CHEM 119 EXAM 2
Draw the Lewis structure for a hydrogen iodide (HI) molecule.
H has 1 valence electron I has 7 valence electrons
Check the box next to each molecule on the right that has the shape of the model molecule on the left:
None of the above.
Draw the molecular orbital (MO) electron diagram for the Be2 +2 molecular ion. Be sure your diagram contains all of the electrons in the ion, including any core electrons.
Refer to Google Doc.
Tag all the sp3 hybridized carbon atoms in this molecule. If there are none, please check the box below.
The first carbon (left) is the only one. We only need the single bonds.
The chemical formula for calcium bicarbonate is: Ca(HCO3)2. How many oxygen atoms are in each formula unit of calcium bicarbonate?
There are 6 O atoms.
Calculate the number of carbon atoms in a 140.0g sample of glucose C6H12O6. Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.
2.808 x 10^24 To go from moles to mass (often measured in grams) , multiply by the molar mass. Conversely, to go from mass to moles, divide by the molar mass. To go from moles to number, multiply by the Avogadro Number (which is usually given the symbol NA). Conversely, to go from number to moles, divide by NA.
Classify each chemical compound:
a. ionic b. ionic c. molecular Molecular compounds are formed between two non-metals. Ionic compounds are formed between metals and non-metals.
What are the angles a and b in the actual molecule of which this is a Lewis structure?
a=120 b= 109.5
What kind of intermolecular forces act between an ammonia molecule and a water molecule? Note: If there is more than one type of intermolecular force that acts, be sure to list them all, with a comma between the name of each force.
dispersion, dipole, hydrogen bonding
Complete this table.
a. Rb2O b. NiI2
Study the following sketch of a molecular orbital (MO) in a homonuclear diatomic molecule. This MO was formed by combining one 1s atomic orbital from each atom.
First, we often classify an MO as bonding or antibonding. A bonding MO tends to keep electrons mostly between neighboring atoms, and an antibonding MO tends to keep electrons mostly away from the space between neighboring atoms. We add an asterisk (*) to the symbol for an MO when it's an antibonding MO. Second, MOs are classified according to how they distribute electrons around the bond axis. The simplest type of MO, which keeps electrons along the bond axis, is called a sigma MO (symbol σ). The next simplest type of MO, called a pi MO (symbol π), keeps electrons on both sides of the bond axis -- but never directly on the bond axis. More complex MOs have more complex patterns. *Antibonding MOs always have a node between the nuclei, bonding MOs never do. Pi MOs always have a node along the bond axis, sigma MOs never do.* Since the MO sketched has no node between the nuclei, it must be a bonding MO. Since the MO sketched has no node along the bond axis, it must be a σ. Since you're told this MO is constructed from 1s AOs, presumably by the LCAO method, the symbol for the MO should be given a subscript 1s. Since this is a bonding MO, its energy will be lower than the energy of either 1s AO.
Write the chemical formula of iron(II) sulfate hexahydrate.
For more search naming hydrates on Google Doc. This uses the rules for naming ionic compounds [cation name][anion name].
Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
For more search naming ionic compounds on Google Doc.
Draw the Lewis dot diagram for a Ge− anion.
1. Count the valence electrons. (Ge has 4 valence electrons plus the negative charge which makes it 5) 2. Write the chemical symbol for the element. 3. Draw the valence electrons as dots around the symbol. 4. Add a charge symbol, if necessary.
Draw the Lewis structure for formic acid (CH2O2). Formic acid molecules have one hydroxy group and one carbonyl group. Be certain you include any lone pairs.
1. Draw a skeleton using just the carbon atoms. 2. Add any heteroatoms to the skeleton. 3. Add the hydrogen skin.
Consider the perchlorate (ClO4-) anion. What is the central atom? Enter its chemical symbol. How many lone pairs are around the central atom? What is the ideal angle between the chlorine-oxygen bonds? Compared to the ideal angle, you would expect the actual angle between the chlorine-oxygen bonds to be ...
1. Lone pairs are fatter than bonding electron groups. 2. Double and triple bonds are fatter than single bonds.
What is the angle between the nitrogen-oxygen bonds in the nitrate (NO3-) ion?
120 degrees Start by drawing the lewis structure. Then refer to the Google Doc to get arrangement angles.
How many valence electrons are in the ammonium ion NH4+?
8 N=5 (1) H=1 (4) Total = 9 - 1 (from + charge) = 8
This is the chemical formula for lead(II) nitrate: PbNO32. Calculate the mass percent of nitrogen in lead(II) nitrate. Round your answer to the nearest percentage.
8%
What is the "AXE" description of the phosphorus tetrachloryl (PCl4+) cation?
AX 4 E 0 There are 4 ligands (other atoms bonded to the central atom) and 0 lone pairs in this example. The picture is a guide.
