Question

How do you write the noble-gas electron configuration for bromine?

Answer 1

Hint First, we have to write the noble gas configuration for bromine as its atomic number is 35 which means there are 35 electrons in the bromine atom so, arrange the electrons according to the increasing energy of the orbitals. Then write the noble-gas which comes before the bromine and write the rest of the configuration.

Complete step by step answer:
Bromine is the element of group 17 of the p-block and comes in the fourth period. So, its atomic number is 35 which means the number of protons and the number of electrons is 35 in the atom.
There are many orbitals that are placed according to their increasing order of energies. The order is$\text{1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p}$. And we know that the s-orbital can accommodate only 2 electrons, the p-orbital can accommodate 6 electrons, the d-orbital can accommodate 10 electrons, and the f-orbital can accommodate 14 electrons. So, the 35 electrons will be arranged as:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^5$
So, this is the ground state electronic configuration therefore, to write the noble-gas electron configuration we have to find the noble gas which comes before the bromine. From the periodic table the noble gas that is before bromine is Argon and the atomic number is 18, so its electronic configuration will be:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^6$
For this configuration in the bromine, we can write [Ar] and then the rest of the

configuration so, the noble-gas electron configuration for bromine will be:
$[Ar]4s^{2}3d^{10}4p^5$

Note: All the noble gas with their atomic number are helium having the atomic number 2, neon having atomic number 10, argon having atomic number 18, krypton having atomic number 36, xenon having atomic number 54, and radon having atomic number 86.