Textbook Problem Chp 4 #9

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

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Jade Fosburgh Discussion 2C
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Textbook Problem Chp 4 #9

Postby Jade Fosburgh Discussion 2C » Sun Nov 12, 2017 3:51 pm

4.9 (a) What is the shape of an ICl3 molecule (iodine is the central atom)

I drew the Lewis structure and made sure the formal charge was equal to zero for all atoms. Iodine had 2 lone pairs of e- and 1 single bond to each Cl. Therefore, since there were 5 regions of electron density, the electron arrangement was trigonal bipyramidal. However, we must account for the two lone pairs. The solutions said the molecule would be T-shaped, with an angle of <90 degrees. However, why wouldn't it be a trigonal planar shape, with a bond angle of <120 degrees? I imagined the lone pairs being arranged on opposite sides of the atom. Wouldn't that be more favorable since the e'-s would be farther apart?

Bansi Amin 1D
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Re: Textbook Problem Chp 4 #9

Postby Bansi Amin 1D » Sun Nov 12, 2017 4:07 pm

Trigonal planar is AX3 while the ICl3 molecule is AX3E2 which is what T- shaped is.

Janine Chan 2K
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Re: Textbook Problem Chp 4 #9

Postby Janine Chan 2K » Sun Nov 12, 2017 4:11 pm

In a trigonal planar shape, there are only 3 bonding pairs surrounding the central atom. As you drew in your Lewis structure, ICl3 has 3 bonding pairs and 2 lone pairs, which would make it T-shaped.

Emily Mei 1B
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Re: Textbook Problem Chp 4 #9

Postby Emily Mei 1B » Sun Nov 12, 2017 4:27 pm

The shape is trigonal planar because you determine molecular shape by the number of bonding pairs on the central atom. When you then factor in the lone pairs, they bonded e- and lone pair e- repel each other and the molecular shape becomes T-shaped, changing the angle to less than 90º.

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