CHEM 2070: Prelim #2 MPS
A rigid stainless steel chamber contains 150 Torr of methane, CH4, and excess oxygen, O2, at 150.0 °C. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.
∆P = 0 torr Looking at the coefficients for the gaseous species, we see that the total moles of gas remains constant throughout this reaction. No change in moles, temperature, or volume means there\'s no change pressure.
During photosynthesis, an electron is believed to travel 25 Å (1Å = 1 x 10¹⁰ m) in one picosecond (1 ps = 10⁻¹² s). The uncertainty in the time measurement is 0.5 ps. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position of the electron at any point during its 25 Å journey. (You can assume that the distance, 25 Å, and the mass of an electron, 9.109 x 10⁻³¹ kg, are known with a relatively high certainty.)
∆x ≥ 116 Å
Select all equatorial atoms in a trigonal bypyramidal structure consisting of no lone pairs and all the same atoms.
Within a trigonal bipyramidal structure, the outer atoms occupy one of two locations, equatorial or axial, around the central atom. Look for the atoms that are in a linear arrangement around the central atom. These atoms are axial, like the north and south pole of the Earth spinning around its axis. In line with this analogy, the other 3 atoms that are separated 120° from each other around the central atom occupy positions that are located on the equator (i.e., equatorial).
50.00 mL of 0.0155 M HI(aq) is mixed with 75.00 mL of 0.0106 M KOH(aq). What are the pH, pOH, [H₃O⁺] and [OH⁻] of the resulting solution? Is this solution acidic, basic or neutral?
[OH⁻] = 0.00016 M OH⁻ [H₃O⁺] = 6.2 x 10⁻¹¹ M H₃O⁺ pOH = 3.80 pH = 10.20
Which of the solutions below have the same molar concentration as Solution X? Check all that apply. A B C Solution X: 4 particles Solution A: 4 particles Solution B: 8 particles Solution C: 2 particles All solutions have the same volume.
solution A is the only one with the same molar concentration as solution X. Concentration expresses the amount of solute per unit volume. You can assume equal volumes for each, so the only difference among the solutions is the number of moles of solute particles. The number of moles is directly proprotional to the number of particles.
According to the CRC Handbook of Chemistry and Physics, the molar volume of O2(g) is 0.2168 L•mol⁻¹ at 280.0 K and 10.00 MPa. The handbook also gives the following values for O2(g): a=1.363 atm•L²•mol⁻² b=0.0319 L•mol⁻¹. You can assume that the CRC Handbook values are correct. (a) Use the ideal gas law to calculate the pressure of 1.00 mole of O2(g) at 280.0 K if the volume is 0.2168 L and determine the percent error of your calculated value using the CRC handbook value as the correct value. (b) Use the van der Waal equation to calculate the pressure of 1.00 mole of O2(g) at 280.0 K if the volume is 0.2168 L and determine the % error of your calculated value using the CRC handbook value as the correct value.
(a) 10.736 MPa; 7.36% error (b) 3.50% error
It takes 23.6 hours for a 1.75 mole sample of helium to effuse through a small hole in a container with a constant temperature of 25.0°C. (a) How long would it have taken for this helium sample to effuse in the same container at 5.00°C? (b) What temperature would be required to reduce the effusion time to 10.0 hours?
(a) 24.4 hours (b) 1.66 x 10³ K or 1.39 x 10³ °C
How many moles are present in 0.722 L of nitrogen gas at 52.2 °C and 771 Torr? Assume ideal behavior.
0.0274 moles
59.0 mL of a 1.30 M solution is diluted to a total volume of 288 mL. A 144-mL portion of that solution is diluted by adding 127 mL of water. What is the final concentration? Assume the volumes are additive.
0.142 M
Calculate the molarity of 0.250 mol of Na2S in 1.70 L of solution.
0.147 M
Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. 0.350 L of 0.420 M H2SO4 is mixed with 0.300 L of 0.290 M KOH. What concentration of sulfuric acid remains after neutralization?