Cobalt(II) chloride forms several hydrates with the general formula CoCl2·xH2O ,where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure CoCl2 behind. Suppose a sample of a certain hydrate is heated until all the water is removed, and it's found that the mass of the sample decreases by 12.% .Which hydrate is it? That is, what is x?
Answer: 1 Solve by plugging in different values for H2O until you are able to find the closest percentage.
A chemist determines by measurements that 0.0650 moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Round your answer to 3 significant digits.
Be sure and use the values given on the ALEKS periodic table.
Below is the Lewis structure of the hydrogen cyanide HCN molecule. H:C:::N: Count the number of bonding pairs and the number of lone pairs around the carbon atom in this molecule.
Bonding pairs: 4 Lone pairs: 0 The highlighted electrons around the carbon atom are forming chemical bonds (H:C:::N:). The carbon atom in this molecule has 4 bonding pairs (bolded). All electrons around the carbon atom are forming chemical bonds. Therefore, the carbon atom in this molecule has 0 lone pairs.
Write the chemical formula for this molecule:
C2H4O2 Write the symbols in the order in which the elements appear in the Periodic Table, going from left to right. Hydrogen is an exception: write H after all elements except those in Groups 6A or 7A.
Write the chemical formula for this molecule:
C4H10O Write the symbols in the order in which the elements appear in the Periodic Table, going from left to right. Hydrogen is an exception: write H after all elements except those in Groups 6A or 7A.
Compound X has a molar mass of 134.12 gmol−1 and the following composition: carbon= 44.77%, hydrogen= 7.52%, oxygen= 47.72%. Write the molecular formula of X.
C5H10O4
This sketch of a neutral molecule is shaded red or blue wherever the electrostatic potential at the molecule's surface isn't zero. What could the chemical formula of the molecule be?
COCl2 Size and shape can be used to determine chemical formula. The color can give you an idea of the chemical identity of the elements. 1. Hydrogen atoms are smaller than almost any other atom. Also, hydrogen is less electronegative than most nonmetals, so in molecules it often has a positive partial charge. 2. Fluorine and oxygen are the most electronegative nonmetals, so in molecules they often have negative partial charges. 3. Most nonmetal atoms can have positive or negative partial charges, depending on whether they are bonded to atoms that are more or less electronegative. For example, Cl in ClF has a positive partial charge, because Cl is less electronegative than F, but Cl in HCl has a negative partial charge, because Cl is more electronegative than H. 4. Identical atoms in symmetric molecules share valence electrons equally, so they will have no partial charge.
There are four sketches below. The first sketch shows a sample of Substance X. The three sketches underneath it show three different changes to the sample. You must decide whether each of these changes is possible. If a change is possible, you must also decide whether it is a physical change or a chemical change. Each sketch is drawn as if the sample were under a microscope so powerful that individual atoms could be seen. Also, you should assume that you can see the entire sample, and that the sample is in a sealed box, so that no matter can enter or leave.
Change 1: A chemical Change Change 2: A physical Change Change 3: impossible The Principle of Conservation of Mass says that matter can't be created or destroyed by ordinary events. ("Ordinary" events are events that don't involve nuclear reactions.)
For each row in the table below, decide whether the pair of elements will form a molecular or ionic compound. If they will, then enter the chemical formula of the compound. If the elements will form more than one compound, enter the compound with the fewest total number of atoms. You may assume all chemical bonds are single bonds, not double or triple bonds.
Molecular compounds are formed between two non-metals while ionic compounds are formed between metals and non-metals. Charge of each element is found on the periodic table.
Complete the table below, which lists information about some diatomic molecules or molecular ions. In particular: Decide whether each molecule is stable or not. Decide whether each molecule would be diamagnetic or paramagnetic. Calculate each molecule's bond order.
Since the bond order is greater than zero, this molecule is stable. Since the B+2 molecule has 1 unpaired electron, it will be paramagnetic.
Which of the highlighted chemical bonds in the molecules below is longest? Shortest? In between? Which highlighted bond requires the highest energy to break? Lowest? In between? Answer these questions by completing the second and third columns in the table.
The smallest atoms will form the shortest bond, with the highest bond energy, and the largest atom will form the longest bond, with the lowest bond energy. You'll also need to remember the fact that the size of atoms increases as we go down a group of the Periodic Table.
A student proposes the following Lewis structure for the oxygen difluoride (OF2) molecule. Assign a formal charge to each atom in the student's Lewis structure.
a. 0 b. 0 c. 0
Arrange the highlighted bonds in the table below in decreasing order of polarity. That is, pick 1 for the most polar bond, pick 2 for the next most polar bond, and so on.
a. 2 b. 1 (most polar) c. 3 (least polar) Bond polarity can be understood in terms of the difference in electronegativities of the partner atoms: Bonds are usually considered nonpolar if the difference in electronegativities is less than about 0.50. The greater the difference in electronegativities, the more polar the bond. The partner with the higher electronegativity will be on the negative end of the bond dipole.
