The percent composition of all elements in hclo3 is:
a. 2% h, 20% cl, 78% o
b. 1% h, 42% cl, 57% o
c. 2% h, 68% cl, 30% o
d. none of the above

The answer is b. 1% H, 42% Cl, 57% O.

Let's first calculate a molar mass (M) of HClO₃. Molar mass is a mass of 1 mole of a substance. It can be expressed as the sum of relative atomic masses (Ar), which are masses of atoms of the elements:
M(HClO₃) = Ar(H) + Ar(Cl) + 3Ar(O)

Ar(H) = 1 u 
Ar(Cl) = 35.5 u
Ar(O) = 16 u
M(HClO₃) = Ar(H) + Ar(Cl) + 3Ar(O) = 1 + 35.5 + 3 · 16 = 84.5 u

Now, we just need to calculate mass percent of each element using the proportion:
Ar : x = M : 100%                   ⇒      x = Ar ÷ M · 100%

Hydrogen: x = Ar(H) ÷ M · 100% = 1 ÷ 84.5 · 100% = 1.2% ≈ 1%
Chlorine: x = Ar(Cl) ÷ M · 100% = 35.5 ÷ 84.5 · 100% = 42%
Oxygen: x = 3Ar(O) ÷ M · 100% = 48 ÷ 84.5 · 100% = 56.8% ≈ 57%

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.