What element has the electron configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$?

Hint: By counting the total number of electrons in the configuration we can find the total number of electrons which will give the atomic number of the element. And from the configuration only, we can find the period and group of the element.

Complete answer:
We know that the elements in nature are differentiated by the number of electrons present in them. These electrons are arranged into a configuration specific to a specific element. So, the elements are arranged in the periodic table according to the last electron entering the last shell.
So, if we are given the electronic configuration of any element, by counting the total number of electrons in the configuration we can find the total number of electrons which will give the atomic number of the element. The given configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$
Now, counting all the electrons in this configuration:
2 + 2 + 6 + 2 + 4 = 16
The last shell in the configuration is 3, this means that the element belongs to the 3rd period. The last electron enters the p-subshell, which means it is the element of the p-block. For finding the group, we have to calculate the number of electrons in the 3rd shell and add it to 10.
So, the group is = 2 + 4 +10 = 16
The group is 16th, the period is 3rd and the number of electrons is 16, so the element is the Sulfur atom.

Note:
There is a specific order in which the electrons are filled in the increasing order of the orbitals, as this order is:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.