0.159 M H2SO4
The amount of I₃⁻(aq) in a solution can be determined by titration with a solution containing a known concentration of S₂O₃²⁻(aq) (thiosulfate ion). The determination is based on the net ionic equation 2S₂O₃²⁻(aq) + I₃⁻(aq) → S₄O₆²⁻(aq) + 3I⁻(aq) Given that it requires 36.2 mL of 0.380 M Na₂S₂O₃(aq) to titrate a 20.0-mL sample of I₃⁻(aq), calculate the molarity of I₃⁻(aq) in the solution.
0.3439 M
1.05 g of an unknown gas at 77 °C and 1.10 atm is stored in a 2.65-L flask. What is the density of the gas? What is the molar mass of the gas?
0.396 g/L 10.345 g/mol
Naturally occurring uranium is mostly ²³⁸U (atomic mass = 238.051u), but it can be enriched to contain more ²³⁵U (atomic mass = 235.054u) by fluorinating the uranium to make UF₆, which is a gas, and then taking advantage of the different effusion rates for compounds of the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6.
0.995734
A 11.0-L helium tank is pressurized to 27.0 atm. When connected to this tank, a balloon will inflate because the pressure inside the tank is greater than the atmospheric pressure pushing on the outside of the balloon. Assuming the balloon could expand indefinitely and never burst, the pressure would eventually equalize causing the balloon to stop inflating. What would the volume of the balloon be when this happens? Assume atmospheric pressure is 1.00 atm. Also assume ideal behavior and constant temperature.
286 L
How many photons are produced in a laser pulse of 0.130 J at 505 nm?
3.3 x 10¹⁷ photons
A mixture of helium and neon gases at 329 K contains three times the number of helium atoms as neon atoms. The concentration (n/V) of the mixture is found to be 0.150 mol/L. Assuming ideal behavior, what is the partial pressure of helium in kPa?
307.74 kPa
How many outer atoms and lone pairs are present in a molecule with a see-saw shape?
4 outer atoms 1 lone pair Steric # of 5
The energy levels of one-electron ions are given by the equation En = (-2.18 x 10⁻¹⁸ J)(Z²/n²) where Z is atomic number and n is the energy level. The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf = 4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ni = 8. What region of the electromagnetic spectrum does this transition fall in?
4.86 x 10⁻⁷ m visible region (found by consulting a chart)
Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). If the position of the car is known to within 5.0 feet at the time of the measurement, what is the uncertainty in the velocity of the car? If the speed limit is 75 mph, could the driver of the car reasonably evade a speeding ticket by invoking the Heisenberg uncertainty principle?
7.6 x 10⁻³⁸ mph ; no
Which atoms have a complete octet? B in BH3 B in BH4 iodine with a +1 charge
B in BH4
A piece of calcium, Ca, is irradiated with light of wavelength 410 nm, ejecting electrons with speed 3.29x10⁵ m/s. This measurement reveals a fundamental property of Ca metal with units of energy. Name this property, define it simply, and calculate its value in J.
Binding energy (Eb) or work function minimum energy required to remove electrons from different subshells within an atom 4.36 x 10⁻¹⁹ J
Sapling #8, Question 5 [copy images below after printing]
CO2 should have 4 2(6)=16 valence electrons. The given structure shows 20 electrons (2 shared pairs and 8 unshared pairs). Thus, the structure is invalid even though each atom has an octet. Each F atom in OF2 is shown with 10 electrons (2 shared pairs and 3 unshared pairs), which is an octet-rule violation. Thus, the structure is invalid even though it has the correct total number of electrons. F2 should have 2(7)=14 electrons. The given structure shows 10 electrons (3 shared pairs and 2 unshared pairs). Thus, the structure is invalid even though each atom has an octet. The N atom in NO3- is shown with 6 electrons (3 shared pairs), which is an octet-rule violation. Thus, the structure is invalid even though it has the correct total number of electrons. Note that octet-rule violations don\'t always indicate an invalid structure. For example, elements in the third period and beyond can have more than 8 electrons. Hydrogen, beryllium, and boron always have less than 8 (2, 4, and 6 electrons respectively). So really, C, N, O, and F are the only atoms that exclusively follow the octet rule, and that\'s only true for molecules with an even number of electrons. For this reason, many instructors prefer to say the \"octet suggestion\" rather than the \"octet rule\".
