Page 1 of 1

### Question 4.13 a

Posted: Mon Nov 20, 2017 3:25 pm
In the homework question 4.13a, why would the shape of (I3)- be linear when it has two bonded pairs and three lone pairs on the central atom? Why would it not be bent?

### Re: Question 4.13 a

Posted: Mon Nov 20, 2017 4:12 pm
This is because its electron shape, taking into account the lone pairs, would be trigonal bipyramidal and i you keep removing the bonded pairs from that original shape we are down to the two opposite ends of the pyramid base giving us a final geometry of linear instead of bent. It has to do with the electron geometry.

### Re: Question 4.13 a

Posted: Mon Nov 20, 2017 4:13 pm
The I3- molecule has the VSEPR notation AX2E3.

Because it has 5 regions of electron density, begin by looking at the trigonal bipyramidal shape.

The bond angles in trigonal bipyramidal are 120o between the equatorial atoms, and only 90o for the axial atoms.
Because the lone pairs have more repulsion than the bonded electrons, the lone pairs go in the equatorial plane. That leaves the other I atoms in the axial position.

The final shape for the molecule is based on the actual bonded atoms (not all the regions of electron density), so the resulting shape of the molecule is linear.

There is one central I atom, two I atoms in the axial positions, and three lone pairs in the equatorial plane.

### Re: Question 4.13 a

Posted: Mon Nov 20, 2017 4:22 pm
if you check VESPR theory diagrams online you can see that if the VESPR notation is AX2E3 (or if it has 2 bonds and 3 lone pairs), the shape is linear.
this is the diagram I'm referring to. http://mrbeckschemistry.weebly.com/uplo ... 79.png?464