CHEM 101 Study Guide #2 PDF - Ionic vs Covalent Bonding

**CHEM 101 Study Guide \#2** **LO 9 and 10: Ionic vs covalent bonding; systematic nomenclature** **Valence electrons:** valence electrons are the electrons that are in the outermost electron shell of an atom. These electrons, being the furthest from the nucleus and thus the least tightly held by t...

**CHEM 101 Study Guide \#2** **LO 9 and 10: Ionic vs covalent bonding; systematic nomenclature** **Valence electrons:** valence electrons are the electrons that are in the outermost electron shell of an atom. These electrons, being the furthest from the nucleus and thus the least tightly held by the atom, are the electrons that participate in bonds and reactions with other atoms and molecules. In **Lewis structures,** valence electrons of main-group elements are represented as dots around the symbol of the element. Lewis structures are used to represent main group elements and their interactions with other atoms and molecules. Common main-group Lewis structures are shown below: To draw a Lewis dot symbol of an element: 1. Determine the number of valence electrons for the element. 2. Write the atomic symbol of the element. 3. Draw dots around the 4 sides of the symbol (1 dot = 1 electron). 4. Do NOT pair electrons until there are 5 or more valence electrons (exception: He) In Lewis Theory, a chemical bond involves the sharing or transfer of electrons between atoms or molecules. If the electrons are transferred, atoms are unbalanced with protons in the nucleus and electrons in outer shells, the atom is called an ion and has charge **(+ or -)** - Metals lose electrons -- Form cations. (Chapter 4.7) - Nonmetals gain electrons -- Form anions. (Chapter 4.7) - Cations and anions stick together -- Form **IONIC** bonds, **BECAUSE** opposite charges attract each other - Atoms will transfer electrons until each atom in the compound has 8 valence electrons (octet rule). **If electrons are shared covalent bonds are formed**: - Nonmetal atoms share electrons with each other. - Covalent bonds are formed. - Atoms will share electrons until each atom in the compound has 8 valence electrons in its outermost shell (octet rule). **Rules for covalent bonding using Lewis Theory:** 1. Count the total valence electrons of all atoms in the compound. 2. Determine the central atom. 3. Bond all the other atoms to the central atom. **1 bond = 2 e^-^** 4. Fill in the valence electrons of terminal atoms. 5. Any left over electrons become lone pairs on the central atom. 6. If the central atom does not have a full octet, make double or triple bonds with terminal O, N, or S atoms. 7. Calculate the formal charge of each atom in the molecule. Example: Water (H~2~O)  - Valence electrons: Hydrogen has 1, Oxygen has 6.  - Skeleton structure:  - H - O - H  - Place remaining electrons:  - H : O : H (with two lone pairs on Oxygen)  *There are a couple of videos in the Modules tab in Canvas (resources) that may be helpful in understanding on how this works.* **POLAR COVALENT** bonds occur when atoms with different **electronegativity** combine to form molecules. Elements vary in **eletronegativity** (a measure of an atom\'s ability to attract shared electrons to itself). This is a characteristic of all elements that varies in periods and is one of the ways the periodic table is organized. A diagram of a periodic table Description automatically generated The difference of EN (∆EN) between two bonded atoms is used to determine the type and polarity of bonds.  **In some molecules the difference in EN between bonded atoms is \< 0.5, the electrons of the bond are equally shared. (Pure or nonpolar covalent bond).** **In some molecules, the EN of the atoms is different. If the EN difference is between 0.5 and 2.0, the electrons of the bond are not equally shared. One atom pulls electrons towards itself. (Polar covalent bond)** A diagram of a colorful oval Description automatically generated **In some molecules, the EN of the atoms is different. If the EN difference is greater than 2.0, one atom steals electrons from the other. *[Will only and always happen with a metal and a nonmetal.]*** **(Ionic bond)** **LO 8 and 9: Lewis structures; Intermolecular forces and states of matter** Intermolecular Forces **Dipole-dipole forces** are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule. Dipole-dipole forces have strengths that range from 5 kJ to 20 kJ per mole. They are much weaker than ionic or covalent bonds and have a significant effect only when the molecules involved are close together (touching or almost touching) A diagram of a molecule Description automatically generated **LO 11: Mole concept calculations: Grams \ Moles \ Particles** The mole is important because it allows chemists to work with the subatomic world with macro world units and amounts. Atoms, molecules and formula units are very small and very difficult to work with usually. However, the mole allows a chemist to work with amounts large enough to use. **A mole of something represents 6.022x10^23^ items. Whether it be atom, molecule or formula unit**. Defining the mole in this way allows you change grams to moles or moles to particles. Even though you can\'t see the particles. 