Indicate the formula of the higher oxide of the element, if the electronic formula of the outer electrons is 4s23d104p1

Let’s compose a complete electronic formula of the element:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1, let’s count the number of electrons – 2 + 2 + 6 + 2 + 6 + 2 + 10 + 1 = 31, since the number of electrons in an atom is equal to the number of protons, that is, the ordinal number of an element in the periodic table, then we have an element with serial number 31 – Galium (Ga), according to the table we find the formula for the higher oxide for the elements of group 3 – R2O3, that is, the formula for the higher oxide is Ga2O3.