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    Can someone explain how to draw AsF5 and ClF2+.

    Like the way I have been taught is to look at the central atom

    For the first one As. Count the number of e- in outer shell.

    Which is 7 + 5( from the electrons fluorine gives),

    12/2 = 6. So square planar?

    But why is it trigonal bipyramidal. Like it works with PCl5 but not this?
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    No thats the completely wrong method
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    Basically firstly As has 5 outer shell electrons so draw the atom with the outer shell electrons somewhere on the page, then draw the Fluorine atoms on the As, simple basic covalent bonding diagram. Hence you can see that there are no free electrons on the atom so there are 5 bonding pairs. Let me know if you dont understand.
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    Use VSEPR.
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    (Original post by Zakria123)
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    Doesn't As have seven electrons on outer shell? And there are 5 bonded so what happens to the lone pair?
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    (Original post by Super199)
    Doesn't As have seven electrons on outer shell? And there are 5 bonded so what happens to the lone pair?
    OH FFS dw I thought it was At
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    (Original post by Super199)
    Doesn't As have seven electrons on outer shell? And there are 5 bonded so what happens to the lone pair?
    As has 5 outer shell electrons
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    Anyways don't use that method you were using, if they give anything even a tad harder I think you will mess up
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    (Original post by Super199)
    Doesn't As have seven electrons on outer shell? And there are 5 bonded so what happens to the lone pair?
    There appears to be a lot of mis-information in this thread.

    If the original molecule is AsF5 then you are dealing with arsenic from group 5 (15) NOT astatine from group 7 (17).

    There are 5 electrons from the central atom plus one from each fluorine = 10 electrons = 5 electron domains, all of which are used for bonding.

    The molecule is trigonal bipyramidal.
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    (Original post by Zakria123)
    Anyways don't use that method you were using, if they give anything even a tad harder I think you will mess up
    With ClF2+ . How do you know there are 2 lone pairs left?
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    (Original post by charco)
    There appears to be a lot of mis-information in this thread.

    If the original molecule is AsF5 then you are dealing with arsenic from group 5 (15) NOT astatine from group 7 (17).

    There are 5 electrons from the central atom plus one from each fluorine = 10 electrons = 5 electron domains, all of which are used for bonding.

    The molecule is trigonal bipyramidal.
    Yeah I misread it as astatine for some reason lol
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    (Original post by Super199)
    With ClF2+ . How do you know there are 2 lone pairs left?
    Your method is fundamentally correct.

    Chlorine = 7 electrons
    2 x F = 2 electrons
    subtract one for the positive charge
    Total electrons = 8 = four pairs, two of which are used for bonding leaving two lone pairs.
    Electron domains = 4 = tetrahedral
    Ionic shape = bent or angular with two lone pairs.
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    (Original post by charco)
    Your method is fundamentally correct.

    Chlorine = 7 electrons
    2 x F = 2 electrons
    subtract one for the positive charge
    Total electrons = 8 = four pairs, two of which are used for bonding leaving two lone pairs.
    Electron domains = 4 = tetrahedral
    Ionic shape = bent or angular with two lone pairs.
    Yeah got it cheers
    Do you mind helping me with another qs:

    An ultrasound imaging agent has the formula C4F10It can be made by the reaction of butane and fluorine as shown in the followingequation.C4H10 + 10F2 →C4F10 + 10HFCalculate the percentage atom economy for the formation of C4F10 in this reaction.Give your answer to three significant figures.
    Where does 238 come from?
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    (Original post by Super199)
    Yeah got it cheers
    Do you mind helping me with another qs:

    An ultrasound imaging agent has the formula C4F10It can be made by the reaction of butane and fluorine as shown in the followingequation.C4H10 + 10F2 →C4F10 + 10HFCalculate the percentage atom economy for the formation of C4F10 in this reaction.Give your answer to three significant figures.
    Where does 238 come from?
    I can't see a 238 in the question... :dontknow:
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    (Original post by charco)
    I can't see a 238 in the question... :dontknow:
    In the ms. I dont see where its from?
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    (Original post by Super199)
    In the ms. I dont see where its from?
    ... there was no mention of a mark scheme!

    238 is the relative mass of C4F10
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    (Original post by charco)
    Your method is fundamentally correct.

    Chlorine = 7 electrons
    2 x F = 2 electrons
    subtract one for the positive charge
    Total electrons = 8 = four pairs, two of which are used for bonding leaving two lone pairs.
    Electron domains = 4 = tetrahedral
    Ionic shape = bent or angular with two lone pairs.
    Hi,

    I use this method myself and works fine but how would you determine the shape if you are given a formula such as NH3? Im unsure how this works since either N or H could be the central element?
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    (Original post by charco)
    I can't see a 238 in the question... :dontknow:
    In the ms?
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    In this case N is the central atom with 3 H bonded to it. It would form 3 bonded pairs and 1 lone pair... So it's either trigonal pyramidal/ terahedral depending on what shapes you learnt.
 
 
 
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