4.8 Bonding in Some Homonuclear Diatomic Molecules

In this section, we shall discuss bonding in some homonuclear diatomic molecules.

Hydrogen molecule \(\left(\mathrm{H}_2\right)\)

It is formed by the combination of two hydrogen atoms. Each hydrogen atom has one electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in \(\sigma 1 \mathrm{s}\) molecular orbital. So electronic configuration of hydrogen molecule is

\(
\mathrm{H}_2:(\sigma 1 s)^2
\)
The bond order of \(\mathrm{H}_2\) molecule can be calculated as given below:
Bond order \(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{2-0}{2}=1\)
This means that the two hydrogen atoms are bonded together by a single covalent bond. The bond dissociation energy of hydrogen molecule has been found to be \(438 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and bond length equal to \(74 \mathrm{pm}\). Since no unpaired electron is present in hydrogen molecule, therefore, it is diamagnetic.

Helium molecule \(\left(\mathrm{He}_2\right)\)

The electronic configuration of helium atom is \(1 s^2\). Each helium atom contains 2 electrons, therefore, in \(\mathrm{He}_2\) molecule there would be 4 electrons. These electrons will be accommodated in \(\sigma 1s\) and \(\sigma^* 1 s\) molecular orbitals leading to electronic configuration:
\(
\mathrm{He}_2:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2
\)
\(
\text { Bond order of } \mathrm{He}_2 \text { is } 1 / 2(2-2)=0
\)
\(\mathrm{He}_2\) molecule is therefore unstable and does not exist.

Similarly, it can be shown that \(\mathrm{Be}_2\) molecule \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\) also does not exist.

Lithium molecule \(\left(\mathrm{Li}_2\right)\)

The electronic configuration of lithium is \(1 s^2, 2 s^1\). There are six electrons in \(\mathrm{Li}_2\). The electronic configuration of \(\mathrm{Li}_2\) molecule, therefore, is
\(
\mathrm{Li}_2:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2
\)
The above configuration is also written as \(\mathrm{KK}(\sigma 2 \mathrm{~s})^2\) where KK represents the closed \(\mathrm{K}\) shell structure \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2\).
From the electronic configuration of \(\mathrm{Li}_2\) molecule it is clear that there are four electrons present in bonding molecular orbitals and two electrons present in antibonding molecular orbitals. Its bond order, therefore, is \(1 / 2(4-2)\) \(=1\). It means that \(\mathrm{Li}_2\) molecule is stable and since it has no unpaired electrons it should be diamagnetic. Indeed diamagnetic \(\mathrm{Li}_2\) molecules are known to exist in the vapour phase.

Carbon molecule \(\left(C_2\right)\)

The electronic configuration of carbon is \(1 s^2 2 s^2 2 p^2\). There are twelve electrons in \(\mathrm{C}_2\). The electronic configuration of \(\mathrm{C}_2\) molecule, therefore, is
\(
\begin{aligned}
& \mathrm{C}_2:(\sigma \mathrm{ls})^2\left(\sigma^* 1 s\right)^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right) \\
& \text { or } K K(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x^2=\pi 2 p_y^2\right)
\end{aligned}
\)
The bond order of \(C_2\) is \(1 / 2(8-4)=2\) and \(C_2\) should be diamagnetic. Diamagnetic \(\mathrm{C}_2\) molecules have indeed been detected in vapour phase. It is important to note that double bond in \(\mathrm{C}_2\) consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other molecules a double bond is made up of a sigma bond and a pi bond. In a similar fashion the bonding in \(\mathrm{N}_2\) molecule can be discussed.

Oxygen molecule \(\left(\mathrm{O}_2\right)\)

The electronic configuration of oxygen atom is \(1 s^2 2 s^2 2 p^4\). Each oxygen atom has 8 electrons, hence, in \(\mathrm{O}_2\) molecule there are 16 electrons. The electronic configuration of \(\mathrm{O}_2\) molecule, therefore, is
\(
\mathrm{O}_2:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^2
\)
\(
\left(\pi 2 p_x{ }^2 \equiv \pi 2 p_y{ }^2\right)\left(\pi^* 2 p_x{ }^1 \equiv \pi^* 2 p_y{ }^1\right)
\)
\(
\mathrm{O}_2:\left[\begin{array}{c}
\mathrm{KK}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_{\mathrm{z}}\right)^2 \\
\left(\pi 2 p_x^2 \equiv \pi 2 p_{\mathrm{y}}^2\right),\left(\pi^* 2 p_{\mathrm{x}}^1 \equiv \pi^* 2 p_{\mathrm{y}}^1\right)
\end{array}\right]
\)
From the electronic configuration of \(\mathrm{O}_2\) molecule it is clear that ten electrons are present in bonding molecular orbitals and six electrons are present in antibonding molecular orbitals. Its bond order, therefore, is Bond order \(=\frac{1}{2}\left[N_{\mathrm{b}}-N_{\mathrm{a}}\right]=\frac{1}{2}[10-6]=2\)
So in oxygen molecule, atoms are held by a double bond. Moreover, it may be noted that it contains two unpaired electrons in \(\pi^* 2 p_x\) and \(\pi^* 2 p_y\) molecular orbitals, therefore, \(\mathrm{O}_2\) molecule should be paramagnetic, a prediction that corresponds to experimental observation. In this way, the theory successfully explains the paramagnetic nature of oxygen.
Similarly, the electronic configurations of other homonuclear diatomic molecules of the second row of the periodic table can be written. In Fig.4.21 are given the molecular orbital occupancy and molecular properties for \(\mathrm{B}_2\) through \(\mathrm{Ne}_2\). The sequence of MOs and their electron population are shown. The bond energy, bond length, bond order, magnetic properties and valence electron configuration appear below the orbital diagrams.