Answer the questions below about the highlighted atom in this Lewis structure:
a. 2 b. 2 c. sp refer to google doc
Molecules of four imaginary substances are sketched in the table below. Each sketch is shaded to show the electrostatic potential at the surface of the molecule. Rank these substances in decreasing order of the strength of the intermolecular forces in them. In other words, choose 1 next to the substance in which the molecules exert the strongest intermolecular forces on each other. Choose 2 next to the substance in which the molecules exert the second strongest intermolecular forces on each other, and so forth. Note: all of the molecules are neutral, and you may assume none of them experience hydrogen bonding.
a. 4 (lowest) b. 2 c. 1 (highest) d. 3 The dispersion force. The dispersion force acts between all molecules. It's stronger between more polarizable molecules, which in general means larger molecules. The dipole force. The dipole force acts only between molecules that have a dipole moment. It's stronger between molecules with bigger dipole moments, which means molecules with bigger or more widely-spaced partial charges. Start by comparing sizes: the bigger molecules will experience the stronger dispersion forces. Next, compare dipole moments using the electrostatic potential maps: molecules with dipole moments will experience dipole forces as well as dispersion forces. Be careful to notice when partial charges cancel out through symmetry, and a molecule ends up with no dipole moment even though its sketch is shaded red and blue. If two molecules both have dipole moments, compare the size of the moments: the size of a dipole moment is proportional to the distance between the partial charges, and the magnitude of the partial charges -- shown here by the intensity of the red and blue shading on the sketches. Molecules with bigger dipole moments experience stronger dipole forces.
Answer the questions in the table below about the shape of the sulfur tetrabromide (SBr4) molecule. How many electron groups are around the central sulfur atom? What phrase best describes the arrangement of these electron groups around the central sulfur atom?
a. 5 b. trigonal bipyramidal 1. Draw the Lewis structure. 2. Count the electron groups around the central atom. 3. Pick the electron-group arrangement. Refer to Google Docs and search arrangement angles.
Re-order each list of elements in the table below, if necessary, so that the elements are listed in order of decreasing electronegativity.
a. Br, I, Te b. F, C, B Basic Rules: 1. Fluorine has the highest electronegativity of any element. 2. Electronegativity usually increases as you go up the Periodic Table. 3. Electronegativity usually increases as you go to the right in the Periodic Table. 4. Noble gases are not normally assigned an electronegativity, because they rarely form chemical bonds.
Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
a. CaS, calcium sulfide b. Ca3N2, calcium phosphide c. Ca3N2, calcium nitride d. CaO, calcium oxide Ionic compounds are named by stating the cation first, followed by the anion. Positive and negative charges must balance. Some anions have multiple forms and are named accordingly with the use of roman numerals in parentheses. Net charge must be 0.
Write the chemical formula of each of the acids listed in the table below.
a. HClO4 b. H2S c. HClO2 d. H2CO3 Perchloric acid is a name in the pattern of an oxyacid. The root perchlor- and the ending -ic tells us the anion left over if all the H+ are removed is perchlorate ClO−4. Since this anion has a charge of −1, there must be one acidic H+ in each molecule of perchloric acid to make the total charge equal to zero. Therefore, the formula of perchloric acid must be HClO4. Hydrosulfuric acid is a name in the pattern of a binary acid. The root sulfur- tells us the anion left over if all the H+ are removed is sulfide S2−. Since this anion has a charge of −2, there must be two acidic H+ in each molecule of hydrosulfuric acid to make the total charge equal to zero. Therefore, the formula of hydrosulfuric acid must be H2S. Chlorous acid is a name in the pattern of an oxyacid. The root chlor- and the ending -ous tells us the anion left over if all the H+ are removed is chlorite ClO−2. Since this anion has a charge of −1, there must be one acidic H+ in each molecule of chlorous acid to make the total charge equal to zero. Therefore, the formula of chlorous acid must be HClO2. Carbonic acid is a name in the pattern of an oxyacid. The root carbon- and the ending -ic tells us the anion left over if all the H+ are removed is carbonate CO2−3. Since this anion has a charge of −2, there must be two acidic H+ in each molecule of carbonic acid to make the total charge equal to zero. Therefore, the formula of carbonic acid must be H2CO3.
In each row, pick the compound with the bigger lattice energy.Note: lattice energy is always greater than zero.
a. Mg 2+ Cl 1- = 2 Mg 2+ S 2- = 4 MgS has a larger charge. b. Rb 1+ F 1- = 1 Cs 1+ F 1- = 1 Same charge, compare radius. RbF has larger lattice energy. c. Be 2+ O 2- = 2 Li 1+ O 2- = 2 Same charge, compare radius. BeO has larger lattice energy. Radius increases to the left and down on the periodic table.
