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?
Textbook Problem Chp 4 #9
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Re: Textbook Problem Chp 4 #9
Trigonal planar is AX3 while the ICl3 molecule is AX3E2 which is what T- shaped is.
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Re: Textbook Problem Chp 4 #9
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.
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Re: Textbook Problem Chp 4 #9
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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