Asked by Hol
I'm drawing a Lewis structure for hydrogen cyanide using nitrogen with a valency of 3. Would I still need 2 dots after the N or would this represent nitrogen valency 5?
You need the two extra electrons for N at the end. Only the electrons involved BETWEEN the elements are bonding electrons; the others are used to fill orbtials but not for bonding purposes. Here is how you determine it.
H has 1 electron in the outer shell.
C Has 4 electrons in the outer shell.
N has 5 electrons in the outer shell.
10 electrons total. So ALL 10 electrons must be used. AND you try to place them such that two electrons are in the H sphere and 8 electrons for the other two elements.
<b>H:C:::N:</b>
The terminal (unpaired) electrons on N are used to make 8 electrons for N but those two are not involved in bonding. You see H has 1 bond, C has 4, and N has 3 (two dots, that is electrons) are equivalent to one (-) bond or 4 electrons are equivalent to a double (=) bond or 6 electrons are equivalent to a triple bond (but I can't draw a triple bond on the computer.
You need the two extra electrons for N at the end. Only the electrons involved BETWEEN the elements are bonding electrons; the others are used to fill orbtials but not for bonding purposes. Here is how you determine it.
H has 1 electron in the outer shell.
C Has 4 electrons in the outer shell.
N has 5 electrons in the outer shell.
10 electrons total. So ALL 10 electrons must be used. AND you try to place them such that two electrons are in the H sphere and 8 electrons for the other two elements.
<b>H:C:::N:</b>
The terminal (unpaired) electrons on N are used to make 8 electrons for N but those two are not involved in bonding. You see H has 1 bond, C has 4, and N has 3 (two dots, that is electrons) are equivalent to one (-) bond or 4 electrons are equivalent to a double (=) bond or 6 electrons are equivalent to a triple bond (but I can't draw a triple bond on the computer.
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