4.13c Bond Angle Question  [ENDORSED]

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Mike Matthews 1D
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4.13c Bond Angle Question

Postby Mike Matthews 1D » Wed Nov 15, 2017 12:59 am

In question 4.13c it asks you to determine the bond angle for IO3-. After determining the Lewis structure, I found the VSEPR formula for this molecule to be AX3E. That being said, the shape of the molecule is trigonal pyramidal. However, I'm confused about how to determine the bond angle. The solutions manual says that the bond angle should be less than 109.5 degrees. However, I thought the 109.5 degree bond angle was only for tetrahedral molecules. Could someone please explain how I would know that the bond angle for this molecule is less than 109.5 degrees.

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Re: 4.13c Bond Angle Question  [ENDORSED]

Postby Chem_Mod » Wed Nov 15, 2017 8:04 am

The trigonal pyramidal shape is like a tetrahedral shape with one position occupied by a lone pair (or radical, as the case may be). Thus, the bonding angles are still approximately 109.5 degrees. To be precise, because the lone pair repels against the nearby bonds, the bond angles are slightly less than 109.5 degrees in this case.

Anna Goldberg 2I
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Re: 4.13c Bond Angle Question

Postby Anna Goldberg 2I » Wed Nov 15, 2017 10:11 am

In question 73, though, how are you able to determine a more exact answer for the bond angle in a shape such as triangular pyramidal or bent? Since it asks you to rank the bond angles in increasing size in various molecules, these approximates are difficult to rank.

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Re: 4.13c Bond Angle Question

Postby Humza_Khan_2J » Thu Nov 16, 2017 6:10 pm

For question 73, I believe it's more qualitative. We know the base geometry, but then have to change it depending on the number of lone pairs(bond angles generally get lesser as lone pairs increase). We can assume that ones with the lesser "initial" bond angles based on the base geometry(assuming there's no lone pairs) will have lesser angles after lone pairs are added, unless the shape turns out to be linear.

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