How do you write the noble-gas electron configuration for bromine?
Answer
583.8k+ views
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:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$
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:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
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:
$[Ar]4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$
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.
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:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$
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:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
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:
$[Ar]4{{s}^{2}}3{{d}^{10}}4{{p}^{5}}$
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

