The heat vaporization of water is 40.66 kj/mol . ? the heat vaporization of water is 40.66 kj/mol. how much heat is absorbed when 3.45g of water boils at atmospheric pressure?

To answer the question above, first convert the given amount of water from grams to mol by dividing it with its molar mass which is 18 g/mol. 
                           amount in mol = 3.45 g x (mol/ 18 g) = 23/120 mol
Then, multiply this amount in mol with the heat of vaporization given.
                        (23/120) mol x (40.66 kj/mol) = 7.793 kJ
Thus, the heat absorbed is approximately equal to 7.793 kJ. 

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The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.