2012 SPRING Semester Midterm Examination

2012 SPRING Semester Midterm Examination

For General Chemistry I

Date: March 28 (Wed), Time Limit: 7:00 ~ 9:00 p.m.

Write down your information neatly in the space provided below; print your Student ID in the upper right corner of every page.

Professor Name / Class / Student I.D. / Name
Problem / points / Problem / points / TOTAL pts
1 / /10 / 6 / /20 / /150
2 / /8 / 7 / /15
3 / /12 / 8 / /25
4 / /8 / 9 / /20
5 / /12 / 10 / /20

** This paper consists of 4 sheets with 10 problems. Please check all page numbers before taking the exam.

Write down your work and answers in the Answer sheet.

Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.

NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER.

(채점답안지 분배 및 이의신청 일정)

1. Period, and Procedure

1) Return and Claim Period: April 2 (Mon), Quiz Session; 7: 00 ~ 7:30

2) Procedure: During the quiz hour, you can take your mid-term paper scored. If you have any claims on it, you can submit a claim paper with your opinion. After writing your opinions on any paper you can get easily, attach it with a stapler to your mid-term paper scored (Please, write your name, professor, and class.). Submit them to your TA. The papers withtheclaimswill be re-examined by TA.

The claim ispermitted only on the period. Keep that in mind!

(A solution file with answers for the examination will be uploaded on 3/30 at the web.)

2. Final Confirmation

1) Period: April 9 (Mon)-10 (Tue)

2) Procedure: During thisperiod, you can check final score of the examination on the website again.

** For further information, please visit a General Chemistry website at www.gencheminkaist.pe.kr.

<The Answers>

Problem / points / Problem / points / TOTAL pts
1 / 5+5 /10 / 6 / 14+6 /20 / /150
2 / 4+4 /8 / 7 / 5+5+5 /15
3 / 3+3+3+3 /12 / 8 / 10+15 /25
4 / 4+4 /8 / 9 / 10+10 /20
5 / 8+4 /12 / 10 / 10+5+5 /20

1. (a) Calculate the maximum wavelength of electromagnetic radiation needed to eject electrons from the surface of tungsten, which has a work function of 7.29 x 10-19 J.

(b) If the maximum speed of electrons emitted from the tungsten surface is 2.00 x 106 m s-1, calculate the wavelength of the incident radiation.

Planck’s constant h = 6.626 x 10-34 J·s; speed of light c = 2.998 x 108 m·s-1

1. (a) This involves the photoelectric equation, Emax = ½meve2 = hn - F, where F is the work function (7.29 x 10-19 J) representing the minimum energy (and hence maximum wavelength) of photons required for photoelectric effect.

Emin = 7.29 x 10-19 J = hc/lmax

lmax = (6.626 x 10-34 J·s) (2.988 x 108 m·s-1) / (7.29 x 10-19 J) = 2.72 x 10-7 m or 272 nm

(b) Maximum energy of electrons ejected by impact of photons of wavelength l is given by,

Emax = ½meve2 = ½ (9.109 x 10-31 kg) (2.00 x 106 m·s-1)2 = 18.2 x 10-19 J = hc/l - F

Hence, hc/l = 25.5 x 10-19 J

l = (6.626 x 10-34 J·s) (2.988 x 108 m·s-1) / 25.5 x 10-19 J = 0.779 x 10-7 m or 77.9 nm

2. (a) Which of the quantum numbers govern the energy of an orbital?

(b) Which of the quantum numbers govern the shape and spatial orientation of an orbital?

2. (a) principal quantum number and orbital angular momentum quantum number

(b) orbital angular momentum quantum number and magnetic quantum number

3. Which of the following statements are true for many-electron atoms? If false, explain why.

(a) The effective nuclear charge Zeff is independent of the number of electrons present in an atom.

(b) Electrons in an s-orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge because an electron in an s-orbital can penetrate to the nucleus of the atom.

(d) Zeff for an electron in a p-orbital is lower than for an electron in as s-orbital in the same shell.

