In the benzene molecule, carbon atoms form a ring with alternating single and double bonds in between each of them. Each carbon atom forms one σ bond with one carbon atom and one σ and one π bond with another. The π bonding electrons are delocalized over the entire molecule. Thus, benzene is a resonance hybrid of two equivalent structures, and the single and double bonds oscillate from one position to the other. Required:
Calculate the bond order for one carbon-carbon bond in the benzene molecule, taking σ and π bonding into consideration.

1.5

Explanation:

From;

BO = ½(B – A)

BO= Bond order

B= Number of bonding electrons

A = Number of antibonding electrons

For the sigma bonds, there are 2 bonding electrons and 0 nonbonding electrons.

σ BO = ½(B – A) = ½(2 – 0) = 1

π bond order for 6 C-C bonds;

π BO = ½(B – A) = ½(6 – 0) = 3

For each  π bond, BO = 3/6 = 0.5.

Hence, for a single C-C bond in benzene, the total Bond order = σ + π = 1 + 0.5 = 1.5.

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."