What is the shape of molecule? Watch

Bustamove
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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?
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EierVonSatan
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It doesn't matter which you draw at the top and bottom, so long as the shape is correct
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LadyEcliptic
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The lone pair of electrons have a greater charge density and therefore are more repulsive, to they will push bond pairs closer together - but as long as you have it in a tetrahedral structure it shouldn't matter
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Bustamove
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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?
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BJack
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(Original post by Bustamove)
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?
As EVS said, it doesn't matter where you draw the lone pairs as long as the structure is clearly tetrahedral.
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EierVonSatan
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(Original post by Bustamove)
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?
You can, But that's a fair bit harder to draw since you can't draw lone pairs going in or out of the plane of paper, unlike bonds.
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Bustamove
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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?
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EierVonSatan
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(Original post by Bustamove)
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?
You don't have to draw the molecule to scale :confused:
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Bustamove
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its kind of hard to describe, but its question 6 jan 2010 AQA Chem unit 1 paper....

question 6
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?
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