Organic Chemistry 9th Edition

Published by Brooks Cole
ISBN 10: 1305080483
ISBN 13: 978-1-30508-048-5

Chapter 1 - Structure and Bonding - Exercises: 24

Answer

(a) $NH_{2}OH$ (b) $AlCl_{3}$ (c) $CF_{2}Cl_{2}$ (d) $CH_{2}O$

Work Step by Step

(a) Nitrogen atom (atomic number 7) has 5 valence electrons and can form 3 covalent bonds (by sharing its valence electrons) can thus complete its octet. In $NH_{?}OH$, Nitrogen is already bonded to one Oxygen molecule and x number of Hydrogens. To determine how many Hydrogen atoms will be bonded to Nitrogen, we subtract the bonds already formed by Nitrogen (except with hydrogen) from the total number of bonds it can form. i.e. 3-1=2. Hence, number of Hydrogen atoms attached to Nitrogen is 2. (b) Aluminium atom (atomic number 13) has 3 valence electrons and thus can form 3 bonds, it also shows an exception to the octet rule as it cannot complete its octet. In $AlCl_{?}$, there can be a maximum of ?=3 bonds that aluminium can form. (c) Carbon atom (atomic number 6) has 4 valence electrons and can thus form 4 covalent bonds to complete its octet. Thus in $CF_{2}Cl_{?}$, the balance number of bonds that Carbon can still form is 4-2=2. Hence, 2 Cl atoms will be bonded to carbon in this molecule. (d)Carbon atom (atomic number 6) has 4 valence electrons and can thus form 4 covalent bonds to complete its octet. Further, the Oxygen atom has a tendency to form 2 bonds to complete its octet. Thus in $CH_{?}O$, Oxygen atom forms a double bond with the Carbon atom and the balance number of bonds that Carbon can still form is 4-2=2. Hence, 2 H atoms will be bonded to carbon in this molecule.
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