Nitric oxide NO consists of one nitrogen atom and one oxygen atom. The bond length is 1.1538 Å.

The molecular orbital Hamiltonian in this case is,

\begin{equation} H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 - \frac{Z^N_{\text{eff}}e^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_N|}- \frac{Z^O_{\text{eff}}e^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_O|}, \end{equation}

where $\vec{r}_N$ and $\vec{r}_O$ are the positions of the carbon atom and the oxygen atom and $Z^N_{\text{eff}}=3.9$ and $Z^O_{\text{eff}}=4.55$ are the effective nuclear charges of nitrogen and oxygen. We consider a trial wavefunction that is a linear combination of 8 atomic orbitals. These are the 2s and 2p orbitals of each atom.

$$\psi_{\text{mo}} = c_1\phi^{\text{N}}_{\text{2s}}+c_2\phi^{\text{N}}_{\text{2px}}+c_3\phi^{\text{N}}_{\text{2py}}+c_4\phi^{\text{N}}_{\text{2pz}}+ c_5\phi^{\text{O}}_{\text{2s}}+ c_6\phi^{\text{O}}_{\text{2px}}+ c_7\phi^{\text{O}}_{\text{2py}}+ c_8\phi^{\text{O}}_{\text{2pz}}.$$

This trial wave function is substituted into the time independent Schrödinger equation,

\begin{equation} H_{\text{mo}}\psi_{\text{mo}}=E\psi_{\text{mo}} . \end{equation}

Multiplying the Schrödinger equation from the left by each of the atomic orbitals and integrating over all space results in the Roothaan equations.

$$\textbf{H}\,\vec{c}=E\,\vec{c},$$

where the matrix elements are,

\[ \begin{equation} H_{ij}=\langle \phi_i|H_{\text{mo}}|\phi_j\rangle, \\ S_{ij}=\langle \phi_i|\phi_j\rangle, \end{equation} \]

and elements $c_i$ of the vector $\vec{c}$ are the coefficients of the trial wave function. The Hamiltonian matrix elements$H_{ij}$ and the overlap matrix elements $S_{ij}$ were determined numerically using the code NOwithLCAO.py. The resulting Hamiltonian matrix in eV and overlap matrix are,

\[ \begin{equation} H= \left[ \begin{matrix} -107.35 & 18.56 & 0.0 & 0.0 & -55.14 & -45.18 & 0.0 & 0.0 \\ 18.56 & -116.07 & 0.0 & 0.0 & 57.06 & 52.22 & 0.0 & 0.0 \\ 0.0 & 0.0 & -103.86 & 0.0 & 0.0 & 0.0 & -27.93 & 0.0 \\ 0.0 & 0.0 & 0.0 & -103.86 & 0.0 & 0.0 & 0.0 & -27-93 \\ -55.14 & 57.06 & 0.0 & 0.0 & -118.73 & -14.27 & 0.0 & 0.0 \\ -45.18 & 52.22 & 0.0 & 0.0 & -14.27 & -124.47 & 0.0 & 0.0 \\ 0.0 & 0.0 & -27.93 & 0.0 & 0.0 & 0.0 & -116.14 & 0.0 \\ 0.0 & 0.0 & 0.0 & -27.93 & 0.0 & 0.0 & 0.0 & -116.14 \\ \end{matrix} \right], \qquad S = \left[ \begin{matrix} 1.00 & 0.0 & 0.0 & 0.0 & 0.43 & 0.32 & 0.0 & 0.0 \\ 0.0 & 1.00 & 0.0 & 0.0 & -0.41 & -0.33 & 0.0 & 0.0 \\ 0.0 & 0.0 & 1.00 & 0.0 & 0.0 & 0.0 & 0.21 & 0.0 \\ 0.0 & 0.0 & 0.0 & 1.00 & 0.0 & 0.0 & 0.0 & 0.21 \\ 0.43 & -0.41 & 0.0 & 0.0 & 1.00 & 0.0 & 0.0 & 0.0 \\ 0.32 & -0.33 & 0.0 & 0.0 & 0.0 & 1.00 & 0.0 & 0.0 \\ 0.0 & 0.0 & 0.21 & 0.0 & 0.0 & 0.0 & 1.00 & 0.0 \\ 0.0 & 0.0 & 0.0 & 0.21 & 0.0 & 0.0 & 0.0 & 1.00 \\ \end{matrix} \right]. \end{equation} \]

Multiplying by the inverse of the overlap matrix allows us to write this in the form of an eigenvalue proble that can be solved for the eight molecular orbitals,

$$\textbf{S}^{-1}\textbf{H}\,\vec{c}=E\,\vec{c}.$$

The coefficients that have to be substituted into $\psi_{\text{mo}}$ to specify the eight molecular orbitals and their corresponding energies are given below.

$$E_1= -138.56\,\text{eV},\quad\vec{c}_1=\left[ \begin{matrix} 0.2153 \\ 0.0 \\ 0.0 \\ -0.4985 \\ -0.3658 \\ 0.0 \\ 0.0 \\ 0.7323 \end{matrix} \right] ;\qquad E_2=-117.26\,\text{eV}\quad\vec{c}_2=\left[ \begin{matrix} -0.4192 \\ 0.0 \\ 0.0 \\ 0.5056 \\ 0.3072 \\ 0.0 \\ 0.0 \\ 0.6730 \end{matrix} \right] ;\qquad E_3=-117.26\,\text{eV}\quad\vec{c}_3=\left[ \begin{matrix} 0.0 \\ -0.1957 \\ -0.1976 \\ 0.0 \\ 0.0 \\ 0.6447 \\ 0.6365 \\ 0.0 \end{matrix} \right] ;\qquad E_4=-107.55\,\text{eV}\quad\vec{c}_4=\left[ \begin{matrix} 0.0 \\ -0.1976 \\ 0.1957 \\ 0.0 \\ 0.0 \\ 0.6366 \\ -0.6448 \\ 0.0 \end{matrix} \right] ;\qquad \\ E_5 = -107.01\,\text{eV}\quad\vec{c}_5=\left[ \begin{matrix} 0.4962 \\ 0.0 \\ 0.0\\ 0.6379 \\ -0.3350 \\ 0.0 \\ 0.0 \\ -0.0013 \end{matrix} \right] ;\qquad E_6= \text{HOMO} = -100.44\,\text{eV}\quad\vec{c}_6=\left[ \begin{matrix} 0.7292 \\ 0.0 \\ 0.0 \\ 0.2982 \\ 0.8121 \\ 0.0 \\ 0.0 \\ 0.1039 \end{matrix} \right] ;\qquad E_7= \text{LUMO} = -100.44\,\text{eV}\quad\vec{c}_7=\left[ \begin{matrix} 0.0 \\ -0.6759 \\ -0.6825 \\ 0.0 \\ 0.0 \\ -0.3011 \\ -0.2972 \\ 0.0 \end{matrix} \right] ;\qquad E_8 = -92.50\,\text{eV}\quad\vec{c}_8=\left[ \begin{matrix} 0.0 \\ -0.6825 \\ 0.6759 \\ 0.0 \\ 0.0 \\ -0.2973 \\ 0.3011 \\ 0.0 \end{matrix} \right]. $$

Nitric oxide has 15 electrons. Four electrons occupy the two molecular orbitals that are mainly comprised of the 1s atomic orbitals but were neglected in this calculation. The other 11 electrons occupy 6 molecular orbitals with the lowest energies that were calculated here. The highest occupied molecular orbital (HOMO) is MO6 and the lowest unoccupied molecular orbital (LUMO) is MO7.