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    Which element has the same oxidation number in all of its known compunds?
    A. berylium
    B. chlorine
    C. nitrogen
    D. sulphur

    the answer is A. why?



    When a hot glass rod is placed in a gas jar of hydrogen iodide, there is an immediate reaction as the hydrogen idodide decomposes
    Which statements about this reaction are correct?

    1. Hydrogen Iodide is purple coloured
    2. The hot rod provides the activation energy
    3. One of the products is a solid

    The answer is 2 and 3

    I can see why the answer is 2, but I don't get 3

    I thought the hydrogen iodide decomposes as 2HI = H2(g) + I2(g)
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    Don't know about your first question but for the second one, I2 is solid, always has been, always will be, look at the size of it. Doesn't it form on the rod or something?
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    oh xhit..yep you're right.

    why did I think I2 was a gas? dumb dumb!!
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    (Original post by rlagksquf)
    Which element has the same oxidation number in all of its known compunds?
    A. berylium
    B. chlorine
    C. nitrogen
    D. sulphur

    the answer is A. why?



    When a hot glass rod is placed in a gas jar of hydrogen iodide, there is an immediate reaction as the hydrogen idodide decomposes
    Which statements about this reaction are correct?

    1. Hydrogen Iodide is purple coloured
    2. The hot rod provides the activation energy
    3. One of the products is a solid

    The answer is 2 and 3

    I can see why the answer is 2, but I don't get 3

    I thought the hydrogen iodide decomposes as 2HI = H2(g) + I2(g)
    Im not sure about this but this is what i think. Since Beryllium is the only metal there, it will always bond ionically. Therefore it always has to lose 2 electrons to form a stable EC. Therefore its oxidation number will always be +2. All the others are non metals and therefore could bond ionically or covalently which result in a different oxidation number each time.
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    thanks , that makes sense
 
 
 
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