For each of the MO valence electronic configurations of a homonuclear diatomic molecule ormolecular ion, identify the element, A, D and E and determine its MO bond order.(a) A2- (σ2s)2 (σ*2s)2 (π2px)2 (π2py)1(b) D2 (σ2s)2 (σ*2s)2(c) E22+ (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π*2px)1 (π*2py)1

(a) A2- (σ2s)2 (σ*2s)2 (π2px)2 (π2py)1 The element for A2- is Oxygen (O). The bond order is calculated as follows: Bond order = ½ [(Number of electrons in bonding MOs) - (Number of electrons in antibonding MOs)] = ½ [(2+2+2+1) - 2] = ½ [7 - 2] = 2.5

(b) D2 (σ2s)2 (σ*2s)2 The element for D2 is Helium (He). The bond order is calculated as follows: Bond order = ½ [(Number of electrons in bonding MOs) - (Number of electrons in antibonding MOs)] = ½ [(2) - 2] = 0

(c) E22+ (σ2s)2 (σ2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π2px)1 (π*2py)1 The element for E22+ is Beryllium (Be). The bond order is calculated as follows: Bond order = ½ [(Number of electrons in bonding MOs) - (Number of electrons in antibonding MOs)] = ½ [(2+2+2+2+2) - (2)] = ½ [10 - 2] = 4

(a) A2-: The molecule A2- has 7 valence electrons. The molecular orbital (MO) configuration suggests that it could be an oxygen molecule (O2-) because oxygen has 6 valence electrons and the extra electron from the negative charge makes it 7. The bond order is calculated as [(number of electrons in bonding MOs) - (number of electrons in antibonding MOs)] / 2 = [(2+2+2+1) - 2] / 2 = 2.5. So, the bond order is 2.5.

(b) D2: The molecule D2 has 4 valence electrons. The MO configuration suggests that it could be a molecule of Beryllium (Be2) because Beryllium has 2 valence electrons and there are two Beryllium atoms. The bond order is calculated as [(number of electrons in bonding MOs) - (number of electrons in antibonding MOs)] / 2 = [(2+2) - 2] / 2 = 1. So, the bond order is 1.

(c) E22+: The molecule E22+ has 8 valence electrons. The MO configuration suggests that it could be a molecule of Oxygen (O2) because Oxygen has 6 valence electrons and there are two Oxygen atoms, but the positive charge means it has lost 2 electrons, making it 8. The bond order is calculated as [(number of electrons in bonding MOs) - (number of electrons in antibonding MOs)] / 2 = [(2+2+2+2+2) - (2+2)] / 2 = 2. So, the bond order is 2.