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When does Boron have less than an octet? In problem 4.81, it asks to draw the lewis structure of borazine (B3N3H6). The solution shows alternating double bonds between boron and nitrogen and a charge of +1 on N and -1 on B. If the structure is drawn with all single bonds and a lone pair on the Nitrogen, the formal charges would be 0, but boron wouldn't have an octet. Would this be an incorrect structure?
The single bond structure is correct, but it is stabilized by its other resonance structures (the double bonded structures). Nitrogen has a lone pair, and Boron has an empty orbital. The N dumps its electrons in the empty orbital of B to form the double bond. The resonance structures helps to stabilize the overall molecule.
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