Draw the Lewis dot structure of HCN (include lone pairs)
H-C≡N:
Consider a single photon with a wavelength of λ, a frequency of ν, and an energy of E. What is the wavelength, frequency, and energy of a pulse of light containing 100 of these photons?
λ, ν, and 100E Electromagnetic energy is quantized, in units of photons. Each photon contributes to the overall energy. The wavelength and frequency, however, are set properties of the pulse of light. Each photon making up the pulse has the same wavelength and frequency. Since energy is proportional to the number of photons, we often see energy of light expressed in joules per mole of photons (J/mol). However, you will never see the wavelength or frequency expressed \"per mole\".
What is the molecular shape of XeF₂?
Linear A xenon atom has 8 valence electrons. A fluorine atom has 7 valence electrons. Thus, an XeF2 molecule has 8 2(7)=22 valence electrons which is equal to 11 electron pairs (in this case 2 bonds and a 9 lone pairs). With two bonded groups and three lone pairs on the central atom, there are five regions of electron density forming a trigonal bipyramidal shape. Since the lone pairs will occupy all of the equatorial positions, the bonds will be in the axial positions at 180 degrees. XeF2 is linear.
Two evacuated bulbs of equal volume are connected by a tube of negligible volume. One of the bulbs is placed in a constant-temperature bath at 225 K and the other bulb is placed in a constant-temperature bath at 350 K. Exactly 1 mole of an ideal gas is injected into the system. Calculate the final number of moles of gas in each bulb.
flask 1 (225 K) holds 0.609 moles gas flask 2 (350 K) holds 0.391 moles gas
The first ionization energy of a sodium atom is 0.830 aJ (attojoules). What is the frequency and wavelength, in nanometers, of photons capable of just ionizing sodium atoms? Assuming an ionization efficiency of 25.0%, how many such photons are needed to ionize 1.00 × 10¹⁶ atoms?
1.25 x 10¹⁵ s⁻¹ 239 nm 4.00 x 10¹⁶ photons
Assuming that the smallest measurable wavelength in an experiment is 0.610 fm (femtometers), what is the maximum mass of an object traveling at 717 m·s-1 for which the de Broglie wavelength is observable?
1.515 x 10⁻²¹ kg
Scientists at SETI (Search For Extraterrestrial Intelligence) receive an extraterrestrial communication. It is from a chemistry student on the planet Oz. The student asks for help with his chemistry homework and gives this problem: "A sealed flask with 0.32 noggins of molecules of an ideal gas is initially at the pressure 12.4 truq and at the temperature of 1452 REM. The sealed flask is heated to a temperature of 1966 REM. No chemical reaction occurs. What is the final pressure?" The student also gives the following information: One noggin is the number of molecules to fill a volume of one frog at one truq and 100 REM. One truq is the atmospheric pressure on Oz. The lowest possible temperature is -200.0 REM. His body temperature is 100.0 REM. You can also assume that ideal gases behave the same on Oz as on Earth. Is it possible to give a logical numerical answer to the chemistry student from Oz? If so what is your answer? If not, why not?
16.3 truq
17.5 L N2 at 25 °C and 125 kPa and 41.9 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 4.00 L. What is the total pressure at 45 °C?
1980.83 kPa
The amount of nitrogen in an organic substance can be determined by an analytical method called the Kjeldahl method, in which all the nitrogen in the organic substance is converted to ammonia. The ammonia, which is a weak base, can be neutralized with hydrochloric acid, as described by the equation NH₃(aq) + HCl(aq) → NH₄Cl(aq) If 27.0 mL of 0.150 M HCl(aq) is needed to neutralize all the NH3(g) from a 2.25-g sample of organic material, calculate the mass percentage of nitrogen in the sample.
