Problem

Explain how ammonia can function both as a Brønst…

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Problem 60 Medium Difficulty

For each of the following, indicate the hybridization of the nitrogen atom (for $\mathrm{N}_{3}^{-},$ the central nitrogen).
(a) $\mathrm{N}_{2} \mathrm{F}_{4}$
(b) $\mathrm{NH}_{2}-$
(c) $\mathrm{NF}_{3}$
(d) $\mathrm{N}_{3}-$

Answer

a. In dinitrogen tetrafluoride, each nitrogen atom shows $s p^{3}$ hybridization. Out of four $s p^{3}$
hybridized orbitals, one $s p^{3}$ orbital is involved in bonding with nitrogen, two $s p^{3}$ orbitals are
involved in bonding with the two fluorine atoms and one $s p^{3}$ orbital contains the lone pair of
electrons.
b. In $\mathrm{NH}_{2}^{-},$ nitrogen atom shows sp'hybridisation, Out of four $s p^{3}$ hybridized orbitals, two $s p^{3}$
orbitals are involved in bonding with nitrogen, one $s p^{3}$ orbital is involved in bonding with the
hydrogen atom and one $s p^{3}$ orbital contains the lone pair of electrons.
c. In nitrogen trifuoride, nitrogen atom shows sp" hybridisation. Out of four $s p^{3}$ hybridized orbitals,
three $s p^{3}$ orbitals are involved in bonding with the three fluorine atoms and one $s p^{3}$ orbital
contains the lone pair of electrons.
d. In azide-ion, central nitrogen atom is found in sp hybridisation. The two sp orbitals are involved
in the bonding with the other two nitrogen atoms.

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Video Transcript

to find the hybridization of the nitrogen atoms and these compounds, we first need to draw the Lewis structures. Remember that nitrogen typically makes three bombs. Um, Hydra. Mr. Bill, you make one the headlights, we'll make one and thing, that's all. We're gonna be staying here. So for 10 to 4, remember, we're gonna have these molecules be symmetrical as possible. We're gonna have to nitrogen each connected two florins, and then each of them is going to each of the nitrogen will have to non run in pairs and then the flooring of all their remaining non bonding electrons. So if we look at each of the nitrogen Hey, each have four electron regions. In each case, we have the three bonding pairs and one on bombing pear. Because we have four electron regions that Mrs necessitates for orbital's. So each nitrogen, his S P B hybridize. Next for any two, we're gonna have nitrogen at the center on there, too. 200 Adam, and then the nitrogen is gonna have to non bony affairs. The extra electron came from the negative charge so that we're all this is Louis structure again. Notice that we have four electron regions again in that means that even though I now have morn on bombing pears, the nitrogen is still at P three. Advertise next for them. Three. We have nitrogen in the center, connected to three foreign items. We're still in the non bombing pears like so who can? The nitrogen has four electron region truth that it is best P three hybridize wants more. Finally, we look at a site which is three night train. Adam, work this one out a little bit more carefully. Each nitrogen bring five aliens electrons. So I have three. Touch five this 15. You know, the electron, The negative charge gives an additional electron. So my structures, we have 16 electrons, place three nights, visions in a row. Connect him with, uh, single bond start. And as I worked through this Louis structural fine that this necessitates double bonds each nitrogen and then to non bonding pairs on the outer nitrogen kind of the electrons. You should have 16 and the structure has in overall charge of minus four. Now look at the nitrogen in the center. It has to electron regions. I know it's still has four bombs, but a double bond only counts as a single electron researcher. For this reason, it wouldn't need to Orbital's in the hybrid or girls, and thus it is S p hybridize.

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