CH 4- 4.1 HW Prob.

(Polar molecules, Non-polar molecules, etc.)

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duenezjuleny1D
Posts: 48
Joined: Fri Sep 28, 2018 12:24 am

CH 4- 4.1 HW Prob.

Postby duenezjuleny1D » Fri Nov 16, 2018 1:14 am

How is it that some shapes can have lone pairs and others must have lone pairs? For example in problem 4.1 (6th edition) there are ball-and-stick models of two molecules, it asks to indicate whether there MUST be, MAY be, or CANNOT be one or more lone pairs of electrons on the central atom. How exactly is that determined?

gwynlu1L
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

Re: CH 4- 4.1 HW Prob.

Postby gwynlu1L » Fri Nov 16, 2018 4:07 am

For shapes like see saw or trigonal pyramidal, you know there has to be a lone pair in order to make those shapes. Without the lone pair, the shape wouldn't make sense since there would basically be a gap in the shape. However, for a shape like bent, it is possible that the VSEPR formula is AX2, or AX2E3 etc. In other words, say the arrangement of the molecule would be trigonal bipyramidal, with 5 regions of electron density. However, there are 3 lone pairs. In this example, the 3 lone pairs would be placed on the plane of the molecule, resulting in a shape that is linear. Therefore, unless you know the formula of the molecule or the VSEPR formula, you can' determine if a linear shape has or does not have lone pairs.

Ashley Kim
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Joined: Fri Sep 28, 2018 12:19 am
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Re: CH 4- 4.1 HW Prob.

Postby Ashley Kim » Fri Nov 16, 2018 10:01 am

Some shapes must have lone pairs because the angle is determined by the lone pair and bonding pair repulsion. Others, like the linear shape, may or may not have lone pairs depending on the structure.


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