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Question: What is a possible set of four quantum numbers (n,l,ml,ms) for the highest energy electron in Ga?...

What is a possible set of four quantum numbers (n,l,ml,ms) for the highest energy electron in Ga?

Explanation

Solution

Hint : For us to be able to fully explain the movement of electrons and their pathways from inside that atom, we make use of four quantum numbers. Here we are given GaGa , that is Gallium as our element so we need to note its electronic configuration; it is on row 44 and column 1313 with 3131 as its atomic value. It will have only 3131 electrons around it and configuration can be: Ga:[Ar]3d104s24p1Ga:[Ar]3{d^{10}}4{s^2}4{p^1} .

Complete Step By Step Answer:
First of all let us see which quantum number can take which value;
Principal (n)(n) - Values taken (1,2,3,...)(1,2,3,...)
Angular Momentum (l)(l) - Values taken (0,1,2,3,...(n1)(0,1,2,3,...(n - 1)
Magnetic (ml)({m_l}) - Values taken (l,...,1,0,1,....,l)( - l,..., - 1,0,1,....,l)
Spin (ms)({m_s}) - Values taken (+12,12)( + \dfrac{1}{2}, - \dfrac{1}{2})
First let us write the electronic configuration of GaGa ;
Element of group 1313 , period 44 ; Ga:[Ar]3d104s24p1Ga:[Ar]3{d^{10}}4{s^2}4{p^1}
The outer electron is one of highest energy so that can be seen on 4p4p orbital so the initial quantum number we can write down directly after observation is;
Principal quantum number: n=4n = 4
Next proceeding with subshells, remember that for ll the angular momentum;
If subshell is sl=0s \Rightarrow l = 0
If subshell is pl=1p \Rightarrow l = 1
If subshell is dl=2d \Rightarrow l = 2
If subshell is fl=3f \Rightarrow l = 3
So we see that the outermost electron lies in subshell pp so l=1\Rightarrow l = 1 (angular momentum).
Moving on with magnetic quantum value, remember here;
If orbital is pxm1=1{p_x} \Rightarrow {m_1} = - 1
If orbital is pym1=0{p_y} \Rightarrow {m_1} = 0
If orbital is pzm1=1{p_z} \Rightarrow {m_1} = 1
There’s only a single electron in pp so pxm1=1{p_x} \Rightarrow {m_1} = - 1 .
To get spin we have two ways of going, that is spin-up or spin-down;
That electron on outer orbit can have;
If spin-up and electron lies on orbital 4px4{p_x} n=4l=1m1=1m2=+12\Rightarrow n = 4 \to l = 1 \to {m_1} = - 1 \to {m_2} = + \dfrac{1}{2}
If spin-down and electron lies on orbital 4px4{p_x} n=4l=1m1=1m2=12\Rightarrow n = 4 \to l = 1 \to {m_1} = - 1 \to {m_2} = - \dfrac{1}{2}

Note :
It is good to remember that electrons that are unpaired are paramagnetic while those that are paired are diamagnetic. Normally the electrons are able to cross each other out so if they are unpaired then there will be a value for that magnetic field, at the same time if the electrons were paired then they are balancing their fields so they will not have a magnetic field around it and so it will not have any magnetic properties.