Complete the following table: Some polyatomic ions
a. NH4 + b. CO3 2- c. PO4 3- d. OH - These should be memorized. There is a list of common polyatomic ions on the Google Doc.
Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. The first row has been completed for you.
a. Na+ Cl- b. Co 2+ O 2- c. Ni 2+ F- d. Cr 3+ I- e. Cu 2+ S 2-
Decide whether the Lewis structure proposed for each molecule is reasonable or not.
a. No, some atoms have the wrong number of electrons around them. O is incorrect. b. Yes, it is a reasonable structure c. Yes it is a reasonable structure
Decide whether each molecule or polyatomic ion is polar or non-polar. If the molecule or polyatomic ion is polar, write the chemical symbol of the atom closest to the negative side. For example, if the molecule were HCl and you decided the hydrogen atom was closest to the negative side of the molecule, you'd enter "H" in the last column of the table.
a. Polar, O b. Polar, Br c. Nonpolar
Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table: Some ionic compounds.
a. Tin (II) nitrite b. Chromium (II) phosphate c. Barium iodate Make sure they are in the simplest empirical formula, get a net charge of 0, and name [cation name][anion name]. For more search naming ionic compounds on Google Doc.
Answer the questions in the table below about the shape of the sulfur pentafluoryl SF5+ cation.
a. Trigonal bipyramidal b. 90 degrees 1. Draw the Lewis structure. 2. Count the electron groups around the central atom. 3. Pick the electron-group arrangement (Google Doc). 4. Assign each lone pair group to a place in the electron-group arrangement. 5. Assign each bonding electron group to a place in the electron-group arrangement. 6. To name the shape of the molecule, name the arrangement of ligands (Google Doc).
Decide whether these proposed Lewis structures are reasonable.
a. Yes. b. No, wrong number of valence electrons. There should be 20 total but only 16 are accounted for. c. No, it doesn't satisfy the octet rule. O does not have enough and one H has two bonds when it can only handle one. Test 1: Does the structure have the correct number of valence electrons? A structure with a different number of valence electrons than are in the molecular formula is simply wrong. Test 2: Does each atom in the structure follow the octet rule?
Fill in the systematic names of the following chemical compounds. Note: for compounds containing hydrogen, you may give the common name instead.
a. bromine dioxide b. dibromine monoxide c. carbon monoxide d. carbon dioxide e. ammonia All of these are molecular, binary (containing two elements) compounds. Use prefixes (mono, di, tri, ect.) 1. If a prefix ending in o or a is used with an element name that begins with an o, we generally drop the o or a from the prefix. 2. If the molecule contains only 1 atom of the first element, the first prefix is left out. 3. On rare occasion two elements form only one compound. If you know this to be the case, you may leave both prefixes out.
Name the highlighted chemical group in each molecule. Given Lewis structure.
a. carbonyl b. methylene c. hydroxyl For more search common chemical groups on the Google Docs.
Sort the molecular compounds in the table into groups with the same empirical chemical formula. That is, select Group 1 next to all the compounds with the first empirical formula. (If all four compounds have the same empirical formula, then they'll all end up in Group 1.) Next, select Group 2 next to all the compounds with the second empirical formula. And so on. If all four compounds have different empirical formulas, then they'll each end up alone in their own group: Compound A will be alone in Group 1, Compound B will be alone in Group 2, and so on. About the sketches: the lines stand for chemical bonds between the atoms. Just ignore the dots. They stand for "lone pairs," and you'll learn about them later. You don't need to know anything about lone pairs to solve this problem.
a. group 4 b. group 4 c. group 3 d. group 2 Get the empirical formula by dividing by a common divisor. For example, C4H4O2 can divide by 2 making it C2H2O. Compare the empirical formulas and separate them into groups.
Name each of the acids listed in the table below.
a. hydrochloric acid b. perchloric acid c. hypochlorous acid d. carbonic acid Hydro + anion root + ic acid For more search naming binary acids on the Google Doc.
Complete the following table: Two polyatomic anions
a. hypochlorite anion b. perbromate anion For more search naming ionic compounds on Google Doc.
What word or two-word phrase best describes the shape of the nitrosyl hydride (HNO) molecule?
bent 1. Draw the Lewis structure of the molecule 2. Count the groups of valence electrons around the central atom. 3. Pick the electron-group structure. 4. Assign each lone pair group to a place in the electron-group structure. 5. Assign each bonding electron group to a place in the electron-group structure. 6. To name the shape of the molecule, consider only the bonding electron groups.
What is the hybridization of the central atom in the water H2O molecule?
sp3 Refer to Hybrid orbitals on the Google Doc
