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Why is it that adding protons has a greater effect than electron-electron repulsion on periodic trends like atomic radius and ionization energy (assuming # of shells constant)? It seems that if protons and electrons have the same charge and electrons are actually closer to the electrons in question (e.g. for ionization energy), that their effect would be the same?

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This is because the attraction is not between each proton and it's corresponding electron, but between each electron and the nucleus. Since the number of protons increases the charge of the nucleus, but the charge of the electron stays constant, the attraction is greater. The attraction of the electron to the nucleus depends on the charge of the nucleus, and it is independent of the number of electrons in "orbit"

A great analogy for this is the solar system. (Gravity works very similarly to electrostatic force.) It won't matter if suddenly there is a mother earth orbiting in the outskirts of the solar system, Earth would be unaffected by that change, and we would still orbit the sun at the sae distance. However, if the sun suddenly becomes noticeably heavier, the planets will each be pulled in with a stronger force, which reduces the orbit radius. This will happen equally to each individual planet. It doesn't matter how many planets we have, The pull of the earth is only affected by the mass of the sun and the mass of earth.

This applies to the atom if you think of the nucleus as the sun and each electron as a planet. The force attracting the electron to the nucleus is only affected by the charge of the nucleus as each electron carries the same charge, and the charge is not the collective charge of all electrons, but the individual charge. Hence all each electron perceives is the increase in the charge of the nucleus that is increasing its attraction to the nucleus.

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  • $\begingroup$ But if electrostatic force is analogous to gravitational force, then adding an electron would be equivalent to adding a planet the mass of the sun that repels (rather than attracts) the other planets. Why would adding one proton to the nucleus exert a greater force than adding an electron if their mass is not what matters -- the charge stays the same? $\endgroup$ – Jess Dec 16 '14 at 15:47
  • $\begingroup$ @Jess You would be adding a proton which attracts the other electrons because the protons are positive whereas the orbiting electrons are all negative. If you are saying that the electron will repel the other electrons in orbit, then you are correct and this also plays a role in determining the size of the orbital. This is why the Ionic radius is different from the atomic radius. $\endgroup$ – Jonathanjaya Dec 17 '14 at 6:12
  • $\begingroup$ So electron repulsion plays a role-- but why is the effect of adding protons greater? i.e when you move across a period, ionization energy increases, justified by the increasing # of protons added to the nucleus. But electrons are also being added simultaneously, with the same charge -- why doesn't the added repulsion cancel out the extra attraction? $\endgroup$ – Jess Dec 18 '14 at 12:48
  • $\begingroup$ Because electron density is diffuse, so the "shielding" of the nuclear charge by other electrons is imperfect. See, for example discussion on the effective nuclear charge. $\endgroup$ – Geoff Hutchison Dec 18 '14 at 23:14
  • $\begingroup$ Jonathanjaya has done convincing explanation; i would just say the success on the part of nucleus is "Unity brings strength" ; protons together build up a heavy positive centre instead of acting as a collection of individual protons, where as electrons remain"divided" and victory is that of the nucleus, which can retain all electrons overcoming the comparatevely weak electron-electron repulsions; now if you wonder how protons can stay together in spite of their repulsion, you have to think about short range forces which are charge independent. $\endgroup$ – Nandakumar U K Jan 24 at 3:40

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