To calculate the bond order and bond length of the N-O bond in nitric oxide NO using molecular orbital theory, we first need to determine the molecular orbital configuration of NO.Nitrogen N has an atomic number of 7, so it has 7 electrons. Oxygen O has an atomic number of 8, so it has 8 electrons. Therefore, the NO molecule has a total of 15 electrons.The molecular orbitals for NO can be represented as follows:1. 1s 2. * 1s 3. 2s 4. * 2s 5. 2pz 6. 2px = 2py 7. * 2px = * 2py 8. * 2pz Now, we can fill the molecular orbitals with the 15 electrons, starting from the lowest energy level:1. 1s - 2 electrons2. * 1s - 2 electrons3. 2s - 2 electrons4. * 2s - 2 electrons5. 2pz - 2 electrons6. 2px = 2py - 4 electrons 2 in each 7. * 2px = * 2py - 1 electron in * 2px The bond order can be calculated using the formula:Bond order = number of electrons in bonding orbitals - number of electrons in antibonding orbitals / 2In this case, there are 10 electrons in bonding orbitals 2 in 1s , 2 in 2s , 2 in 2pz , and 4 in 2px and 2py and 5 electrons in antibonding orbitals 2 in * 1s , 2 in * 2s , and 1 in * 2px . So, the bond order is:Bond order = 10 - 5 / 2 = 5 / 2 = 2.5The bond order of the N-O bond in nitric oxide NO is 2.5.As for the bond length, it is experimentally determined and cannot be directly calculated using molecular orbital theory. However, we can infer that a higher bond order generally corresponds to a shorter bond length and a stronger bond. The experimental bond length of the N-O bond in nitric oxide NO is approximately 115 picometers pm .