CHEM 115 EXAM #2 Practice Fall 2013

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1 Name CHEM 115 EXAM #2 Practice Fall 2013 Circle the most correct answer for the following. (3 points each) 1. Which of the following electron configurations is not allowed by the Pauli exclusion principle? a. 1s 2 2s 2 2p 2 b. 1s 2 2s 2 2p 6 3s 2 3p 1 c. 1s 2 2s 2 2p 6 3s 2 3p 4 d. 1s 2 2s 2 2p 6 3s 2 3p 6 e. none of the above [an example of a bad e configuration that violates the Pauli Excl. Prin. would be 1s 2 2s 2 2p 6 3s 3 3p 1 ] 2. What is the ground state electron configuration of argon (Ar)? a. 1s 2 2s 2 2p 6 3s 3 b. 1s 2 2s 2 2p 6 3s 2 3p 1 c. 1s 2 2s 2 2p 6 3s 1 3p 4 d. 1s 2 2s 2 2p 6 3s 2 3p 6 e. none of the above 3. How many neutrons, protons, and electrons are present in? a. 48 neutrons, 22 protons, 22 electrons b. 22 neutrons, 48 protons, 22 electrons c. 24 neutrons, 24 protons, 22 electrons d. 26 neutrons, 26 protons, 26 electrons e. none of the above, the correct answer is: 26 neutrons, 22 protons, 22 electrons 4. Of the species listed below, which should be the smallest? a. Ca 2+ b. K + c. S 2- d. Cl - e. N 3-5. Which of the following elements has chemical properties MOST SIMILAR to those of chlorine, Cl? a. Iron, Fe b. sulfur, S c. magnesium, Mg d. fluorine, F e. potassium, K 6. Which of the following elements has chemical properties MOST SIMILAR to those of sodium, Na? a. Iron, Fe b. oxygen, O c. magnesium, Mg d. chlorine, Cl e. potassium, K 7. Which of the elements below, in the ground electronic state, has one (1) 5s electron? a. Na b. K c. Al d. Ga e. none of the above 8. Which of the following orbitals has the highest energy in a multielectron atom? a. 3s b. 3p c. 4s d. 3d e. 4p (6 points) 9. Write the chemical symbols for six ions (ex. Mg 2+ ) that are isoelectronic with Ne AND arrange them in order of increasing size. Al 3+ < Mg 2+ < Na + < Ne < F - < O 2- < N 3- (over)

2 2 Solve the following problems. Show all your work, always show your units, and express your answers in the proper number of significant figures. (point values as indicated) 10. The equation below is the Bohr equation for the energy levels in any hydrogen-like atom. a. Calculate the first three energy levels for the Li 2+ ion. The atomic number (Z) for Li is 3. Substitution of 3 for Z in the above equation yields: E n = -9 B /n 2 = x10-18 J /n 2 For n = 1 E 1 = x J For n = 2, E 2 = x J For n = 3, E 3 = x J b. Calculate the energy lost by an electron dropping from level 3 to level 1 in Li 2+. ΔE = E f E i = E 1 E 3 = x J (-2.18 x J) = x J c. Calculate the frequency and wavelength of a photon of light that corresponds to the energy lost by the electron in part b. [h = 6.626x10-34 Js and c = 3.00x10 +8 m/sec] 11. a. Write the electron configuration for the following two species (you may use the shorter version... you know, the one that will start with [Kr]):... sorry for the error, I was editing an old question that asked about Pb! Sn [Kr] 5s 2 4d 10 5p 2 or [Kr] 4d 10 5s 2 5p 2 Sn 2+ [Kr] 5s 2 4d 10 or [Kr] 4d 10 5s 2 b. Define the terms paramagnetic and diamagnetic AND use Sn and Sn 2+ as examples to more clearly demonstrate the meaning of these terms. Paramagnetic species have one, or more, unpaired electrons, Sn is paramagnetic b/c the two electrons in the 5p level are unpaired (per Hund s Rule of Maximum Multiplicity). Diamagnetic species have all electrons paired, Sn 2+ is diamagnetic b/c the electrons in the 4d, 5s, and all lower E orbitals are paired. c. Which electron(s) has(have) the highest energy in Sn? The 5 p electrons. d. Which is larger Sn or Sn 2+? Explain your choice. Sn is larger b/c it has 50 protons attracting 50 electrons. Sn 2+ is smaller b/c the 5p level is empty and the same 50 protons are now attracting only 48 electrons.

3 3 For numbers answer in concise complete sentences as appropriate and provide the specific details requested in the other questions. (6 points each) 12. What are the three types of radioactivity (discussed in this course) that helped to reveal the structure of the atom. Give the mass number, atomic number, and charge on each. The three were alpha (a), beta (b), and gamma (g) rays. Mass number (A) Atomic number (Z) Charge alpha (α) beta (β) gamma (γ) What are cathode rays? Are the cathode rays produced by all metals the same? Explain why. Cathode rays are actually streams of electrons that were ejected from the cathode (the negative electrode) in a vacuum tube. Since all elements possess identical electrons (same charge and mass), it does not matter what type of metal the electrode was made out of, the cathode rays had the same properties. 14. Explain in some detail our current understanding of the nature of isotopes. Providing specific examples will enhance your explanation. Isotopes are variants of atoms of a particular chemical element (which by definition must contain the same number of protons), that have different numbers of neutrons. The number of nucleons (protons and neutrons) in the nucleus, known as the mass number, is not the same for two isotopes of any element. (example: C-12 has 6 protons and 6 neutrons while C-13 has 6 protons and 7 neutrons). 15. Identify each of the following as either molecular or ionic compounds and provide the name of the compounds (0.5 points each blank)? Compound Ionic or Molecular Compound name CuBr 2 ionic copper (II) bromide Fe 2 (SO 4 ) 3 ionic iron (III) sulfate P 2 O 5 molecular diphosphorus pentoxide FeCl 3 6 H 2 O ionic iron (III) chloride hexahydrate CO molecular carbon monoxide Fe 2 O 3 ionic iron (III) oxide Ni 3 (PO 4 ) 2 ionic nickel (II) phosphate H 2 O molecular water (or dihydrogen monoxide) SF 6 molecular sulfur hexafluoride KNO 3 ionic potassium nitrate

4 4 (6 points) 16. Use dot symbols to show the formation of NaCl (g) and MgO (g). 17. Use the following reaction diagram (a Born-Haber cycle) to find the lattice energy for MgO(s). Sum of E loop = 0 = (+411 kj kj kj kj 355 kj + lattice E) Lattice E = -786 kj / mole NaCl

5 5 BONUS (for up to ten points) Avogadro s number is huge. To try to understand how big this number really is consider the following: An average M&M has a volume of 1.1 cm 3. Calculate the volume of a mole of M&Ms. You may want to calculate this in m 3 for easier use in the next step. 1.1 cm 3 = 1.1x10-6 m 3 One mole of these occupies a volume = (6.022x10 23 ) X(1.1x10-6 m 3 ) = 6.62x10 17 m 3 The surface area of the 48 contiguous states is x m 2. Calculate the depth (in meters) to which one mole of M&Ms would cover the 48 contiguous states. Volume / area = depth 6.62x10 17 m 3 / x m 2 = ~84,654 m Finally, to put this into a number which might be more meaningful to a U.S. citizen, convert this depth to miles (1 mile = meters). ~52.6 miles!!!!

CHEM 115 EXAM #1 Practice Fall 2007

Name CHEM 115 EXAM #1 Practice Fall 007 Circle the correct answer or fill in the blanks. (numbers 1 -,.5 points each) 1. The term used to describe a substance composed of two or more elements combined