In what block, group, and period on the periodic table can the atom with a 1s22s22p63s23p64s23d104p7 ground-state electron configuration be found?

the configuration given is wrong.

Justification.

1) The configuration given is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁷

This configuration is impossible. The orbitals p can have a maximum of 6 electrons, so 4p⁷ is wrong.

2) Just to show how the system works, I will modifiy the electron configuration to this:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Look at the outer shell; the valence electron shell. It is 4s² 4p⁶.

i) The number 4 is the main quantum number, i.e. the main energy level. It is also the number of the row (period) in the periodic table. So, the element is in the period 4.

ii) The number of valence electrons is 2 + 6 = 8

That means that the element is in the column 18, or group 18.

That is the group of the noble gases.

iii) Since the last electron is in a p orbital, the block is the p.

There are 4 blocks: s, p, d, and f.

The elements of the block s are those whose last shell is ns¹ or ns² (columns 1 and 2); the elements of the group p are those whose last shell is ns² np¹, ns² np², ns² np³, ns² np⁴, ns² np⁵, or ns² np⁶. Those are the columns 13, 14, 15, 16, 17, and 18, respectively.

The elements of group d are the transition metals. They have partially filled a d orbital, or maximum 10 eletrons with this for ns² (n-1) d¹ ⁻ ¹⁰.

The elements of group f are the inner transition metals.

3) With the group and period you can find the element in the periodic table.

In my example, group 10 and period 4, you find the noble gas Kr.

You can check that the atomic number of Kr is 36 and the total number of electrons is also 36: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 = 36.

That tells you that you have not make a mistake.