tetravalence of carbon and the formation of covalent bonds by it are explained in terms of its electronic configuration and the hybridisation of \(s\) and \(p\) orbitals. It may be recalled that formation and the shapes of molecules like methane \(\left(\mathrm{CH}_4\right)\), ethene \(\left(\mathrm{C}_2 \mathrm{H}_4\right)\), ethyne \(\left(\mathrm{C}_2 \mathrm{H}_2\right)\) are explained in terms of the use of \(s p^3, s p^2\) and \(s p\) hybrid orbitals by carbon atoms in the respective molecules.

Hybridisation influences the bond length and bond enthalpy (strength) in compounds. The \(s p\) hybrid orbital contains more \(s\) character and hence it is closer to its nucleus and forms shorter and stronger bonds than the \(s p^3\) hybrid orbital. The \(s p^2\) hybrid orbital is intermediate in \(s\) character between \(s p\) and \(s p^3\) and, hence, the length and enthalpy of the bonds it forms, are also intermediate between them. The change in hybridisation affects the electronegativity of carbon. The greater the \(s\) character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an \(sp\) hybrid orbital with \(50 \%\) s character is more electronegative than that possessing \(s p^2\) or \(s p^3\) hybridised orbitals. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned, about which you will learn in later units.

Some Characteristic Features of \(\pi\) Bonds

In a \(\pi\) (pi) bond formation, parallel orientation of the two \(p\) orbitals on adjacent atoms is necessary for a proper sideways overlap. Thus, in \(\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2\) molecule all the atoms must be in the same plane. The \(p\) orbitals are mutually parallel and both the \(p\) orbitals are perpendicular to the plane of the molecule. Rotation of one \(\mathrm{CH}_2\) fragment with respect to other interferes with maximum overlap of \(p\) orbitals and, therefore, such rotation about carbon-carbon double bond \((\mathrm{C}=\mathrm{C})\) is restricted. The electron charge cloud of the \(\pi\) bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general, \(\pi\) bonds provide the most reactive centres in the molecules containing multiple bonds.

Example 8.1: How many \(\sigma\) and \(\pi\) bonds are present in each of the following molecules?
(a) \(\mathrm{HC} \equiv \mathrm{CCH}=\mathrm{CHCH}_3\)
(b) \(\mathrm{CH}_2=\mathrm{C}=\mathrm{CHCH}_3\)

Answer:

\(
\text { (a) } \sigma_{\mathrm{C}-\mathrm{c}}: 4 ; \sigma_{\mathrm{C}-\mathrm{H}} \text { : }6 ; \pi_{\mathrm{C}=\mathrm{C}}: 1 ; \pi \mathrm{C} \equiv \mathrm{C}: 2
\)
\(
\text { (b) } \sigma_{\mathrm{C}-\mathrm{C}}: 3 ; \sigma_{\mathrm{C}-\mathrm{H}}: 6 \text {; }\pi_{\mathrm{C}=\mathrm{C}}: 2 .
\)

Example 8.2: What is the type of hybridisation of each carbon in the following compounds?
(a) \(\mathrm{CH}_3 \mathrm{Cl}\),
(b) \(\left(\mathrm{CH}_3\right)_2 \mathrm{CO}\),
(c) \(\mathrm{CH}_3 \mathrm{CN}\),
(d) \(\mathrm{HCONH}_2\),
(e) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCN}\)

Answer: \(\text { (a) } s p^3, \text { (b) } s p^3, s p^2 \text {, (c) } s p^3, s p \text {, (d) } s p^2 \text {, (e) }s p^3, s p^2, s p^2, s p\)

Example 8.3: Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.
\(
\text { (a) } \mathrm{H}_2 \mathrm{C}=\mathrm{O} \text {, (b) } \mathrm{CH}_3 \mathrm{~F} \text {, (c) } \mathrm{HC} \equiv \mathrm{N} \text {. }
\)

Answer: (a) \(s p^2\) hybridised carbon, trigonal planar;
(b) \(s p^3\) hybridised carbon, tetrahedral;
(c) \(s p\) hybridised carbon, linear.