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    from 9701_w11_qp_11 (international a level):
    Q4.Methyl isocyanate, CH3NCO, is a toxic liquid which is used in the manufacture of somepesticides.In the methyl isocyanate molecule, the sequence of atoms is H3C—N C O.What is the approximate angle between the bonds formed by the N atom?A104°B109°C120°D180°

    MS says it's C, however i don't quite understand.
    I figured that Nitrogen would have 2 lone pairs left over, therefore it will have a tetrahedral shape (repulsion between the 2 C's and the 2 lone pairs), resulting in 104 degrees.
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    (Original post by TeachChemistry)
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    The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.
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    (Original post by Jpw1097)
    The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.
    Since 117 is not an option in the given answers your reasoning is, although correct, irrelevant.
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    (Original post by Jpw1097)
    The angle wouldn't be 120° though and the shape wouldn't be trigonal planar. The lone pair would repel the electrons in the bonding pairs more and so the angle would be around 117/118° and the shape would be considered bent.
    Your suggestion seems sensible, but it turns out that angle is actually 125o

    Int VSEPR fab?
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    (Original post by TeachChemistry)
    Since 117 is not an option in the given answers your reasoning is, although correct, irrelevant.
    I realised...
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    (Original post by Pigster)
    Your suggestion seems sensible, but it turns out that angle is actually 125o

    Int VSEPR fab?
    Perhaps that's due to the high e- density in the N=C bond? The electrons in the double bond might repel the electrons in the lone pair and the bonding pair, resulting in an angle of 125°.
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    (Original post by Jpw1097)
    Perhaps that's due to the high e- density in the N=C bond? The electrons in the double bond might repel the electrons in the lone pair and the bonding pair, resulting in an angle of 125°.
    Clearly the inter-domain repulsion depends on the relative electron densities of the domains and their relative distance from the nucleus.

    This makes applying VSEPR a bit of a hand wavy exercise, particularly coupled with the fact that it falls apart in period 3 (and above) when the domains used in bonding are now much further from the nuclear center AND there is often contribution from 'd' orbitals.

    .. is all.
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    The equivalent molecule: SiH3NCO has a bond angle of 180o.

    The lone pair on the N becomes a second bond to the Si, in a p(pi)d(pi) style.

    H3Si-=N+=C=O
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    (Original post by Pigster)
    The equivalent molecule: SiH3NCO has a bond angle of 180o.

    The lone pair on the N becomes a second bond to the Si, in a p(pi)d(pi) style.

    H3Si-=N+=C=O
    interesting.
 
 
 
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