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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Capital on October 07, 2012, 11:18:11 PM

Title: Bond angles
Post by: Capital on October 07, 2012, 11:18:11 PM

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My answer to part b is 109.5 degrees because there are two lone pairs (not shown) and two attached atoms to the central atoms. The parent shape of H-O-C is tetrahedral and the shape is bent and all the angles in tetrahedral are 109.5 degrees

My answer to part c is also 109.5 degrees based on the same reasoning (one lone pair and three attached atoms to C because C must follow the octet rule) Parent shape of O-C-O must also be tetrahedral and the shape is trigonal pyramidal.

However, my answer to part b is correct but my answer to part c is wrong. According to the answer key, the bond angle of O-C-O is 120 degrees. Why? Is my reasoning incorrect?

Title: Re: Bond angles
Post by: Hunter2 on October 08, 2012, 01:03:45 AM

The carbon in COOH  is sp² hybridized. So it is 120 °.

Title: Re: Bond angles
Post by: Borek on October 08, 2012, 03:52:50 AM

Note that carbon marked as C is only trivalent, and the oxygen attached to it is monovalent (the one without H atom next to it). That should tell you the formula is incorrect. It should be a double bond between C and O.

Title: Re: Bond angles
Post by: AWK on October 08, 2012, 07:51:46 AM

Quote from: Hunter2 on October 08, 2012, 01:03:45 AM

The carbon in COOH  is sp² hybridized. So it is 120 °.

This is only an approximation.
The real angles may differ from idealized ones  even by 5-6 degrees (angle O-C-O usually is about 123-124 degree, the angle C-C=O is greater than C-C-O).
The angle at O in C-O-H is about 105 degres, and the angle at protonized tertiary nitrogen is usually a bit different than 109 degree.