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Electron Spin or Spin Quantum Number is the fourth quantum number for electrons in atoms and molecules. Denoted as \(m_s\), the electron spin is constituted by either upward (\(m_s=+1/2\)) or downward (\(m_s=1/2\)) arrows.
In 1920, Otto Stern and Walter Gerlach designed an experiment, which unintentionally led to the discovery that electrons have their own individual, continuous spin even as they move along their orbital of an atom. Today, this electron spin is indicated by the fourth quantum number, also known as the Electron Spin Quantum Number and denoted by m_{s}. In 1925, Samuel Goudsmit and George Uhlenbeck made the claim that features of the hydrogen spectrum that were unexamined might by explained by assuming electrons act as if it has a spin. This spin can be denoted by an arrow pointing up, which is +1/2, or an arrow pointing down, which is 1/2.
The experiment mentioned above by Otto Stern and Walter Gerlach was done with silver which was put in an oven and vaporized. The result was that silver atoms formed a beam that passed through a magnetic field in which it split in two.
An explanation of this is that an electron has a magnetic field due to its spin. When electrons that have opposite spins are put together, there is no net magnetic field because the positive and negative spins cancel each other out. The silver atom used in the experiment has a total of 47 electrons, 23 of one spin type, and 24 of the opposite. Because electrons of the same spin cancel each other out, the one unpaired electron in the atom will determine the spin. There is a high likelihood for either spin due to the large number of electrons, so when it went through the magnetic field it split into two beams.
Note: In this module, capital "L" will be used instead of small case "l" for angular momentum quantum number.
A total of four quantum numbers were developed to better understand the movement and pathway of electrons in its designated orbital within an atom.
(For more information about the three quantum numbers above, see Quantum Number.)
The lines represent how many orientations each orbital has, (e.g. the sorbital has one orientation, a porbital has three orientations, etc.) and each line can hold up to two electrons, represented by up and down arrows. An electron with an up arrow means it has an electron spin of +\(\frac{1}{2}\), and an electron with a down arrow means it has an electron spin of \(\frac{1}{2}\).
Significance: determines if an atom will or will not generate a magnetic field (For more information, scroll down to Magnetic Spin, Magnetism, and Magnetic Field Lines). Although the electron spin is limited to +½ or –½, certain rules apply when assigning electrons of different spins to fill a subshell (orientations of an orbital) . For more information, scroll down to Assigning Spin Direction.
An atom with unpaired electrons are termed as paramagnetic
An atom with paired electrons are termed as diamagnetic
Applying concepts of magnetism with liquid nitrogen and liquid oxygen:
{{media("http://www.youtube.com/watch?v=KcGEev8qulA\)
The magnetic spin of an electron follows in the direction of the magnetic field lines as shown below.
An effective visual on how to assign spin directions can be represented by the orbital diagram (shown previously and below.) Restrictions apply when assigning spin directions to electrons, so the following Pauli Exclusion Principle and Hund's Rule help explain this.
When one is filling an orbital, such as the p orbital, you must fill all orbitals possible with one electron spin before assigning the opposite spin. For example, when filling the Flourine, which will have a total of two electrons in the s orbital and a total of five electrons in the p orbital, one will start with the s orbital which will contain two electrons. So, the first electron one assigns will be spin up and the next spin down. Moving on to the three p orbitals that one will start by assigning a spin up electron in each of the three orbitals. That takes up three of the five electrons, so with the remaining two electrons, one returns to the first and second p orbital and assigns the spin down electron. This means there will be one unpaired electron in Flourine so it will be paramagnetic.
The Pauli exclusion principle declares that there can only be a maximum of two electrons for every one orientation, and the two electrons must be opposite in spin direction; meaning one electron has \(m_s = +\frac{1}{2}\) and the other electron has \(m_s = \frac{1}{2}\).
Hund's Rule declares that the electrons in the orbital are filled up first by the +\(\frac{1}{2}\) spin. Once all the orbitals are filled with unpaired +\(\frac{1}{2}\) spins, the orbitals are then filled with the \(\frac{1}{2}\) spin. (See examples below, labeled electronic configuration.)
(For more information on Pauli Exclusion Principle and Hund's Rules, see Electronic Configuration.)
Example 1 

Sulfur  S (16 electrons) Electronic Configuration: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{4} OR [Ne] 3s^{2} 3p^{4} As shown in the following image, this is a demonstration of a \(\frac{1}{2}\) and a +\(\frac{1}{2}\) Electron Spin.

