Fall 2015 Practice Midterm

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Maverick Tan 1F
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Joined: Fri Oct 07, 2016 3:02 am

Fall 2015 Practice Midterm

Postby Maverick Tan 1F » Tue Nov 01, 2016 4:05 pm

On pg. 216, Question 7, where it asks how many lone pairs are present in the structure of Ioratadine, how did we get 9 lone pairs?

Anmol Dhaliwal 2C
Posts: 25
Joined: Wed Sep 21, 2016 2:57 pm

Re: Fall 2015 Practice Midterm

Postby Anmol Dhaliwal 2C » Tue Nov 01, 2016 4:29 pm

In the structure, there is 2 Nitrogens and each have 3 bonds connected to it but Nitrogen's valence electrons is 5 not 3 so it would take a lone pair for both Nitrogens to fill their valence electrons (Thats 2 lone pairs). The 2 Oxygens connected only have 2 electrons connected to each and the valence of oxygen is 6 so it would take 2 lone pairs to fix both of them (So thats another 4 lone pairs (2 lone pairs for each) that becomes 6 total so far). Then the Cl has 7 valence electrons and only has one connected to it and needs 6 more which would be 3 pairs so thats the 6 lone pairs we have from the other Nitrogens and Oxygens combined with the lone pairs of Chlorine would be 9 lone pairs.

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Re: Fall 2015 Practice Midterm

Postby MayaKhalil_1L » Tue Nov 01, 2016 4:33 pm

Each atom in the molecule structure should have an octet( excluding the hydrogens). However, the carbons reach their octet by bonding with hydrogen atoms in the molecule, while the rest of the atoms need lone pairs to reach an octet.

Both the oxygens, each having only two bonds, need 2 sets of lone pairs, so that's 4
Then both the nitrogens, having three bonds, need one set of lone pairs so thats another 2
And finally the chlorine only has one bond and therefore needs 3 sets of lone pairs to reach its octet.

so in total 4+2+3= 9

Hope this helps!:)

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