1mole of an element = **6.022x10^23^ ** atoms = Molar / average atomic / mass in grams **\***Which you look up in the Periodic Table  A diagram of a mole Description automatically generated - How many aluminum atoms are there in an aluminum can with a mass of 25.0 g? 25 g / 26.98 = 0.92 moles 0.92 X 6.022x10^23^ = 5.54x10^23^ particles (atoms) **LO 13 and 14: Classify chemical reactions; balance chemical equations.** **What is a Chemical Reaction?** A **chemical reaction**, a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds. A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.  Evidence of a chemical reaction include: - Unexpected color change. - Formation of a solid. - Formation of a gas. - Change of temperature. - Emission of light. A chemical change may not show any change to the naked eye. An observation of one of these changes does not prove that a chemical change has occurred. But these changes do provide clues that a chemical reaction might have occurred. **If you observe one of these changes, chemical analysis must be conducted to provide conclusive evidence that a chemical reaction has occurred.** A table of chemical reactions Description automatically generated  A screenshot of a computer Description automatically generated Part of the lure of chemistry is that things don't always work out the way you expect. You plan a reaction, anticipate the products and, quite often, the results astound you! The exercise, then, is trying to figure out what was formed, why, and whether your observation leads to other useful generalizations. The first step in this process of discovery is anticipating or predicting the products which are likely to be formed in a given chemical reaction. **To predict products in a chemical reaction,** you need to first identify the type of reaction (synthesis, decomposition, single replacement, double replacement, combustion) and then use the known patterns for each type to determine what new compounds will form, considering the charges of the ions involved and ensuring the overall charge of the product is neutral; finally, write the balanced chemical equation with the predicted products on the right side of the arrow. **Different types of reactions require different approaches to solve or predict products. There are two videos in the "Resources" Module that show some ways to predict products.** **Balancing chemical equations** **Step 1: Writing the imbalanced chemical equation:-** **Step 2: Comparing the number of atoms on both reactant and product side :-** **Step 3: Balancing the Carbon atoms:-** **Step 4: Balancing the Hydrogen and Oxygen atoms:-** **Step 5: Balanced chemical equation:-** The "longer video" shows how to balance chemical equations in the several practice problems. **LO 14: Mass and moles stoichiometry, limiting reactants, theoretical yield, percent yield** - It tells us how much of the reactants to mix together. - It tells us how much of the products to expect. Stoichiometry measures these quantitative relationships and is used to determine the amount of products and reactants that are produced or needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as **reaction stoichiometry.** Using mole-to-mole factors we can convert between moles of one compound in the equation to moles of another compound in the equation. - **We cannot measure moles in the lab.** - **We can measure grams in the lab.** - **So determine the mass of each substance to use or the mass of the product(s) to expect we will use molar mass and mole-to-mole factors.**  Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called ***composition stoichiometry***. The following steps would be used: 1. Write and balance the equation 2. Mass to moles: Convert grams of Cu to moles of Cu 3. Mole ratio: Convert moles of Cu to moles of Ag produced 4. Mole to mass: Convert moles of Ag to grams of Ag produced **A worked example:** Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid [copper](https://en.wikipedia.org/wiki/Copper) (Cu) were added to an aqueous solution of silver nitrate (AgNO~3~), the [silver](https://en.wikipedia.org/wiki/Silver) (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO~3~)~2~) and solid silver. How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate? The complete balanced equation would be: **Cu + 2 AgNO~3~ → Cu(NO~3~)~2~ + 2 Ag** For the mass to mole step, the mass of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its [molar mass](https://en.wikipedia.org/wiki/Molar_mass): 63.55 g/mol. A black text on a white background Description automatically generated(16.00 g Cu1)(1 mol Cu63.55 g Cu)=0.2518 mol Cu Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio. (0.2518 mol Cu1)(2 mol Ag1 mol Cu)=0.5036 mol Ag Now that the moles of Ag produced is known to be 0.5036 mol, we convert this amount to grams of Ag produced to come to the final answer: A black text with black numbers Description automatically generated with medium confidence(0.5036 mol Ag1)(107.87 g Ag1 mol Ag)=54.32 g Ag This set of calculations can be further condensed into a single step:  mAg=(16.00 g Cu1)(1 mol Cu63.55 g Cu)(2 mol Ag1 mol Cu)(107.87 g Ag1 mol Ag)=54.32 g