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Reactivity of halogens

  1. Apr 27, 2016 #1
    The bond enthaply of halogens decreases down the group so why is the reactivity of halogens with hydrogens decreases down the group?
     
  2. jcsd
  3. Apr 27, 2016 #2
    Bond always involves two atoms. When you say enthalpy which is measure of the bond energy decreases along the group means the hydrogen halogen bond only I think. More the binding energy more reactive it will be because by bonding more energy will be released. So everything is in order in HF binding energy is greatest and F is most reactive with H to form HF. Take the analogy of an iron piece and attach it to a weak magnet and then to a strong magnet. where the binding energy is more that is stronger magnet attracts the piece more strongly than the weak magnet, hence it is more reactive. Think it over.
     
  4. Apr 28, 2016 #3
    Oh so you are saying that hydrogen will be attracted more towards an electro negative atom. Can you explain me in terms of bond strength of H-X and how will it affect the reactivity?
     
  5. Apr 28, 2016 #4

    DrDu

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    The first link I found when looking for the bond enthalpies of the halogens contains both the bond enthalpies of the halogens and the halogen hydrides. You can see that both decrease, so supposing that the enthalpy difference is a proxy for the driving force of the reaction, a decrease in bond strength of the halogens isn't made up by the bond formed in the hydrides.


    http://www.chemguide.co.uk/inorganic/group7/properties.html
     
  6. Apr 29, 2016 #5

    James Pelezo

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    Bond Energy trends are a function of electronegativity in series and of ionic size in groups. In series (consider Series II => Li, Be, B, C, N, O, F & Ne). Electronegativity increases with increasing atomic number due to increasing number of electrons and protons making up the atomic and ionic structures.
    The effect of increasing electronegativity of neutral elements in series functions to decrease atomic radius b/c of increased force of electrostatic attraction. Ionizations follow the same trends in series for the Group I - III elements, but C, N, O & F tend to gain electrons causing a large in crease in ionic radius after Boron because of e-/e-repulsion, but electronegative effects still increase from Carbon through Fluoride. The order of increasing electronegativity is C <N<O<F. Applying this concept to chemical bonds (Covalent bonds) consider the C - H bond 1st. The low electronegativity of Carbon does not distort the chemical bond as much as the same bonds in N, O and F leaving a high electron cloud density in the Carbon to Hydrogen bond and results in a reasonably strong bond as compared to N - H, O - H or F - H which are polar covalent in nature. Distortion of the bonds by the highly electronegative N, O and F ions weakens the covalent bonds in these systems.

    In groups or families of elements, size of central element plays a greater factor in strength of bond. This is especially true in the halogen family. Considering the compounds H - F, H - Cl, H - Br and H - I, bond strength decreases with increasing atomic number in the order (H - F) > (H - Cl) > (H - Br) > (H - I) due to the addition of principle energy levels (rings) creating a 'shielding effect', 'Shielding' decreases the electrostatic influence of the nucleus on the bond between the valence of the anion and the attached substrate. The acidity of the Hydrogen - Halide group is also referred to as the 'Hydrogen Halide Paradox'
     
  7. Apr 29, 2016 #6
    X-X + H-H -->2H-X
    Can you please also tell me what role does the bond enthalpy of H-X plays in the reactivity trends of halogens towards hydrogen?
     
  8. Apr 29, 2016 #7

    James Pelezo

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    I couldn't structure the energy flow diagram directly, so I typed it out in word and posted in jpeg. Hope this helps.
    Energy Factors in Hydrogen Halide Formation016.jpg
     
  9. Apr 29, 2016 #8

    James Pelezo

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    ΔHaX(g) = Heat of atomization for halogen, not hydrogen. My mistake.
     
  10. Apr 30, 2016 #9
    So is this reasoning correct?
    Down the group the strength of H-X bonds increases and to form a strong bond more energy is required. As a result due to difference in energy requirements the reactivity of H-X is higher up the group
     
  11. Apr 30, 2016 #10

    James Pelezo

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    Not exactly ... 'Down' the group the strength of H-X bonds decrease and form weaker bonds giving up 'less' energy on formation... The bonding energies of the H-X series are 'size of halide ion' dependent, not 'electronegativity' dependent as in series substances. Such is why the trend is called 'The Hydrogen Halide Paradox'. As atomic number increases additional energy levels are present causing a shielding effect that reduces the influence of the nucleus on the valence level electrons. As a result the bond energy trends, from strongest to weakest follow H-F > H-Cl > H-Br > H-I ... The numerical values (supported by many online references) are ...
    • ΔHb(H-F) = 135 Kj/mol
    • ΔHb(H-F) = 105 Kj/mol
    • ΔHb(H-F) = 87.5 Kj/mol
    • ΔHb(H-F) = 71 Kj/mol
    One practical result of this trend is ...
    H-F is a weak acid (~2.5% ionized) in aqueous media where as HCl, HBr & HI are strong acids ionizing 100% in aqueous media b/c of decreasing H/X bond strengths.
     
  12. May 2, 2016 #11

    James Pelezo

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    Oops in last post, correction ...
    • ΔHb(H-F) = 135 Kj/mol
    • ΔHb(H-Cl) = 105 Kj/mol
    • ΔHb(H-Br) = 87.5 Kj/mol
    • ΔHb(H-I) = 71 Kj/mol
     
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