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    q) state and explain whether the electronegativity of flourine is greater than,, similar to or less than that of bromine..

    hence explain why hydrogen flouride can form a hydrogen bond but hydrogen bromide cannot........
    the answer in the markscheme is mainly about the less sheilding in the flourine atom and then says that HF has a greater dipole moment (HF is more polar)

    first why doesnt it mention that the hydrogen bond is formed due to the highh electronegativity difference.. isnt it the main reason for the fomation of the hydrogen bond??? plus i do not understand the second point concerning the dipole moment
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    (Original post by NoorL)
    q) state and explain whether the electronegativity of flourine is greater than,, similar to or less than that of bromine..

    hence explain why hydrogen flouride can form a hydrogen bond but hydrogen bromide cannot........
    the answer in the markscheme is mainly about the less sheilding in the flourine atom and then says that HF has a greater dipole moment (HF is more polar)

    first why doesnt it mention that the hydrogen bond is formed due to the highh electronegativity difference.. isnt it the main reason for the fomation of the hydrogen bond??? plus i do not understand the second point concerning the dipole moment

    I think the first point can either mean two things:
    • It is insufficient to simply state that fluorine is more electronegative than bromine. You must state why, which is because there's less shielding as fluorine has less energy levels.
    • It's referring to the availability of the lone pair of electrons on fluorine. A hydrogen bond needs to things; a hydrogen atom with a high delta charge, and a lone pair of electrons to attract the hydrogen atom. The lone pair of electrons is more dense at the 2nd energy level. In higher energy levels, (Bromine is period 4), the electron pair is more diffuse and so there is a lower charge density. Thus it's less likely to attract a positive charge.
    You're right though, hydrogen bonding is formed when there is high electronegativity between the two atoms in a bond, but perhaps you need to expand.
    In fact I think that relates more to the second point concerning the dipole moment. Dipole moment is just another term for describing the separation of charge in a bond.

    Because of fluorine's electronegativity, the dipole moment is greater. In other words, there is a greater separation of charge of positive at one end and negative at the other end. Thus it is able to form a hydrogen bond.

    In a H-Br bond, there still exists a dipole moment (or dipole, you're pretty much saying the same thing) because bromine is more electronegative than hydrogen. However, the separation of charge is smaller so it cannot form the electrostatic force of attraction required in hydrogen bonding.
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    (Original post by RMNDK)
    I think the first point can either mean two things:
    • It is insufficient to simply state that fluorine is more electronegative than bromine. You must state why, which is because there's less shielding as fluorine has less energy levels.
    • It's referring to the availability of the lone pair of electrons on fluorine. A hydrogen bond needs to things; a hydrogen atom with a high delta charge, and a lone pair of electrons to attract the hydrogen atom. The lone pair of electrons is more dense at the 2nd energy level. In higher energy levels, (Bromine is period 4), the electron pair is more diffuse and so there is a lower charge density. Thus it's less likely to attract a positive charge.
    You're right though, hydrogen bonding is formed when there is high electronegativity between the two atoms in a bond, but perhaps you need to expand.
    In fact I think that relates more to the second point concerning the dipole moment. Dipole moment is just another term for describing the separation of charge in a bond.

    Because of fluorine's electronegativity, the dipole moment is greater. In other words, there is a greater separation of charge of positive at one end and negative at the other end. Thus it is able to form a hydrogen bond.

    In a H-Br bond, there still exists a dipole moment (or dipole, you're pretty much saying the same thing) because bromine is more electronegative than hydrogen. However, the separation of charge is smaller so it cannot form the electrostatic force of attraction required in hydrogen bonding.
    thank u so much!
    i didn't fully understand the point concerning charge density though...
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    (Original post by NoorL)
    thank u so much!
    i didn't fully understand the point concerning charge density though...
    Ah, charge density refers to the amount of charge in a given volume. Treat it like concentration. (I believe you're in AS so you won't have come across this term, or maybe you will depending on your exam board)

    So compare the lone pair of electrons on a fluorine atom to a bromine atom.

    In this case, you shouldn't think of electrons as two particles. Instead, it helps to think of them as a cloud of electric charge.

    In the bromine atom the cloud is more diffuse, i.e. more spread out.
    This is because
    • Bromine is a larger atom. This increased distance between the nucleus and outer electrons weakens the electrostatic force of attraction so the electron cloud is more loosely bound. It's more free to disperse.
    • There are more energy levels and so increased shielding further weakening the electrostatic force of attraction. (These first two points are pretty much interlinked, but my exam board treated them as two separate points)
    • The outer energy energy level is larger in diameter so the orbitals will be much large. Since each orbital only contains two electrons/two negative charges, the distribution of this negative charge will be much greater.
    All of this means that the charge density, i.e. the amount of charge per volume is much lower (but remember the charge itself doesn't change).

    Spoiler:
    Show
    I tried with an analogy. If it works it works...

    It's like if I dropped a ten pound note on the floor, and I scattered ten £1 coins on the floor, you'd probably pick up the tenner, but you probably wouldn't pick up every single coin. In each case there's still £10 worth of money, but the density of the value has decreased with the coins.
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    (Original post by RMNDK)
    Ah, charge density refers to the amount of charge in a given volume. Treat it like concentration. (I believe you're in AS so you won't have come across this term, or maybe you will depending on your exam board)

    So compare the lone pair of electrons on a fluorine atom to a bromine atom.

    In this case, you shouldn't think of electrons as two particles. Instead, it helps to think of them as a cloud of electric charge.

    In the bromine atom the cloud is more diffuse, i.e. more spread out.
    This is because
    • Bromine is a larger atom. This increased distance between the nucleus and outer electrons weakens the electrostatic force of attraction so the electron cloud is more loosely bound. It's more free to disperse.
    • There are more energy levels and so increased shielding further weakening the electrostatic force of attraction. (These first two points are pretty much interlinked, but my exam board treated them as two separate points)
    • The outer energy energy level is larger in diameter so the orbitals will be much large. Since each orbital only contains two electrons/two negative charges, the distribution of this negative charge will be much greater.
    All of this means that the charge density, i.e. the amount of charge per volume is much lower (but remember the charge itself doesn't change).
    Spoiler:
    Show
    I tried with an analogy. If it works it works...

    It's like if I dropped a ten pound note on the floor, and I scattered ten £1 coins on the floor, you'd probably pick up the tenner, but you probably wouldn't pick up every single coin. In each case there's still £10 worth of money, but the density of the value has decreased with the coins.
    soo the larger the atom,, the less the charge density ( and the less effect the lone pairs have??)
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    (Original post by NoorL)
    soo the larger the atom,, the less the charge density ( and the less effect the lone pairs have??)
    Correct, although you might not want to use the word 'effect' when you're answering this kind of question, but I think you understand it anyway.
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    (Original post by RMNDK)
    Correct, although you might not want to use the word 'effect' when you're answering this kind of question, but I think you understand it anyway.
    thanks a lot
 
 
 
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