2.52%
Container A holds 747 mL of ideal gas at 2.60 atm. Container B holds 144 mL of ideal gas at 4.20 atm. If the gases are allowed to mix together, what is the resulting pressure?
2.86 atm Use Boyle's law to find partial pressures of each gas, then add to find total pressure.
Indium has a threshold frequency of 9.96x10⁴ s⁻¹. Light from a laser shines on an indium surface and electrons are emitted with a speed of 5.23x10⁵ m/s. What is the wavelength of the light being emitted from the laser?
253 nm
The fastest recorded airplane was an SR-71 Blackbird flying at a speed of 2,193 miles per hour. At what temperature (in degrees Celsius) does the urms speed of hydrogen gas, H2(g), reach the same speed?
77.67 K = -195.5°C
Number of valence electrons Xe
8
What is the pressure of the gas in this mercury manometer if h = 83 mm and atmospheric pressure is 763 mmHg? manometer open @ one end, attached to gas @ the other
846 mmHg P(gas) = P(height) + P(atm) The pressure unit known as \"millimeters of mercury\", or mmHg, originated with instruments like barometers and manometers where the the height of the mercury was a direct measurment of the pressure. In other words, the value of h (in mm) is equal to the difference between the gas pressure (in mmHg) and atmospheric pressure (in mmHg). The pressure of the gas pushes the mercury out of the tube while atmospheric pressure pushes the mercury into the tube. Since the mercury level on the left side of the tube is lower, we can tell that the gas is pushing harder than the atmosphere. So the pressure of the gas is 83 mmHg higher than atmospheric pressure.
Which of the following electron configurations of neutral atoms represent excited states? [Kr]5s¹4d⁵ 1s²2s²2p⁶3s²3p⁶3d² [Ar]4s²3d³ 2s² [Xe]6s²4f¹
A neutral atom has the same number of protons and electrons. To identify the neutral atoms, determine the number of electrons, which is also the number of protons and corresponds to the atomic number on the periodic table. For example, [He]2s23s1 contains 2 2 1=5 electrons, and therefore 5 protons, which corresponds to Boron. When an electron is promoted to a higher energy orbital, the atom is said to be in an excited state, instead of the ground state. 2s2 is an excited state of helium, He. The ground state is 1s2. 1s22s22p63s23p63d2 is an excited state of calcium, Ca. The ground state is 1s22s22p63s23p64s2. [Ar]4s23d3 is the ground state configuration of vanadium, V. [Kr]5s14d5 is the ground state of molybdenum, Mo. Note that this deviates from the regular filling order, from which we would predict the configuration [Kr]5s24d4. [Xe]6s24f1 is an excited state of lanthanum, La. The ground state is [Xe]6s25d1.
The letters A through F represent atomic orbitals. Provided the information below, fill in the blanks. A, Total Nodes: 5 Radial Nodes: 3 Angular Nodes: Orbitals in a Degenerate Set: Name of Orbital: B, Total Nodes: Radial Nodes: 3 Angular Nodes: Orbitals in a Degenerate Set: 9 Name of Orbital: C, Total Nodes: 5 Radial Nodes: Angular Nodes: Orbitals in a Degenerate Set: 1 Name of Orbital: D, Total Nodes: Radial Nodes: 2 Angular Nodes: 1 Orbitals in a Degenerate Set: Name of Orbital: E, Total Nodes: 3 Radial Nodes: Angular Nodes: 2 Orbitals in a Degenerate Set: Name of Orbital: F, Total Nodes: Radial Nodes: Angular Nodes: Orbitals in a Degenerate Set: Name of Orbital: 4s
A, Total Nodes: 5 Radial Nodes: 3 Angular Nodes: 2 Orbitals in a Degenerate Set: 5 Name of Orbital: 6d B, Total Nodes: 7 Radial Nodes: 3 Angular Nodes: 4 Orbitals in a Degenerate Set: 9 Name of Orbital: 8g C, Total Nodes: 5 Radial Nodes: 5 Angular Nodes: 0 Orbitals in a Degenerate Set: 1 Name of Orbital: 6s D, Total Nodes: 3 Radial Nodes: 2 Angular Nodes: 1 Orbitals in a Degenerate Set: 3 Name of Orbital: 4p E, Total Nodes: 3 Radial Nodes: 1 Angular Nodes: 2 Orbitals in a Degenerate Set: 5 Name of Orbital: 4d F, Total Nodes: 3 Radial Nodes: 3 Angular Nodes: 0 Orbitals in a Degenerate Set: 1 Name of Orbital: 4s
Indicate the electron pair geometry and the molecular geometry for each of the six compounds listed below by completing the following table. CO₂ SO₃ O₃ CH₄ PF₃ OF₂
