Question

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

Answer 1

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 $ Ga $ , that is Gallium as our element so we need to note its electronic configuration; it is on row $ 4 $ and column $ 13 $ with $ 31 $ as its atomic value. It will have only $ 31 $ electrons around it and configuration can be: $ Ga:[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) $ - Values taken $ (1,2,3,...) $
Angular Momentum $ (l) $ - Values taken $ (0,1,2,3,...(n - 1) $
Magnetic $ ({m_l}) $ - Values taken $ ( - l,..., - 1,0,1,....,l) $
Spin $ ({m_s}) $ - Values taken $ ( + \dfrac{1}{2}, - \dfrac{1}{2}) $
First let us write the electronic configuration of $ Ga $ ;
Element of group $ 13 $ , period $ 4 $ ; $ Ga:[Ar]3{d^{10}}4{s^2}4{p^1} $
The outer electron is one of highest energy so that can be seen on $ 4p $ orbital so the initial quantum number we can write down directly after observation is;
Principal quantum number: $ n = 4 $
Next proceeding with subshells, remember that for $ l $ the angular momentum;
If subshell is $ s \Rightarrow l = 0 $
If subshell is $ p \Rightarrow l = 1 $
If subshell is $ d \Rightarrow l = 2 $
If subshell is $ f \Rightarrow l = 3 $
So we see that the outermost electron lies in subshell $ p $ so $ \Rightarrow l = 1 $ (angular momentum).
Moving on with magnetic quantum value, remember here;
If orbital is $ {p_x} \Rightarrow {m_1} = - 1 $
If orbital is $ {p_y} \Rightarrow {m_1} = 0 $
If orbital is $ {p_z} \Rightarrow {m_1} = 1 $
There’s only a single electron in $ p $ so $ {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 $ 4{p_x} $ $ \Rightarrow n = 4 \to l = 1 \to {m_1} = - 1 \to {m_2} = + \dfrac{1}{2} $
If spin-down and electron lies on orbital $ 4{p_x} $ $ \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.