account for this difference
2006-12-28 08:29:59 · 6 answers · asked by chanti 3 in Science & Mathematics ➔ Chemistry
Because the two electrons lost to form the Ca(2+) ion are in a higher energy level with a larger radius than the rest of the electrons. When the 2 ions are lost, the remaining electrons are all in the lower energy shells with the smaller radius.
2006-12-28 08:38:42 · answer #1 · answered by Bigfoot 7 · 0⤊ 0⤋
The outer electrons in the valence shell are stripped off to make the Ca2+. Those extra electrons in their outer orbits account for the greater effective radius. (Note: Protons stay the same...they have to for it to be Ca.)
2006-12-28 16:34:23 · answer #2 · answered by Jerry P 6 · 0⤊ 0⤋
two reasons:
1) Ca2+ does not have the two outermost electrons which means automatically the radius is smaller
2) in Ca2+ as the electrons are lesser the attracting force experienced by the remaining electrons in relatively more and hence there is some shrinkage
2006-12-28 16:37:02 · answer #3 · answered by mandeep 3 · 1⤊ 0⤋
because the Ca (2+) has two more protons then it does electrons. This causes the electrons to move closer in to the stronger positive force.
2006-12-28 16:33:38 · answer #4 · answered by sikla_of_dragga 2 · 0⤊ 0⤋
The less electrons in the ion, the less space is taken up by electrons
2006-12-28 16:34:09 · answer #5 · answered by S 3 · 0⤊ 0⤋
there is no electrons in 4s so its smaller
2006-12-28 16:35:43 · answer #6 · answered by ? 3 · 0⤊ 0⤋