Carbon dioxide has two electron groups around the central atom, so the electron pair geometry is linear. There are no lone pairs, so the molecular geometry is also linear. Sulfur trioxide has three electron groups around the central atom, so the electron pair geometry is trigonal planar. There are no lone pairs, so the molecular geometry is also trigonal planar. Ozone has three electron groups around the central atom, so the electron pair geometry is trigonal planar. There is one lone pair, so the molecular geometry is bent. Methane has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There are no lone pairs, so the molecular geometry is also tetrahedral. Phosphorus triflouride has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There is one lone pair, so the molecular geometry is trigonal pyrimidal. Oxygen difluoride has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There are two lone pairs, so the molecular geometry is bent. The electron pair geometry is determined from the arrangement of all the electron groups around the central atom. An electron group consists of a lone pair of electrons, a single bonded atom, a double bonded atom, or a triple bonded atom. There are a total of four electron groups around the central atom of which one is a lone pair of electrons. To minimize repulsion, the groups are arranged as far away from each other as possible. The four electron groups in PF3 are therefore arranged 109.5° from each other. If there are no lone pairs of electrons, the molecular geometry is the same as the electron pair geometry. With one or more lone pairs of electrons, the molecular geometry is determined by the positions of the bonded atoms only.
Complete these structures by adding electrons, in the form of dots, as needed. F-H H-O-H O=N-H
Each atom, except for hydrogen, follows the octet rule which states that the atoms will combine in such a way that the valence shell of each atom has eight electrons. A hydrogen atom\'s valence shell is filled by only two electrons. Each line (bond) between two atoms represents two electrons, which both count towards the valence shell of each atom involved in the bond. To complete the octet of each atom, electrons represented by dots, are added to each atom so that there are eight total electrons around each atom. In the first structure, the F atom has two electrons from the single bond. Thus, it needs six more electrons to complete its octet. The H atom does not require additional lone pair electrons as its valence shell is filled by the two electrons in the single bond. In the second structure, the O atom has four electrons, two from each single bond. Thus, it needs four more electrons for an octet. Again, the H atoms do not require additional lone pair electrons as their valence shells are filled by the two electrons in each of the single bonds. In the third structure, the O atom has four electrons from the double bond. Thus, it needs four more electrons to complete its octet. The N atom has six electrons, four from the double bond and two from the single bond. Thus, it needs two more electrons to complete its octet. Once again the H atom does not require additional lone pair electrons as its valence shell is filled by the two electrons in the single bond.
Rank the main group elements in the third period by ionization energy.
Noble gases are very stable. Of the elements in a given period of the periodic table, the noble gas will require the most energy to ionize because of this stability. Alkali metals tend to form 1 ions. Of the elements in a given period of the periodic table, the alkali metal will require the least energy to ionize because of the stability of the resulting ion. Start by putting the noble gas at the top, the alkali metal at the bottom, and the remaining elements filled in to match the periodic table. Then, the atom with the ns² electron configuration (in this case Ca) is particularly stable and should move up one spot in the ranking. Similarly, the atom with the ns²np³ electron configuration (in this case As) is particularly stable and should move up one spot in the ranking.