When given a principle quantum number, n, with either the s, p, d or forbital, identify all the possibilities of L, m_{L} and m_{s}.
Example 2 

Given 5f, identify all the possibilities of the four quantum numbers. SOLUTION In this problem, the principle quantum number is n = 5 (the subshell number placed in front of the orbital, the forbital in this case). Since we are looking at the fobital, therefore L = 3. (Look under "Subshells" in the module Quantum Numbers for more information) Knowing L = 3, we can interpret that m_{L} = 0, \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 since m_{L} = L,...,1, 0, 1,...+L. As for m_{s}, since it isn't specified in the problem as to whether it is \(\frac{1}{2}\) or +\(\frac{1}{2}\), therefore for this problem, it could be both; meaning that the electron spin quantum number is \(\pm\)\(\frac{1}{2}\). 
Example 3 

Given 6s and m_{L}= +1, identify all the possibilities of the four quantum numbers. SOLUTION The principle quantum number is n = 6. Looking at the sorbital, we know that L = 0. Knowing that m_{L} = L,...,1, 0, 1,...+L, therefore m_{L} = +1 is not possible since in this problem, the interval of m_{L} can only equal to 0 according to the angular momentum quantum number, L. 
Example 4 

Given 4d and m_{s} = +\(\frac{1}{2}\), identify all the possibilities of the four quantum numbers. SOLUTION The principle quantum number is n = 4. Given that it is a dorbital, we know that L = 2. Therefore, m_{L} = 0, \(\pm\) 1, \(\pm\) 2 since m_{L} = L,...,1, 0, 1,...+L. For m_{s}, this problem specifically said m_{s }= +\(\frac{1}{2}\); meaning that the electron spin quantum number is +\(\frac{1}{2}\). 
First, draw a table labeled n, L, m_{L}and m_{s}, as shown below:
n  L  m_{L}  m_{s} 
Then, depending on what the question is asking for, fill in the boxes accordingly. Finally, determine the number of electrons for the given quantum number, n, with regards to L, m_{L} and m_{s}.
Example 5  

How many electrons can have n = 5 and L = 1? 6
This problem includes both \(\frac{1}{2}\) , +\(\frac{1}{2}\) , therefore the answer is 6 electrons based on the m_{L}. 
Example 6  

How many electrons can have n = 5 and m_{s }= \(\frac{1}{2}\)? 5
This problem only wants the Spin Quantum Number to be \(\frac{1}{2}\) , the answer is 5 electrons based on the m_{L}. 
Example 7  

How many electrons can have n = 3, L = 2 and m_{L} = 3? zero SOLUTION
Since m_{L} = L...1, 0, +1...+L (See Electronic Orbitals for more information), m_{L} is not possible because L = 2, so it is impossible for m_{L} to be equal to 3. So, there is zero electrons. 
Example 8  

