all one, draw arrows, answer question EnergyEIIIIdentify the groups of orbitals in the molecular orbital energy level diagram ofbromomethane, in the order | < II < III < IV Curved arrows are used to illustrate the flow of electrons. Using theprovided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic steps.Be sure to account for all bond-breaking and bond-making steps.:OH:Select to Add Arrows:Br:HH₂OH₂OVersion: 1.177.0+ production0:0HHHHSelect to Add Arrows:Br:H₂OBr:HHxxxXxHHHHSelect to Add ArrowsProbl

Note: Since you have posted multiple questions, we will provide the solution only to the first question as per our Q&A guidelines. Please repost the remaining questions separately.The molecular orbital diagram is a representation showing the distribution of electrons with the molecular orbitals of a molecule.The molecular orbital is formed when the atomic orbitals of the constituent atoms undergo an overlapping interaction. The relative energy values of molecular orbitals are based on the nature of the constituent atoms.The general energy order of the molecular orbitals is represented as follows-Antibonding molecular orbital >Non-bonding molecular orbital > Bonding molecular orbitalThe structure of bromomethane is shown as follows-During the formation of the given molecule, the atomic orbitals (sp3 hybrid orbitals) of carbon (C) combine with atomic orbitals of H to form three sigma bonding C-H molecular orbitals (σC-H) and three sigma antibonding C-H molecular orbitals (σ*C-H).Similarly, one atomic orbital (sp3 hybrid orbital) of carbon (C) combines with an atomic orbital of Br to form one sigma bonding C-Br molecular orbitals (σC-Br) and one sigma antibonding C-Br molecular orbital (σ*C-Br).The three lone pairs on the Br atom occupy the three non-bonding orbitals of Br (nBr)σC-Br orbitals have more energy than σC-H orbitals as the C-H bond is a more stable bond. On the other hand, the energy of σ*C-H is higher than that of σ*C-Br. Based on this, I, II, III, IV, and V in the given system can be identified as follows-I- σC-HII- σC-Br III- nBrIV- σ*C-BrV- σ*C-HThe energy order of the molecular orbitals can be represented as follows-σC-H< σC-Br < nBr< σ*C-Br < σ*C-HThe energy order of the molecular orbitals is-σC-H< σC-Br < nBr< σ*C-Br < σ*C-HOption D is the correct choice.