Natural gas is a mixture of hydrocarbons, primarily methane, CH4, and ethane, C2H6. A typical mixture might have χmethane = 0.895 and χethane = 0.105. Assume that you have this mixture in a 10.50 g sample of a natural gas in a volume of 5.00 L at a temperature of 293 K. What is the partial pressure of each component in the sample (in atm)?
Partial Pressure of CH₄: 2.58 atm Partial Pressure of C₂H₆: 0.303 atm
Which of these molecules are polar? Check all that apply. SO₂ CO₂ CH₂Cl₂ PCl₃
SO2 and CO2 each have two polar bonds of equal magnitude. The only way for such a molecule to be nonpolar is if those two bonds are directly across from each other. This is the case for the linear CO2 molecule, but not for the bent SO2 molecule, therefore, SO2 is polar. CH2Cl2 has four bonds in a tetrahedral shape, but the C-H bonds have a different polarity than the C-Cl bonds. This molecule would be nonpolar if the C-H bonds were directly across from each other and if the C-Cl bonds were directly across from one another, so the dipole moments canceled. In a tetrahedral molecule, none of the bonds are directly across from each other, so this molecule is polar. PCl3 has three polar bonds of equal magnitude. Since those bonds are arranged asymmetrically in a trigonal pyramidal shape, this molecule is polar.
73.0 mL of a 1.70 M solution is diluted to a volume of 208 mL. A 104-mL portion of that solution is diluted using 109 mL of water. What is the final concentration?
STRATEGY 1. Determine the number of moles of solute at each step in the process. 2. Determine the final volume. 3. Calculate the concentration using the number of moles and volume of the solution. 0.2915
Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760 torr? Assume ideal behavior.
Somehow you need to deal with the fact that your percentages don't add up to 100%. One way is like this: Take a hypothetical sample consisting of 78.1 moles of N2, 20.9 moles of O2, and 0.934 moles of Ar: 78.1 mol + 20.9 mol + 0.934 mol = 99.934 mol total V = nRT / P = (99.934 mol) x (62.36367 L Torr/K mol) x (22 + 273 K) / (760 torr) = 2419 L total (78.1 mol N2) x (28.01344 g N2/mol) = 2187.85 g N2 (20.9 mol O2) x (31.99886 g O2/mol) = 668.78 g O2 (0.934 mol Ar) x (39.9481 g Ar/mol) = 37.31 g Ar 2187.85 g + 668.78 g + 37.31 g = 2893.94 g total (2893.94 g) / (2419 L ) = *1.20 g/L*
Write Lewis formulas for the reactant and product species in the following chemical equation. Include nonbonding electrons. 2OH(g) → H₂O₂(g)
The OH molecule has 7 valence electrons. Since 7 is an odd number, there will have to be a radical electron in this molecule. Hydrogen can have a maximum of 2 electrons (one bond) so all the nonbonding electrons must go on the O atom. H2O2 is produced by combining two OH molecules at the radical. Thus, the structure should have the general arrangement H-O-O-H. Each oxygen atom gets two lone pairs to complete the octet.
Select the true statements regarding these resonance structures of formate. The actual structure of formate switches back and forth between the two resonance forms. Each oxygen atom has a double bond 50% of the time. Each carbon-oxygen bond is somewhere between a single and double bond. The actual structure of formate is an average of the two resonance forms.
The actual structure of formate is an average of the two resonance forms. Thus, each carbon-oxygen bond is somewhere between a single and a double bond.
How many structures are possible for a trigonal bipyramidal molecule with a formula of AX₃Y₂?
The atoms in the axial positions can either be X and X, X and Y, or Y and Y, allowing for three unique isomers.
Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n = 5 to n = 1.
The change in energy for the electron is calculated by ∆E = -2.179 x 10⁻¹⁸ J (1/n₁² - 1/n₂²) The energy of the photon is the absolute value of this change in energy, which is 2.092 x 10⁻¹⁸.
Determine the element of lowest atomic number that contains a complete d subshell in the ground state.
