## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals - 9-3 Molecular Orbital Theory - Applying Chemical Principles - Questions - Page 367: 4

#### Answer

See the answer below.

#### Work Step by Step

From the text: $IE=h\nu-KE$ $h.c/\lambda=3.40×10^{-18}\ J$ $IE=2.98×10^{-18}\ J$ $2.98×10^{-18}\ J\div 1.602×10^{-19}\ J/eV=18.6\ eV$ $2.98×10^{-18}\ J×6.022×10^{23}\ 1/mol=$ $1794.0\ kJ/mol$

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