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Sodium in water

  1. May 3, 2005 #1


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    This isn't really homework, but I'm trying to fully understand the reaction between sodium metal (Na) and water (H2O). Here's the equation:

    2Na + 2H2O -> 2NaOH + H2

    And here's the steps I think the process goes through:

    1) Na has a single electron in its valence shell. This electron is quite a long way from the nucleus so it has a weak attraction to the Na atom.

    2) H2O is a polar molecule with the O atom being slightly negative and the 2 H atoms being slightly positive.

    3) I've read that when Na is placed in water it becomes a positive ion (Na+). I assume this is because it loses its valence electron, but why? What pulls the valence electron away? I guess its the slightly positive H atoms in the H2O molecule. If this is true, does this cause one of the H atoms in the H2O molecule to break free? If so, does the now free H atom have 0, 1 or 2 electrons? I'm thinking it has just the 1 electron leaving the OH molecule a negative ion (OH-).

    4) Now at some point the Na+ ion bonds with OH to form NaOH. I'm thinking that when the H atom breaks free from the H2O molecule (point 3) it leaves OH-, which then bonds with the Na+ ion. When they bond you end up with a neutral NaOH molecule.

    5) The H atom, which broke free in step 3, has a single electron and forms a bond with another H atom (H2).

    6) The heat from the reaction is enough to ignite the hydrogen and oxygen mix that forms above the above.

    Does this sound correct or have I totally missed the point?

    Thanks for any guidance.
  2. jcsd
  3. May 3, 2005 #2
    The Sodium wants to lose the electron to stabilize it's outer electron shell. The shell for that electron wants to have either 0 electrons or 8 to satisfy it. It is easier for Sodium to give up, an electron than it is to gather 7 more. Because sodium has one electron in it's outer shell it is highly reactive.

    When the sodium reacts with the water it takes the place of one of the hydrogen atoms. This happens because sodium is more reactive than the hydrogen it is replacing. Reactivity is largely due to the atomic radius of an element and the valence. Larger metals lose their outer electrons more easily. If the sodium was less reactive than the hydrogen then there probably would be no chemical reaction in this case.

    The hydrogen would form dimers, bonded pairs. This is why your equation has 2Na and 2HOH. Each hydrogen atom in the dimer would contain 1 electron (that they share with the other.)

    Sounds like you pretty much got it.
    Last edited: May 4, 2005
  4. May 4, 2005 #3


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    Screw sodium, I want to see the cesium.

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