Ch.8 - Periodic Properties of the ElementsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Imagine a universe in which the value of ms can be +1/2, 0, and -1/2. Assuming that all the other quantum numbers can take only the values possible in our world and that the Pauli exclusion principle applies, give the following.The number of unpaired electrons in fluorine.

Solution: Imagine a universe in which the value of ms can be +1/2, 0, and -1/2. Assuming that all the other quantum numbers can take only the values possible in our world and that the Pauli exclusion principle

Problem

Imagine a universe in which the value of ms can be +1/2, 0, and -1/2. Assuming that all the other quantum numbers can take only the values possible in our world and that the Pauli exclusion principle applies, give the following.

The number of unpaired electrons in fluorine.

Solution

We have to determine the number of unpaired electrons in an alternate universe whether the spin quantum number can have three possible values (-½, 0 and +½).


To solve this problem, we will first have to write the electron configuration of fluorine.

Fluorine belongs to the group 7A and has the atomic number 9.

 


Before reaching F, you will pass through the atomic orbitals 1s and 2s.

• In a neutral atom:

Atomic number= # of protons = # of electrons

F: atomic number = 9 → 9 protons & 9 electrons

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