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In a given Period,the atomic radius decreases from left to right? Why?

OK, I have a very inquisitive general chem student and we were discussing atomic radii trends in the Periodic Table. We know that the radius decreases from L to R in a given period and my explanation was that the number of protons increases and that electrons being added are being done so into regions of low density from previous occupied orbitals...in other words something like close packing...can anyone make this a little clearer?? Better analogy??

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  • 10 years ago
    Favorite Answer

    The sizes of the atoms, and therefore, the atomic radius decreases from left to right in a period because of the increase in the effective nuclear charge. Effective nuclear charge is the charge the outermost electron "sees" after subtracting the shielding effect of the inner electrons. As protons and electrons are added to the nucleus (as atomic number increases) the added electrons (in the same energy level) do little to shield the nucleus and so the charge on the nucleus appears to be greater. The increasing charge on the nucleus "pulls" the outermost electrons closer, thus reducing the atomic radius.

    The decrease in atomic radisu follows from Coulomb's law:

    F = kq1q2 / r^2 .... where F is the force of attraction, k is a constant, q1 and q2 are the charges of the nucleus and electron and r is the distance between them. The increasing force of attraction results in a smaller value of r.

  • ?
    Lv 6
    10 years ago

    Well, there are 3 variables you have to account for when determining the trend for atomic radii:

    1. The change in the number of protons

    2. The inner shells of electrons

    3. The valence shell

    As you said, the number of protons increases across the row, which means the net positive charge, or net positive electric field from the protons increases in magnitude. The other charged particles, the electrons, more specifically, the inner core of electrons, become more easily attracted to the net electric field created by the nucleus' increasing number of protons (because the number of core electrons on a row remains constant). The inner core effectively 'shuts out' the influence of the nucleus on the valence shell electrons because the valence shell is in an energy level higher than that of the inner core (and hence farther away from the nucleus). In essence of what I'm trying to say, there is this 'fight' between the constriction of radii between inner shell/nucleus attraction, and the influence of the nucleus on the valence shell, as well as the repulsion of inner core and valence electrons (which tries to increase radii). In summation, the nuclear charge continuously increases, while the inner core of electrons remains the same, so the net electric field, or rather we call it 'effective nuclear charge' in some textbooks, serves to decrease the radius as we move across a row on the periodic table.

    As for analogies, the closest I can think of would be a nut. At the center of the nut is the nucleus. Between it and the hard shell are the core electrons, and the core electrons are being held more tightly with an increasing nuclear charge. On the other hand, imagine the valence shell electrons floating around the 'shell' of the nut, which are effectively shut out by the hard shell of the nut. In summation then, the inner core/valence shell repulsion, nucleus/valence shell attraction, and inner core/nucleus attraction, all due to the effects of the net electric field created by the nucleus and electrons (both inner and valence shell electrons), serve to decrease the radius as we move across the row.

  • Cazley
    Lv 5
    10 years ago

    Don't try to make it too tough - as you go across the periodic table, atoms in the same period have electrons in the same energy levels. These electrons will be more attracted to a nucleus containing more protons, and will be found closer to that nucleus. A valence electron of neon, for example, attracted by the ten protons in the nucleus, will be closer to the nucleus (on average) than a valence electron of fluorine, which is being attracted by only nine protons. It is true that there is more electron-electron repulsion, but the presence of an extra electron somewhere in the energy level will not outweigh the increased attraction of another proton in the nucleus.

  • 4 years ago

    through fact the fee of the nucleus will enhance, so electrons are drawn nearer. although, atomic radius additionally will enhance from ideal to backside - this is by technique of the fact with each era you're including a clean electron shell that's lots extra from the nucleus than shells present day interior the previous era.

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