2C. 3 part b

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Isaac Wen
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2C. 3 part b

Postby Isaac Wen » Wed Nov 11, 2020 10:07 pm


So problem 2c.3 part B wants us to draw Lewis structures for a hydrogen phosphate ion and the instruction manual shows four different structures (three with one double bond between O and P). My question is this: why can't we have three double bonds instead of only having one? The formal charges of all the O atoms would drop to 0 and P would have -2.

Yuelai Feng 3E
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Re: 2C. 3 part b

Postby Yuelai Feng 3E » Wed Nov 11, 2020 10:36 pm

Hi! So in the lewis structure you suggested, the sum of |formal charge| is 2, which is the same as the 3 structures in the solution manual. In this case, we need to compare the electronegativity of the atoms to determine which structure is more valid. Here, O is more electronegative than P, so O will attract more electrons and have a more negative formal charge than P. Therefore it's very unlikely that P has FC=-2 while O has FC=0. Hope it helps!

Jeffrey Hablewitz 2I
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Re: 2C. 3 part b

Postby Jeffrey Hablewitz 2I » Wed Nov 11, 2020 10:37 pm

While you could technically have double bonds between the oxygens and phosphorus, it is far less energetically favorable. Because oxygen is more electronegative than phosphorus the negative formal charges should be placed on the oxygen atoms. In general, when a molecule must have negative formal charges they should be placed on the most electronegative atoms. Hope this helps!

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