3. (a) false. Zeff is considerably affected by the total number of electrons present in the atom because the electrons in the lower energy orbitals will “shield” the electrons in the higher energy orbitals from the nucleus. This effect arises because the e-e repulsions tend to offset the attraction of the electron to the nucleus.

(b) true.

(d) true.

4. (a) Which has the larger second ionization energy, B or C, and why?

(b) Which is the largest and which is the smallest among K+, Cl-, and Ca2+, and why?

4. (a) The first ionization of each removes a 2p electron, but the second removes a 2s electron from B, but only another 2p electron from C. The change in subshell with B means it has the higher second ionization energy.

(b) All three have the same number of electrons, but Cl– has the fewest number of protons in its nucleus to hold these electrons in place (making it the largest) while Ca2+ has the largest number of protons (making it the smallest). K+ is in between.

5. (a) We learned that the Coulomb potential energy of the interaction of two individual ions is , where and are the charges of the two ions and is the distance between two ions. Consider the following one-dimensional nanorod made of only four ions. The distance between two adjacent ions is d. The charge of each ion is indicated (2e or -2e). Calculate the total Coulomb potential energy of this system.

(b) In which of the nanorods Mg2+Se2-Mg2+Se2- and Ca2+Se2-Ca2+Se2- are the interactions between the ions stronger? Why?

5. (a)

(b) Mg2+Se2-Mg2+Se2-, the same charge, smaller distance, smaller positive ions

6. (a) Write the Lewis structures of SO2 and N2O. Indicate the most stable structure for each molecule and explain the reason.

(b) Write the possible Lewis structures of HNO and HON molecules. Which one would be energetically favored? Why?

6. (a) SO2

10 electrons on S stable resonance structure

N2O

10 electrons on middle N stable most stable least stable

 Negative charge

on more negative atom

(b)

HNO is energetically favored because of the neutral formal charge.

7. Place the following molecules or ions in order of decreasing bond length, and explain the reason.

(a) the CO bond in CO, CO2, CO32-

(b) the SO bond in SO2, SO3, SO32-

7. (a) CO32- > CO2 > CO

CO32- : resonance between two single bonds and one double bond

CO2 : two double bonds; CO : triple bond

(b) SO32- > SO2 ~ SO3

SO32- : resonance between two single bonds and one double bond

SO2 : double bonds; SO3 : double bonds

CH3NH2 : single bond; CH2NH : double bond; HCN : triple bond

8. (a) Use the VSEPR theory to sketch and name the molecular geometries of the molecules, SCl2, SCl4, and SCl6, and state whether each is polar or nonpolar.

(b) Bromine can form compounds or ions with any number of fluorine atoms from one to five (BrF, BrF2-, BrF3, BrF4-, BrF5). Describe their geometries based on the VSEPR theory.

8. (a)

(b) BrF(linear), BrF2-(linear), BrF3 (T-shaped), BrF4-(square planar), BrF5 (square pyramidal)

9. Use valence (VB) theory to predict the hybridization in formaldehyde and allene. Sketch the hybrid atomic orbitals depicting their overlap and s and p-bonding.

(a) formaldehyde H2C=O

(b) allene H2C=C=CH2

9. (a) C atom is sp2-hybridized, as is the O atom. These trigonal planar sp2 orbitals s-bonded with 1s orbital of H atom and 2p orbital of O atom. Unhybridized 2p orbital of C atom and O atom form a p-bond which is orthogonal to the s-bond.

(b) The central C atom forms one p-bond on each side. These p-bonds are orthogonal to s-bond. The central C atom must be sp-hybridized. The terminal C atoms are sp2 hybridized. These sp2 orbitals are s-bonded with 1s orbitals of H atoms and the sp hybrid orbital of the central C atom.

10. (a) Draw the molecular orbital energy-level diagram for N2 and label the energy levels according to the type of orbitals from which they are made, whether they are s or p-orbitals, and whether they are bonding or antibonding.

(b) The orbital structure of the heteronuclear diatomic ion NO+ is similar to that of N2. How will the fact that the electronegativity of N differs from that of O affect the molecular orbital energy-level diagram of NO+ compared with that of N2? Use this information to draw the energy-level diagram for NO+.

10.