Using electronic configuration, please explain the charge 1+ and 3+ on Ti ions.
When determining electron configuration, you have to first understand that the configuration builds upon itself. It all starts with the first row being 1s (hydrogen). From there it builds upon itself in the order of the periodic table going across (by rows).
The first number determines which row it is in. We use letters to describe its location pretty much. Since the periodic table is organized according to traits, the letters represent the location and how the energy levels are organized. Finally there is an exponent which tells you how many electrons are exactly in that energy level. The levels go in this order:
Alkali & Alkaline: s
So the general order is s, p, f, d...
Hydrogen is 1s^1 (because it has one electron and has only one energy level)
Helium is 1s^2 (because it has two electrons and two energy levels)
Lithium would be 1s^2 2s^1... Another way to write this is (He) 2s1 since essentially Lithium is just a Helium atom with 1 extra electron (for configuration purposes).
When you make one with Ion is a little different however. If you are looking at Titanium you first need to know Ti's regular electron configuration which is: (Ar) 4s^2 3d^2
When you have a positive ion, you have lost electrons, so you will manipulate the exponent which represents the electron count. Now atoms and electrons are lazy, and like to bond, so they get rid of electrons from easier, less complicated energy levels and leave at least one or more on the outer shell.
So, the confituration for Ti +3 would lose three electrons: 1 from the 3d^2 level turning it into 3d^1 (so it has one in its outter shell; then it'll lose the other two (totaling to 3) from the 4s^2 level turning it into 4s^0.
The configuration for Ti+1 would just lose the one from 3d^2, turning it into 3d^1
Ions can not lose the last energy level the normal atom has. It is still technically that atom so it has to keep that classification as much as possible.