00:01
Here the three questions are given.
00:02
So explanation one by one is as in question number 17, we have to arrange according to their first ionization energy that is represented by an aberrated form ie of neon, sodium, potassium and air.
00:28
So the ionization energy, that is ie, increases as we move from left to right, across a row or we can say that cleared if it decreases as we move from top to bottom in a group we expect that it is vary in the order like this sodium greater than less than potassium less than ar because ne is above ar in group 8a we accept ne to exhibit the greater first ie such that ar is less than an e similarly the k is the alkali metal directly below and a in group first a, so we expect i1 for k be less than that of sodium.
02:44
From these observations we conclude that ie order of the following is as such that so the answer is in the following order k is less than na which is less than p which is less than a .r.
03:05
And the highest is of neon.
03:15
So the next 18 questions are saying that arrange the elements, helium, calcium and sulfur in the increasing ionization energy.
03:28
So the i .e is the energy required to remove one electron from a neutral atom.
03:44
So s and s -e belong to 6a group elements and down the group ie decreases.
04:05
So s have more ie value than s e and the sulfur and chlorine are the elements of the same tier...