HW 4.9

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

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alexagreco1A
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HW 4.9

Postby alexagreco1A » Mon May 28, 2018 6:13 pm

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

I thought that this molecule would have a trigonal planar arrangement, as it is AX3, but the answer key says that it is t-shaped. How could this be?

Solene Poulhazan
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Re: HW 4.9

Postby Solene Poulhazan » Mon May 28, 2018 7:40 pm

This is t-shaped because there are 5 electron groups, 3 bonding groups and 2 lone pairs. A trigonal planar arrangement only has 3 electron groups.

Gianna Graziano 1A
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Re: HW 4.9

Postby Gianna Graziano 1A » Tue May 29, 2018 9:16 pm

Hi Alexa,
If this molecule were trigonal planar, it would have only 3 bonds and zero LP. As Sloene stated, molecules that are t-shaped have either 5 bonds and 2 lone pairs OR 6 bonds and 3 lone pairs. In both of these cases the angle is less than 90 degrees.

Jordanmarshall
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Re: HW 4.9

Postby Jordanmarshall » Wed May 30, 2018 11:58 am

Why does ICl3 have 5 electron groups?

RubyLake1F
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Re: HW 4.9

Postby RubyLake1F » Wed May 30, 2018 1:29 pm

The I in ICL3 has 5 regions of electron density around it because the 3 chlorines each contribute 7 valence e-, and the I also contributes 7e- for a total of 28 e-. When you distribute electrons around the chlorines to fill their valence shells, you have only used 24 and have 4 leftover which go as two lone pairs on the central Iodine. This means that I has three bonding pairs (to the three chlorines) and two lone pairs, giving it a formal charge of zero (7-3-4=0).
Draw the Lewis structure to find the number of regions of electron density around the central atom.

Isabelle De Rego 1A
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Re: HW 4.9

Postby Isabelle De Rego 1A » Wed May 30, 2018 1:30 pm

Jordanmarshall wrote:Why does ICl3 have 5 electron groups?

ICl3 has 5 electron groups because of the two lone pairs. When we are counting the groups we don't only count the atoms. Remember that we are counting regions of "high electron density", so atoms and lone pairs are both included in this description. Draw out the Lewis structure, that will give you a better visual!

Sarah Brecher 1I
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Re: HW 4.9

Postby Sarah Brecher 1I » Wed May 30, 2018 6:08 pm

I thought it was trigonal bipyramidal because there are 5 regions of electron density. What is the difference between trigonal bipyramidal and T-shaped then?

Chem_Mod
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Re: HW 4.9

Postby Chem_Mod » Wed May 30, 2018 6:23 pm

Trigonal bipyramidal shape results from a VSEPR formula of AX5

T-shaped results from a VSEPR formula of AX3E2

Both have 5 regions of electron density

Corryn Doll
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Re: HW 4.9

Postby Corryn Doll » Wed May 30, 2018 6:33 pm

due to there being 5 bonding pairs, it would seem as though the shape would result in a triagonal bipyramidal, however, when looking at the VSEPR formula AX3E2, we find that the shape is really T-shaped.

If your not sure about the shape of something, make it a habit to always check the VSEPR model. If the shape was really triagonal bipyramidal, the VSEPR formula would be AX5.

Nicole Shak 1L
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Re: HW 4.9

Postby Nicole Shak 1L » Wed May 30, 2018 7:16 pm

Are we expected to memorize these bonding angles? How do you know that the bonding angle is slightly less than 90 degrees?

AnnaYan_1l
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Re: HW 4.9

Postby AnnaYan_1l » Thu May 31, 2018 11:23 am

Nicole Shak 1L wrote:Are we expected to memorize these bonding angles? How do you know that the bonding angle is slightly less than 90 degrees?


If there are lone pairs in a molecule, this creates repulsion, pushing certain atoms in the molecule "away," which slightly alters the bond angles that would exist if the lone pairs were replaced with actual atoms.
That is, in a tetrahedral shaped molecule (4 regions of electron densities: 4 bonding pairs) like , all H-C-H angles are 109.5 degrees. In a trigonal pyramidal shaped molecule (4 regions of electron densities: 3 bonding pairs, 1 lone pair) like , the bond angles are less than 109.5 degrees (specifically, 107 degrees.

However, we won't be expected to know the exact bond angles if we get a molecule with long pairs, as it differs slightly between molecules (for example, has bond angles of 107 degrees between H-N-H bonds, while has bond angles of 106 degrees between the O-S-O bonds). Instead, we'll just be expected to know that the angles are less than what you'd expect for a molecule with some number of electron densities but lone pairs than a molecule with the same number of electron densities but atoms.

Hope that helps! Let me know if what I said was confusing at all.


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