Chemistry Chapter 6 Focus Study

24 July 2022
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question
The hydrides of group 5A are NH3, PH3, AsH3, and SbH3. Arrange them from highest to lowest boiling point.
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Step 1: To rank the molecules from highest boiling point to lowest boiling point, first consider the molar mass of the molecules. The greater the molar mass, the greater the strength of the London dispersion forces (a type of intermolecular force of attraction between two molecules). Consequently, the boiling point will also be higher. For example, consider group 6A hydrides: H2O, H2S, H2Se, and H2Te. Among these hydrides, the molar mass increases in the following order: H2O < H2S < H2Se < H2Te Therefore, the boiling point of H2O is expected to be the lowest. However, it is observed that the boiling point of H2O is highest among these hydrides. This is due to the hydrogen bonding exhibited by the molecules of water. Thus, if the molecule shows hydrogen bonding, the boiling point of the substance will be higher than expected owing to the strength of the hydrogen bond relative to London dispersion forces. For group 5A hydrides, first determine their relative molar mass, then determine whether each molecule exhibits hydrogen bonding. Considering the molar mass and hydrogen bonding, rank the hydrides from highest boiling point to lowest boiling point.
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The hydrides of group 5A are NH3, PH3, AsH3, and SbH3. Arrange them from highest to lowest boiling point. Steps 2 and 3
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Step 2: Arrange the group 5A hydrides PH3, SbH3, NH3, and AsH3 from highest to lowest molar mass. Highest to lowest - SbH3, AsH3, PH3, and NH3 Step 3: The boiling point of a compound increases with an increase in molar mass.
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The hydrides of group 5A are NH3, PH3, AsH3, and SbH3. Arrange them from highest to lowest boiling point. Step 4
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Identify the group 5A hydrides that exhibit hydrogen bonding opened hint: NH3 The NH3 molecule exhibits hydrogen bonding, so more energy is required to separate NH3 molecules from one another. You can relate this to boiling points of group 5A hydrides. Hydrogen bonding requires a highly electronegative atom and a hydrogen atom. It is often easiest to remember that molecules with an N, O, or F atom bonded to a hydrogen atom may form hydrogen bonds. Based on this information, identify which of the following group 5A hydrides exhibits hydrogen bonding
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Rank the hydrides of group 5A are NH3, PH3, AsH3, and SbH3. Arrange them from highest to lowest boiling point. FINAL STEP
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London dispersion forces increase with increasing molar mass. How do London dispersion forces relate to the boiling point of a compound? SbH3, NH3, AsH3, PH3
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What intermolecular forces are present in each of the substances?
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DISPERSION ONLY: C6H14, C4H10 Dispersion forces and dipole-dipole forces: HCN Dispersion forces, dipole-dipole forces, and hydrogen bonding: CH3OH, HF
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Arrange the molecules in order of decreasing boiling point.
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Highest to Lowest: C2H5OH HBr H2 Strong intermolecular forces tend to result in liquids and solids at room temperature (high melting and boiling points), while weak intermolecular forces tend to result in gases at room temperature (low melting and boiling points).
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Classify these molecules as polar or nonpolar
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Polar: H2O CH3Cl HBr Nonpolar BBr3 H2 CCl4
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The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.
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Ionic NaCl H bonding H20 Dipole HBr Dispersion N2 The strength of these attractive forces is directly proportional to the charge separation that occurs in the molecules. Ionic bonds involve oppositely charged ions, which have the greatest charge separation. Nonpolar molecules, which form momentary dipole moments, have dispersion forces and the smallest separation of charge. Ionic bonds are the strongest of these attractive forces, and dispersion forces are the weakest.
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Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.
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*Dispersion forces are the weakest, so their boiling points are the lowest * Ionic forces are the strongest, so their boiling points are higher The effect of hydrogen bonding can be seen in the striking difference in boiling points of similar compounds. Consider the approximate boiling points of the following polar compounds that all have the same shape: H2Te H2Se H2S H2O 0 ?C ?40 ?C ?60 ?C 100 ?C The boiling point of H2O is drastically higher than the other similar compounds. The difference is due to the hydrogen bonding in H2O.
