Question

What is the noble gas configuration for the following: fluorine, phosphorous, bromine, manganese, copper $?$

Answer 1

Hint : First we have to understand the electron configuration. Then understand the noble gas electron configuration and identify the noble gas in the period before your element. Substitute the noble gas for the same number of electrons the noble gas has in the given element.

Complete Step By Step Answer:
Electron configurations of atoms follow a standard notation in which all electron-containing atomic subshells (with the number of electrons they hold written in superscript) are placed in a sequence.
The three rules that must be followed while writing the electronic configuration of elements are:
The Aufbau principle: electrons must completely fill the atomic orbitals of a given energy level before occupying an orbital associated with a higher energy level.
Pauli’s exclusion principle: states that no two electrons can have equal values for all four quantum numbers.
Hund’s rule of maximum multiplicity: All the subshells in an orbital must be singly occupied before any subshell is doubly occupied. Furthermore, the spin of all the electrons in the singly occupied subshells must be the same.
The noble gas electron configuration is a type of shortcut to writing out the full electron
configuration of an element. The noble gas is substituted to represent all of the electrons that aren’t valence electrons. The period of an element is the horizontal row that the element is located in. If the element is in the fourth row of the periodic table, it is in period four. The noble gas you will use will be located in period three. The noble gases and their periods are Helium, Neon, Argon, Krypton, Xenon, Radon and $1,2,3,4,5,6$ respectively.
For example, sodium is in period three. We will use neon for the noble gas configuration because it is in period $2$ . Replace the electron configuration of the noble gas by removing the same number of electrons the noble gas has from the element you get the noble gas configuration of the given element.
Fluorine $F$ , $Z = 9$ , $1{s^2},2{s^2},2{p^5}$
Phosphorus $P$ , $Z = 15$ , $1{s^2},2{s^2},2{p^6},3{s^2},3{p^3} = \left[ {Ne} \right]3{s^2},3{p^3}$
Bromine $Br$ , $Z = 35$ , $1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^{10}},4{s^2},4{p^5} = \left[ {Ar} \right]3{d^{10}},4{s^2},4{p^5}$
Manganese $Mn$ , $Z = 25$ , $1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^5},4{s^2} = \left[ {Ar} \right]3{d^5},4{s^2}$
Copper $Cu$ , $Z = 29$ , $1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^{10}},4{s^1} = \left[ {Ar} \right]3{d^{10}},4{s^1}$ .

Note :
Note that the electron configuration of an element describes how electrons are distributed in its atomic orbitals. In the Aufbau principle, electrons occupy orbitals in the increasing order of orbital energy level. In Pauli’s exclusion principle each subshell of an orbital can accommodate a maximum of $2$ electrons and both these electrons must have opposite spins.