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For elements with the same number of electrons, the atom with more protons in the nucleus or the higher effective nuclear charge has the smaller atomic radius. In the case of Na+ and F-, Na+ has the smaller ionic radius because it has a stronger pull from the nucleus, causing electrons to be pulled tighter in. Both atoms have the same number of shells, so the only difference is the effective nuclear charge.
Last edited by Sabina House 2A on Tue Nov 03, 2020 5:17 pm, edited 1 time in total.
The radius at that point depends on the nuclear charge. That is, the more protons in the nucleus, the greater the attraction between the nucleus and the individual electrons. If there are more protons in the nucleus, the distance between the valence electrons and the nucleus decreases, and hence the ionic radius decreases as well. In the example you have provided, Na+ would have the smaller ionic radius because of the fact that it has more protons than F-. Hope this helps!
If you are considering Na+ and F-, an easy way to think is to imagine that sodium has lost an outermost electron on its outermost shell, so it loses one shell (3s) and because of this, the electrostatic attraction is stronger and the nucleus will pull the other shells closer to it. so the radius decreases. On the other hand, F- gains an extra electron on its 2p shell, with the extra electron, the nuclear charge is relatively weaker and the electrons expand out more, pulling away from the nucleus, and the radius increases.
Hope it helps!
Hope it helps!
Last edited by Lorraine Jiang 2C on Wed Nov 04, 2020 1:18 am, edited 1 time in total.
Na+ and F-. One is a cation and one is an anion. Na+ has lost an electron, and F- has gained an electron. This electron gain in F- has caused electron repulsion and will expand the size of the atom. The loss of the electron in Na+ removes an electron from the n = 3 level, so now the same number of protons remain, but the electrons are held more tightly. The ionic radius of Na+ is therefore smaller than that of F-. Hope this helps!
To add on to previous replies, most of the time anions (negatively charged ions) will have a greater atomic radius than cations (positively charged ions). This is because anions have gained an electron, which contributes to greater electron-electron repulsions within the atom, therefore increasing the radius of the atom because the electrons want to be far away from each other. In a cation however, an electron has been lost, so the protons can actually hold the remaining electrons more closely and therefore the atomic radius is less.
In general, if the ions are isoelectronic it means they have the same number of electrons, so their ionic radius will depend on the nuclear charge which is affected by the number of protons. The greater the number of protons, the more attracted the electrons are to the nucleus, thus the smaller the ionic radius.
The ionic radius has to do with the distance from the nucleus, and when there are more protons, this leads to a greater attraction to the nucleus, causing the distance to be smaller, which equates to a smaller ionic radius. An ion with a negative charge would not be as attracted to the nucleus as the one with a positive charge, which means the ionic radius is larger due to distance from the nucleus.
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