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When determining between polar and nonpolar
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To determine whether a bond is polar, compare the electronegativity values between the two atoms. If the difference between the electronegativity values is between 0.5 and 1.8, the bond is polar. For example, the difference between electronegativity value of O and Cl atoms is 0.5(3.5?3.0=0.5), which is between 0.5 and 1.8, thus O?Cl bonds is polar covalent. This molecule has a bent shape in which the dipoles of the O?Cl bonds do not cancel each other out. Thus, the OCl2 molecule is polar. However, the molecule may still be nonpolar if the dipole from each bond cancels in a symmetrical molecule, such as in the tetrahedral molecule CCl4. The difference between the electronegativity value of the two atoms is 0.5 (3.0 - 2.5 = 0.5), but the bond dipoles all cancel in the nonpolar molecule.
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Identify the major type of attractive force between the particles of each of the following: BrF
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dipole-dipole attraction
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Identify the major type of attractive force between the particles of each of the following: KCl
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ionic bond
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Identify the major type of attractive force between the particles of each of the following: NF3
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dipole-dipole attraction
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Identify the major type of attractive force between the particles of each of the following: Cl2
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dispersion forces
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Identify the strongest intermolecular forces between the particles of each of the following: PCl3
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dipole-dipole attraction
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Identify the strongest intermolecular forces between the particles of each of the following: I2
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dispersion forces
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Identify the strongest intermolecular forces between the particles of each of the following: CF4
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dispersion forces
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Identify the strongest intermolecular forces between the particles of each of the following: HF
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hydrogen bond
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Identify the strongest attractive forces between the particles of each of the following
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Dipole-dipole attraction SiH3Br Dispersion forces H2 CBr4 Hydrogen bonding CH3NH2 Nonpolar molecules, in which there are little to no differences in electronegativity between the atoms, are only capable of temporary dipoles, and the resulting weak attractions between nonpolar molecules are considered dispersion forces. Permanent dipoles in polar molecules allow for stronger attractions between the partial positive charges and partial negative charges of molecules, which are referred to as dipole-dipole interactions. An even stronger type of dipole-dipole interaction can occur when the dipole involves a bond between hydrogen and fluorine, oxygen, or nitrogen, and that is referred to as a hydrogen bond
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Identify the major intermolecular forces between each of the following atoms or molecules:
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Dispersion forces H2 SiH4 Dipole-dipole attractions HBr Hydrogen bonding CH3?CH2?OH Dispersion forces are the only intermolecular forces that occur between nonpolar molecules. In these molecules, all the bonds are nonpolar or the polar bonds cancel each other out. Since SiH4 is tetrahedral in shape, the dipole moment from each bond cancels out and the molecule is nonpolar even if each bond is itself polar. Polar molecules are attracted to each other by dipole-dipole attractions or hydrogen bonding. As the strongest attractive force between polar covalent molecules, hydrogen bonding occurs only between hydrogens bonded to partially negative nitrogen, fluorine, or oxygen atoms. Since this intermolecular attraction is the strongest, if it is present between molecules it would be considered to be the major force.
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Indicate the major type of intermolecular forces between particles in each of the following compounds.
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Dipole-dipole attractions CH2F2 OBr2 HCl Dispersion forces IBr Hydrogen bonds H2O Ionic bonds NaF After drawing the Lewis structure of each molecule, the polarity of each bond must be calculated by finding the difference in electronegativity between atoms. If there is a difference of between 0.5 and 1.8, then a bond is polar, while a difference of greater than 1.9 indicates an ionic bond. If the bonds are nonpolar and the overall molecule is nonpolar, then the major attractive forces are dispersion forces. If the bonds are polar and the polarity of the bonds cancel each other out, then the molecule is also nonpolar and the major attractive forces are dispersion forces. If the bonds are polar but the polarity of the bonds do not cancel each other out owing to their orientation about the center atom, then the forces are dipole-dipole attractions. Hydrogen bonding is a stronger force that occurs if a highly electronegative atom, such as O, N, or F, is bound to a hydrogen atom. If the molecule is composed of ions, then the major forces are ionic bonds
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Ionic compounds
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metals and nonmetals
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Molecular (covalent) compound
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nonmetals
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Atomic elements
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Have single atoms as their basic units. Most elements fall into this category.