The d-block contains some anomalous electron configurations, particularly in groups 6 and 11. The term anomalous is simply the adjective form of anomaly, which means \"a deviation from the common rule\". Normally, the 4s subshell would be full before any electrons are placed in the 3d subshell. But a full or half-full d subshell is particularly stable, and so one electron can be \"borrowed\" from 4s to achieve that stability. The first element to have a full d subshell is Cu, 4s¹3d¹⁰.
From the following Lewis structures and what you know about VSEPR theory, rank the molecules in order of increasing X-Y-X bond angle. YX₄ YX₃E YX₂E₂
The electron groups (lone pairs and bonded atoms) around a central atom will distribute themselves to be as far apart as possible. For an sp3 hybridized molecule with four bonded atoms the X-Y-X bond angles will be that of an ideal tetrahedron, 109.5°. Methane, CH4, is an example of an ideal tetrahedral molecule. Lone pairs of electrons occupy slightly more space than bonded atoms, so an sp3 hybridized molecule with three bonded atoms and one lone pair will have X-Y-X bond angles that are slightly less than 109.5°. For example, ammonia, NH3, with three bonded hydrogen atoms and one lone pair of electrons on the nitrogen atom, has H-N-H bond angles of ~107°. An sp3 hybridized molecule with two bonded atoms and two lone pairs of electrons will have X-Y-X bond angles that are less than both YX4 and YX3 molecules. Water, H2O, has two hydrogen atoms and two lone pairs of electrons on the oxygen atom and has an H-O-H bond angle of ~104°.
Predict the molecular shape of methane, the carbonate ion, carbon dioxide, and the sulfite ion.
The most stable geometry is an arrangement that keeps the atoms or electrons bonded to the central atom as far apart as possible. Methane, CH4, has a central carbon atom surrounded by four hydrogen atoms. It has tetrahedral geometry. The bond angle is 109.5°. The carbonate ion, CO₃²⁻, has a central carbon atom bonded to three oxygen atoms. CO₃²⁻ is trigonal planar, with a bond angle of 120°. Carbon dioxide, CO₂, has a central carbon atom bonded to two oxygen atoms. Carbon dioxide is linear, with a bond angle of 180°. The sulfite ion, SO₃²⁻, has a central sulfur atom surrounded by three oxygen atoms and a lone pair of electrons. Owing to the lone pair of electrons, the sulfite ion has a trigonal pyramidal molecular geometry - not trigonal planar - with bond angles less than 109.5°.
Draw a Lewis structure for SO2 in which all atoms obey the octet rule. Show formal charges. Do not consider ringed structures. Draw a Lewis structure for SO2 in which all atoms have a formal charge of zero. Explicitly showing the zero charges is optional. Do not consider ringed structures.
The octet rule states that atoms in a molecule form covalent bonds with the other atoms in the molecule in a way that each atom has eight electrons in its valence shell, resulting in the same electron configuration as a noble gas. An SO2 molecule has three atoms, each with six valence electrons, for a total of 18 valence electrons. With 18 electrons, the only Lewis structure that obeys the octet rule is ::O:(-)-S:(+)=O::(0) In this structure, the single-bonded oxygen atom donates seven electrons, which is one more than its expected valence of six; thus, its formal charge is -1. The double-bonded oxygen atom donates six electrons in this structure, which is equal to its expected valence of six; thus, its formal charge is zero. The sulfur atom donates five electrons in this structure, which is one less than its expected valence of six; thus, its formal charge is 1. Atoms located below the second period on the periodic table, such as sulfur, are capable of forming compounds that deviate from the octet rule by having more than eight electrons in the valence shell. With 18 electrons, the only Lewis structure in which each atom has a formal charge of zero is :O:=S:=:O: In this structure, each atom donates six electrons, and so all formal charges are zero. The central S atom is surrounded by a total 10 electrons, which is more than an octet. So which resonance structure accurately represents the actual structure of SO2? In reality, both of these resonance forms contribute to the overall structure of SO2. To determine which structure is the largest contributor to the actual structure of the molecule, the bond lengths of the actual structure, determined through experimentation, would need to be measured. If the measured bond length is closer to the length of a double bond, then the second resonance structure is the largest contributor. If the measured bond length is closer to the length of a one-and-a-half bond, then the first resonance structure is the largest contributor. In the absence of these measurements, the resonance structure that represents the actual structure of SO2 is determined based on either minimized formal charge or completed octets. Some instructors teach that the actual structure is represented by the resonance structure in which the formal charges are minimized on all atoms. Others teach that the actual structure is represented by the resonance structure in which each atom has a complete octet in its valence shell. Check with your instructor to see which is preferred for your course.
