Question

Which element has the electron configuration \[\left[ {Xe} \right]6{s^2}4{f^{14}}5{d^{10}}6{p^2}\]?

Answer 1

Hint: We have to know that the atomic number represents the number of protons in an atom. The number of protons and number of electrons are equal in a neutral atom. Since, they carry opposite charges making an atom neutral. Neutrons are neutral species since they carry no charge on them.

Complete answer:
The electronic configuration is there to distract you: $Z = 54 + 2 + 14 + 10 + 2 = 82$ and, therefore, the element is LEAD.
The element is defined by Z, the atomic number, which is the number of protons, positively charged, massive nuclear particles. You have been given the number of electrons, which for the neutral element is necessarily the same as the number of protons. Look up the Periodic Table, and for $Z = 82$.
The element having atomic number 82 is lead having an atomic symbol Pb. It is a heavy metal that is denser than most common materials. Lead is soft and malleable, and also has a relatively low melting point. When freshly cut, lead is silvery with a hint of blue; it tarnishes to a dull gray color when exposed to air. Lead has the highest atomic number of any stable element and three of its isotopes are endpoints of major nuclear decay chains of heavier elements.

Note:
We have to know that the sum of lead's first and second ionization energies—the total energy required to remove the two 6p electrons—is close to that of tin, lead's upper neighbor in the carbon group. This is unusual; ionization energies generally fall, going down a group, as an element's outer electrons become more distant from the nucleus, and more shielded by smaller orbitals.