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Molecular elements
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diatomic molecules, two atoms of that element bonded together as their basic unit.
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When writing IONIC FORMULAS
When writing IONIC FORMULAS
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Lithium oxide elements -> Li O charge ---->Li+ O^-2 balance --> Li+ balanced ionic formula: Li2O
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When writing Formulas with POLYATOMIC IONS
When writing Formulas with POLYATOMIC IONS
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Calcium nitrate elements -> K N charge ---> K^+1 N^-3 balance --> K^+1 balance --> K^+1 balanced ionic formula: K3N (allows us to look at common polyatomic ions during exam)
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How to write the formular for COPPER (II) PHOSPHATE
How to write the formular for COPPER (II) PHOSPHATE
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Cu^2+Po4^3- use the criss-cross method Cu3(PO4)2
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Naming Covalent Molecular Compounds
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1) for 1st element, start with element name 2) for 2nd element, start with -ide name 3) use prefies to show how many atoms of each type there are N2O3 nitrogen oxide dinitrogen trioxide
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Prefixes
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1-mono 2-di 3-tri 4-tetra 5-penta 6-hexa 7-hepta 8-octa 9-nona 10-deca
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Common Polyatomic Ions
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? NO3- Nitrate ? OH- Hydroxide ? SO42- sulfate ? CO32- carbonate ? PO43- phosphate ? NH4+ ammonium ? HCO3- hydrogen carbonate
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How to draw LEWIS STRUCTURES
How to draw LEWIS STRUCTURES
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1) Find the total number of valence electrons (group number + group number) 2) Put the least electronegative atom in the center 3) Put 2 electrons between atoms to form a chemical bond 4) Complete octets on the outside atoms 5) If central atom doesn't have an octet, move electrons from outer atoms to form double or triple bonds
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VSEPR theory
VSEPR theory
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Valence-shell electron-pair repulsion theory; because electron pairs repel, molecules adjust their shapes so that valence electron pairs are as far apart as possible
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Steric number
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the number of domains/things (atoms and electron pairs) bonded to the central atom (unbonded/unshaded atoms)
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Linear molecular geometry
Linear molecular geometry
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2 electron groups, 2 bonding groups, 0 lone pairs
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Trigonal planar molecular geometry
Trigonal planar molecular geometry
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3 electron groups, 3 bonding groups, 0 lone pairs
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Bent molecular geometry
Bent molecular geometry
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3 electron groups, 2 bonding groups, 1 lone pairs
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tetrahedral molecular shape
tetrahedral molecular shape
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4 electron groups, 4 bonding groups, 0 lone pairs
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pyramidal molecular shape
pyramidal molecular shape
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4 electron groups, 3 bonding groups, 1 lone pairs
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bent molecular shape (2)
bent molecular shape (2)
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4 electron groups, 2 bonding groups, 2 lone pairs
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Electron groups: 2 Bonding groups: 2 Lone pairs: 0 Shape: Bond angle:
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Linear 180 degrees
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Electron groups: 3 Bonding groups: 3 Lone pairs: 0 Shape: Bond angle:
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trigonal planar 120 degrees
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Electron groups: 3 Bonding groups: 2 Lone pairs: 1 Shape: Bond angle:
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bent 120 degrees
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Electron groups: 4 Bonding groups: 4 Lone pairs: 0 Shape: Bond angle:
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tetrahedral 109 degrees
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Electron groups: 4 Bonding groups: 3 Lone pairs: 1 Shape: Bond angle:
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pyramidal 109 degrees
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Electron groups: 4 Bonding groups: 2 Lone pairs: 2 Shape: Bond angle:
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bent 109 degrees