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: 23

Answer

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Work Step by Step

$O_2^{2-}$ has 14 valence electrons so the configuration is: $[core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^4$ $O_2$ has 12 valence electrons so the configuration is: $[core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^2$ A) Since there are no unpaired electrons, the ion is diamagnetic while the molecule has two and is paramagnetic. B) Net bonds 1 $\sigma$ for both and 1 $\pi$ for the molecule. C) Since $B.O.=(N.\ Bonding -N.\ Antibonding)/2$ This ion has a bond order if 1 and the molecule, 2. D) Since bond length decreases with bond order, the molecule has a shorter bond.
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