### Solutions: Check your work!

**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 principal quantum number is n = 5. Given that it is a p-orbital, 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 principal quantum number is n = 6. Given that it is a f-orbital, 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.