Question

In the electronic configuration of \[Br\], where are the valence electrons? Are they in the \[3d\]?

Answer 1

Hint: Bromine lies in the \[{4^{th}}\] period of the periodic table. Its atomic number is \[35\]. It is present in group number \[17\] of the periodic table and is considered a halogen. It is a reddish-brown liquid at room temperature. The electronic configuration of the entire group is similar, as they have the same valence electrons.

Complete answer: Electronic configuration- the electronic configuration of an element is the distribution of electrons in its atomic orbitals or it describes the position of the electrons inside the orbitals. The structure of the periodic table of an element is partly based on electronic configuration.
The electronic configuration of Br is. We can see here that \[3d\] orbital is completely filled. When the \[5\] orbitals of \[3d\] subshell have 10 electrons in them they are completely filled and now it cannot take mare electrons. This makes the d subshell stable and it falls into the lower third electron shell. The \[4s\] and \[3d\] subshells have similar kinetic energy. The \[3d\] subshell is nearest to the nucleus than \[4s\] but the structure of \[3d\] is a little complicated with the presence of four lobes. Whereas when we look into the \[4s\] subshell it is far from the nucleus but its structure is very simple consisting of only one spherical orbital. Initially the electrons fill up \[4s\] subshell after that they enter into \[3d\], as this happens the \[3d\] comes closer to the nucleus and becomes a favored place for new electrons.
Therefore there is a total of \[7\] valence electrons in bromine which are present in \[4s\] and \[4p\] rather than \[3d\].

Note:
When any orbital is filled it will not accommodate extra electrons as it will be completely stable. Valence electrons are electrons present in the outermost shell of an atom. Since the outer shells of Br are \[4s\] and \[4p\] and they in total have \[7\] electrons in them. Thus the valence electron in Br is \[7\].