7th Edition 2E #21 part d

(Polar molecules, Non-polar molecules, etc.)

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Adrian C 1D
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Joined: Fri Sep 28, 2018 12:19 am

7th Edition 2E #21 part d

Postby Adrian C 1D » Tue Nov 13, 2018 6:19 pm

Part D requires you to draw the lewis structure and give the approximate bond angle of the molecule N2H4.
I drew the correct lewis structure but I could not figure out how the bond angle would be 107 degrees and have the molecular geometry of a trigonal pyramidal.
Can someone please explain.

Andonios Karas 4H
Posts: 30
Joined: Fri Sep 28, 2018 12:27 am

Re: 7th Edition 2E #21 part d

Postby Andonios Karas 4H » Tue Nov 13, 2018 7:35 pm

Each N atom, which has 5 valence electrons, is bonded to 3 atoms: 2 H and 1 N (each with only single bonds).
Now, we must count how many lone pairs and bonding pairs there are for each N atom which comes out to be 3 bonding pairs and 1 lone pair.
The 1 lone pair will repel the bonding pairs away from it.
The 3 bonding pairs will arrange themselves symmetrically.
Looking at Figure 2E.7, we can use this information to get the correct shape.

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