Question 3.55 (Sixth Edition)

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Steve Magana 2I
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Question 3.55 (Sixth Edition)

Postby Steve Magana 2I » Sun Nov 04, 2018 12:11 pm

Question: Which of the following species are radicals? (a) NO2; (b) CH3; (c) OH; (d) CH2O.

Can someone explain to me what are radicals please? Thank you!

Maria Solis Disc 1G
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Re: Question 3.55 (Sixth Edition)

Postby Maria Solis Disc 1G » Sun Nov 04, 2018 12:25 pm

A radical is basically a molecule that has more than one unpaired electron. In this case the answer is OH since there can only be a single bond in OH, O must have 5 electrons, resulting in one being unpaired.

Maria Solis Disc 1G
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Re: Question 3.55 (Sixth Edition)

Postby Maria Solis Disc 1G » Sun Nov 04, 2018 12:26 pm

When you draw the lewis structures other of the answers given also seem to be radicals.

jessicahe4Elavelle
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Re: Question 3.55 (Sixth Edition)

Postby jessicahe4Elavelle » Sun Nov 04, 2018 12:28 pm

Radicals are molecules that contain one unpaired electron. To answer this question it would be best if you draw the Lewis structures for which one and see which molecule has an unpair electron. You can also identify them by the fact that their number of valence electrons is odd (meaning that at least one electron does not come in a pair).

angelagd3l
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Re: Question 3.55 (Sixth Edition)

Postby angelagd3l » Sun Nov 04, 2018 12:28 pm

Besides drawing the lewis structure is there another way to find a radical?

armintaheri
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Re: Question 3.55 (Sixth Edition)

Postby armintaheri » Sun Nov 04, 2018 12:36 pm

A radical is any molecule/atom that has unpaired valence electrons. Normally you want electrons to be either in bonds or in lone pairs. When an electron is not in a bond or a pair and it's just hanging out by itself, it's really reactive, and you call the molecule/atom a radical. You can draw Lewis structures to answer the question, or just think about it conceptually:


A) Nitrogen has 3 unpaired valence electrons. In NO2 all of them are in bonds with oxygen. So NO2 is not a radical.

B) Carbon has 4 unpaired valence electrons. Only three of them are in bonds with hydrogen, which leaves one unpaired and unbonded valence electron. So CH3 is a radical.

C) Oxygen has two unpaired valence electrons. Hydrogen can only form one bond. So there is going to be a single bond between O and H, which means one of oxygen's two valence electrons is unbonded and unpaired. So OH is a radical.

D) CH2O is formaldehyde. If you know what formaldehyde looks like then it's easy. If you don't, you should still be able to figure out that the most stable form is with carbon the middle, single-bonded to the hydrogens and double-bonded to the oxygen. Hydrogen has one unpaired valence electron, oxygen has two, carbon has four. With this arrangement, all of those valence electrons are in bonds. So CH2O is not a radical.

Alexa_Henrie_1I
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Re: Question 3.55 (Sixth Edition)

Postby Alexa_Henrie_1I » Sun Nov 04, 2018 7:08 pm

A faster way to find a radical is by counting how many electrons a molecule has. An odd number of electrons would signify a radical whereas an even number of electrons would signify paired electrons.


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