hey guys, I'm doing a old past paper for chem unit 1 (jan 2010) and i came across a questions for drawing CLF2+
i just wanna know why do you draw the 2 lone pairs at the top of the tetrahedral shape, and 2 flourine bond pairs at the bottom?
how do you determine where to draw the 2 lone pairs?
okay, but why do you draw the 2 lone pairs together at the top? why can't you have one lone pair on one side, and another lone pair on the other side, and have the bond pairs on the top and bottom?
but if i draw it like that (a lone pair on the left, lone pair on the right, and a bond pair on the top and a bond pair at the bottom) , then would the bond angles still be 104.5?
because if i draw the lone pairs on different sides, wouldn't they balance out and stay as 109.5?
its kind of hard to describe, but its question 6 jan 2010 AQA Chem unit 1 paper....
A molecule of ClF3 reacts with a molecule of AsF5 as shown in the following equation.
ClF3 + AsF5 → ClF2+ + AsF6-
Use your understanding of electron pair repulsion to draw the shape of the AsF5 molecule
and the shape of the ClF2+ ion. Include any lone pairs of electrons.
Name the shape made by the atoms in the AsF5 molecule and in the ClF2+ ion.
Predict the bond angle in the ClF2+ ion
In the mark scheme, they've only shown CLF2+ drawn one way (which is two lone pairs at the top and two bond pairs at the bottom), and the bond angle that they predicted is 104.5 which would make sense as the two lone pairs at the top causes the molecule to be bent, so 109.5 - 2.5 -2.5 = 104.5
but the thing i don't understand is how you know that the two lone pairs are at the top and bond pairs at the bottom (or the other way round)... why is it not a lone pair on the left side, lone pair on the right and a bond pair at the top and a bond pair at the bottom? i don't know if thats incorrect and i don't know how you know where to draw the lone pairs...
is there a rule to where you draw the lone pairs? and if CLF3 looses a F, which one of the flourines would you loose?