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Chemistry 13 Online
OpenStudy (anonymous):

Draw Lewis structures for each of the following moecules. Show resonance structures, if they exist A) O2 B) N2 C)CO D)SO2 Draw lewis structures for each of the fllowing polyatomic ions. Show resonanace structures, if they exist. A)OH- B)H3C2O2- C)BrO3- Determine the type of hybrid orbitals form by the boron atom in a molecule of boron fluoride, BF3.

OpenStudy (anonymous):

@Rea201 @red12534

OpenStudy (anonymous):

fllowing is following

OpenStudy (abb0t):

|dw:1384825442293:dw|

OpenStudy (abb0t):

Draw the bonds for each, making sure to follow the octet rule!

OpenStudy (anonymous):

the octet rule is 8 dots? Not sure

OpenStudy (abb0t):

Those little dots are electrons, e\(^-\)

OpenStudy (anonymous):

Do you do that to all of them.

OpenStudy (abb0t):

Yes, it means that each atom has 8 electrons. So, draw a bond

OpenStudy (abb0t):

|dw:1384825557842:dw|

OpenStudy (anonymous):

So like O2 will have 6? and 2 oxide with 6 ?

OpenStudy (anonymous):

which makes 12?

OpenStudy (abb0t):

drawing a bond between them, they went from six electrons each to now 7!

OpenStudy (abb0t):

And, remember octet rule? They must have 8 e\(^-\). Meaning, what do you need? Well, that means you need another bond!

OpenStudy (abb0t):

|dw:1384825639766:dw|

OpenStudy (abb0t):

Therefore, O\(_2\) should look like this \(\sf \huge\color{red}{:ö=ö:}\)

OpenStudy (abb0t):

Now, follow the same process with the other compounds, making sure that each element has 8 e\(^-\)

OpenStudy (anonymous):

so then for N2 it would be like this |dw:1384825809477:dw|

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