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 3.32 × 10¹⁴ s⁻¹. With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of λ = 275 nm?
The threshold frequency, ν0, of a metal is the minimum frequency of the light needed to eject electrons. Since energy is equal to hν, it can be found to be 2.20 x 10⁻¹⁹ J. The maximum energy of the emitted electrons is the difference between the applied energy (from the light) and the minimum photon energy, E0. Also note that 275 nm is equal to 275×10⁻⁹ m and that E=hc/λ. This difference is 5.02 x 10⁻¹⁹ J.
Data for the photoelectric effect of silver are given below: Frequency of incident radiation (10¹⁵ s⁻¹) and KE of ejected electrons (10⁻¹⁹ J) f = 2, KE = 5.9 f = 2.5, KE = 9.21 f = 3, KE = 12.52 f = 3.5, KE = 15.84 f = 4, KE = 19.15 Using these data, find the experimentally determined value of Planck\'s constant, h, and the threshold frequency, ν0, for silver.
The value for Planck's constant is found by dividing ∆KE by ∆frequency, which is constant throughout the table. This value is 6.626 x 10⁻³⁴ J ∙ s. To find the threshold frequency, you must plot the graph to find its equation. The data are plotted below. The equation, Ek = hν - hν0, tells us that a plot of Ek versus ν is a straight line whose slope is the Planck constant. The intercept of the straight line with the horizontal axis (where Ek = 0) gives the minimum energy required to eject an electron, hν0. From the graph we find that hν0 = 7.354 x 10⁻¹⁹ J, so ν0 is 1.110 x 10⁵ s⁻¹.
What information is needed to determine the energy of an electron in a many-electron atom? Check all that apply: n, l, m(l), m(s) What information is most important in determining the size of an orbital?: n, l, m(l), m(s)
The value of n indicates the relative energy level of a shell and the size of an orbital. For example, 2s is both higher in energy and larger than 1s. The value of ℓ indicates the shape of an orbital (s, p, d, or f) and is used to differentiate the energy levels of subshells in the same shell (e.g., 2s vs. 2p). Within a given subshell, the orientation of the orbitals is indicated by mℓ. For example, the three p orbitals in the 2p subshell are distinguished by x, y, and z orientations, as indicated by the value of mℓ. The value of ms indicates the spin on the electron but says nothing about the shape, size, orientation, or energy of the orbital.
A line in the Paschen series of hydrogen has a wavelength of 1090 nm. From what state did the electron originate? In what region of the electromagnetic spectrum is this line observed?
The wavelength, λ, is related to the energy states of the transition via the Rydberg equation. Because we are told the emission line is in the Paschen series, we know that nf = 3. Be sure to convert the wavelength to meters: 1090 nm = 1.09 × 10⁻⁶ m. Consulting a chart will show that this wavelength falls in the infrared region.
A metal foil has a threshold frequency of 5.45× 10¹⁴ Hz. Which of the colors of visible light have enough energy to eject electrons from this metal?
green, blue, indigo, violet
For each electron in a ground-state Be atom, select the set of quantum numbers that represents it.
n = 1, ℓ= 0, mℓ= 0, ms= -1/2 n = 1, ℓ= 0, mℓ= 0, ms= 1/2 n = 2, ℓ= 0, mℓ= 0, ms= 1/2 n = 2, ℓ= 0, mℓ= 0, ms= -1/2
