To calculate the bond order of the nitric oxide molecule NO , we need to consider its molecular orbital diagram and electron configuration. Nitrogen N has 7 electrons, and oxygen O has 8 electrons. In total, there are 15 electrons in the NO molecule.The molecular orbital diagram for NO is similar to that of N2 and O2, with the sigma and pi orbitals in the following order:1, 2, 3, 1, 4, 2Now, let's fill in the 15 electrons in the molecular orbitals:1: 2 electrons2: 2 electrons3: 2 electrons1: 4 electrons 2 in each degenerate orbital 4: 2 electrons2: 3 electrons 2 in one degenerate orbital and 1 in the other The bond order can be calculated using the formula:Bond order = number of electrons in bonding orbitals - number of electrons in antibonding orbitals / 2In the case of NO:Bond order = 10 - 5 / 2 = 5 / 2 = 2.5The bond order of the nitric oxide molecule NO is 2.5.To calculate the bond length, we can use the empirical relation between bond order and bond length, which states that as bond order increases, bond length decreases. However, we don't have a specific formula to calculate the exact bond length of NO. Instead, we can refer to experimental data.The experimental bond length of NO is approximately 1.15 angstroms .