a) Draw the Lewis structures for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). b) Explain how these Lewis structures are consistent with the relative enthalpies of atom combination for \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\).

Short Answer

Expert verified
The Lewis structure of \(\mathrm{N}_{2}\) is \(\mathrm{::N::=::N::}\) and of \(\mathrm{O}_{2}\) is \(\mathrm{::O::=::O::}\). These structures indicate a triple bond in \(\mathrm{N}_{2}\) and a double bond in \(\mathrm{O}_{2}\). The greater bonding energy in \(\mathrm{N}_{2}\) explains the higher enthalpy of atom combination for nitrogen as compared to oxygen.

Step by step solution

01

Draw the Lewis Structure for N2

Each nitrogen atom has 5 valence electrons. In \(\mathrm{N}_{2}\), each nitrogen atom shares 3 electrons with the other, forming a triple bond. The Lewis structure can be shown as: \\[ \mathrm{::N::=::N::} \\] Each line represents a pair of shared electrons, the triple bond represents 6 shared electrons, and the remaining 4 electrons (2 on each atom) are unshared.
02

Draw the Lewis Structure for O2

Each oxygen atom has 6 valence electrons. In \(\mathrm{O}_{2}\), each oxygen atom shares 2 electrons with the other forming a double bond. Each atom also has two lone pairs. The Lewis structure can be shown as: \\[ \mathrm{::O::=::O::} \\] Each line represents a pair of shared electrons, the double bond represents 4 shared electrons, and the remaining 8 electrons (4 on each atom) are lone pairs.
03

Explain the Relation to Enthalpies of Atom Combination

The enthalpy of atom combination is related to the energy required to break the bonds between the atoms. The triple bond in \(\mathrm{N}_{2}\) is stronger and requires more energy to break than the double bond in \(\mathrm{O}_{2}\). Thus, the enthalpy of atom combination for \(\mathrm{N}_{2}\) is higher than for \(\mathrm{O}_{2}\).

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