Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals - Study Questions - Page 369b: 25


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Work Step by Step

Molecular orbital configuration: $Li_2:\ [core]\ (\sigma_{2s})^2$ $B_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\pi_{2p})^2$ $C_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\pi_{2p})^4$ $N_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4$ $O_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^2$ Since $B.O.=(N.\ Bonding -N.\ Antibonding)/2$, the bond orders are, respectively: 1,1,2,3,2, thus the shortest bond is in $N_2$ because it has the highest bond order.
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