How many electrons can have n = 3, m_{L} = +2 and m_{s }= +\(\frac{1}{2}\)? 1 SOLUTION
This problem only wants the Spin Quantum Number to be +\(\frac{1}{2}\) , the answer is 1 electrons based on the m_{L}. 
Problem (1): Sodium (Na) > Electronic Configuration [Ne] 3s^{1}
Spin direction for the valence electron or m_{s} = +\(\frac{1}{2}\)
Sodium (Na) with a neutral charge of zero is paramagnetic, meaning that the electronic configuration for Na consists of one or more unpaired electrons.
Problem (2): Chlorine (Cl) > Electronic Configuration [Ne] 3s^{2} 3p^{5}
Spin direction for the valence electron or m_{s} = +\(\frac{1}{2}\)
Chlorine (Cl) with a neutral charge of zero is paramagnetic.
Problem (3): Calcium (Ca) > Electronic Configuration [He] 4s^{2}
Spin direction for the valence electron or m_{s} = \(\pm\)\(\frac{1}{2}\)
Whereas for Calcium (Ca) with a neutral charge of zero, it is diamagnetic; meaning that ALL the electrons are paired as shown in the image above.
Problem (4): Given 5p and m_{s} = \(\frac{1}{2}\), identify all the possibilities of the four quantum numbers.
The principle quantum number is n = 5. Given that it is a porbital, we know that L = 2. And based on L, m_{L} = 0, \(\pm\) 1, \(\pm\) 2 since m_{L} = L,...,1, 0, 1,...+L. As for m_{s}, this problem specifically says m_{s} = \(\frac{1}{2}\), meaning that the spin direction is \(\frac{1}{2}\), pointing downwards ("down" spin).
Problem (5): Given 6f, identify all the possibilities of the four quantum numbers.
The principle quantum number is n = 6. Given that it is a forbital, we know that L = 3. Based on L, m_{L} = 0, \(\pm\) 1, \(\pm\) 2, \(\pm\) 3 since m_{L} = L,...,1, 0, 1,...+L. As for m_{s}, since it isn't specified in the problem as to whether it is \(\frac{1}{2}\) or +\(\frac{1}{2}\), therefore for this problem, it could be both; meaning that the electron spin quantum number is \(\pm\)\(\frac{1}{2}\).
Problem (6): How many electrons can have n = 4 and L = 1? 6
n  L  m_{L}  m_{s} 
4  1  \(\pm\) 1  \(\frac{1}{2}\) , +\(\frac{1}{2}\) 
0 
This problem includes both \(\frac{1}{2}\) , +\(\frac{1}{2}\) , therefore the answer is 6 electrons based on the m_{L}.
Problem (7): How many electrons can have n = 4, L = 1, m_{L} = 2 and m_{s} = +\(\frac{1}{2}\)? zero
n  L  m_{L}  m_{s} 
4  1  NOT POSSIBLE  +\(\frac{1}{2}\) 
Since m_{L} = L...1, 0, +1...+L, m_{L} is not possible because L = 1, so it is impossible for m_{L} to be equal to 2 when m_{L} MUST be with the interval of L and +L. So, there is zero electron.
Problem (8): How many electrons can have n = 5, L = 3, m_{L} = \(\pm\) 2 and m_{s} = +\(\frac{1}{2}\)? 2
n  L  m_{L}  m_{s} 
5  3  \(\pm\) 2  +\(\frac{1}{2}\) 
\(\pm\) 1  
0 
This problem only wants the Spin Quantum Number to be +\(\frac{1}{2}\) and m_{L} = \(\pm\) 2, therefore 2 electrons can have n = 5, L = 3, m_{L} = \(\pm\) 2 and m_{s} = +\(\frac{1}{2}\).
Problem (9): How many electrons can have n = 5, L = 4 and m_{L} = +3? 2
n  L  m_{L}  m_{s} 
5  4  3, +3  \(\frac{1}{2}\) , +\(\frac{1}{2}\) 
\(\pm\) 2  
\(\pm\) 1  
0 
This problem includes both \(\frac{1}{2}\) and +\(\frac{1}{2}\) and given that m_{L} = +3, therefore the answer is 2 electrons.
Problem (10): How many electrons can have n = 4, L = 2 and m_{L} = \(\pm\) 1? 4
n  L  m_{L}  m_{s} 
4  2  \(\pm\) 1  \(\frac{1}{2}\) , +\(\frac{1}{2}\) 
0 
This problem includes both \(\frac{1}{2}\) and +\(\frac{1}{2}\) and given that m_{L} = \(\pm\) 1, therefore the answer is 4 electrons.
Problem (11): How many electrons can have n = 3, L = 3, m_{L} = +2 and m_{s} = \(\frac{1}{2}\)? zero
n  L  m_{L}  m_{s} 
3  3 (NOT POSSIBLE)  \(\pm\) 2  \(\frac{1}{2}\) 
\(\pm\) 1  
0 
Since L = n  1, there is zero electron, not possible because in this problem, n = L = 3.
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