what is the basic factor which determines the order of ionization energies of K,Li,Ca,S?
Looked up ionization energies (kJ/mol) from Wikipedia


K  = 418.8
Li = 520.2
Ca = 589.8
S = 999.6


so the book answer is correct...



Ok, here is best hand waving explanation that I can give. The general trend...



Li and K are both alkali metals. As the atomic number goes up the shells get further away from the nucleus and the inner electrons shield the valence electron. So in order of increasing ionization energy

Li > Na > K > Rb > Cs > Fr

So to start we easily get Li > K.  

Ca and K are in the same period in the periodic table. Generally as you go up in atomic number in a period the ionization energy will increase. So Ca > K.



Sulfur is a non-metal in the upper right hand of the periodic table so it should have the highest ionization energy. 

So we're down to either S > Ca > Li > K or S > Li > Ca > K. 


  To me giving either of these answers wouldn't be embarrassing. 


If you go to the Bohr model of the hydrogen atom then energy of the $n$th shell is 

$\mathrm{E}_n  \propto \dfrac{Z^2}{n^2} $  

If we assume that the 2s and 3s electrons are perfectly shielded by all the electrons in the lower shells, then

$ \mathrm{E_{Li}} \propto \dfrac{1}{2^2} = 0.25$

The two Ca 3s electrons being equal the charge is 2

$ \mathrm{E_{Ca}} \propto \dfrac{2^2}{4^2} = 0.25$

So I would have guessed S > Li > Ca > K

Now I know that is wrong. I know that the "problem" is with the shielding of the electrons. Below is a section of the periodic table with ionization energies listed. 

I just don't know how to explain the "right" answer without hand waving about shielding. 

I don't know how you'd rationalize all the relationships between the group 1 and 2 elements. 



If you assume the Bohr model as noted above then you end up with this which correctly predicts the trend down a group but overestimates shielding of the inner electrons.