101
HOUR CHEM EXAM IV
November 18, 1997
1.
A.
Write the complete electronic configuration for an atom having 62 protons. (For full credit you must show the configuration in the same form as the periodic table - just like we do in class.) Write the vse configuration for the +1 ion of the element having at.no. 81. Write the vse configuration for the +3 ion of the element with 28 protons. Predict the structure of the SnF4 2- anion (Sn is central atom) and: i. iii. name the structure show bonding angles ii. iv. diagram the structure is the anion polar?
B. C.
2.
A.
B.
Predict the structure of the NO2 1- anion (N is central atom) and: i. iii. name the structure show bonding angles ii. iv. diagram the structure is the anion polar?
3.
A.
In the 6f subshell, what is the: i. maximum number of electrons ii. minimum value for quantum number n In the third shell, what is the maximum number of: i. subshells ii. orbitals In the R = 5 subshell, what is the maximum value of: i. ms ii. mR
B.
C.
4.
When an electron in hydrogen atom moves from some outer shell to the third shell it emits a photon with wavelength of 1283.44 nm. What is the value of quantum number n for the outer shell? h = 6.626 x 10 -34 JCsec -18 2 2 )E = ( ! 2.1874 x 10 J ) [ ( 1/n final ) - ( 1/n initial ) ] c = 3 x 10 8 cm/sec Apply the concept of formal charge to the proposed linkage structures shown below, and thereby determine which structure would be the more acceptable. str. A H - Cl - O str. B H - O - Cl
5.
6.
Apply the valence bond approach to the nitrite ion, NO2 1- , and answer the following questions: A. What is the type of orbital hybridization and geometry for the nitrogen? B. What is the type of orbital hybridization and geometry for the oxygen (both oxygen atoms have same hybridization) C. Diagram the ion showing sigma and pi bonding orbitals for all atoms .
7. F 1-
Balance the following equation: + Au + HXeO6 3= AuF6 1+ HXeO4 1(base)
8.
Use the molecular orbital approach to compare the bonding and properties for CO and CN. A. B. C. D. E. Determine the bond order for each and show its value. Which one will be paramagnetic (if any)? Which one will have the longer bond length? Which one will be diamagnetic (if any)? Which one will have the higher bond dissociation energy?
9.
For a mole of photons to possess a total energy of 1000 kJ, what must be their corresponding wavelength (in nm)?
10.
What are the numerical values of the four quantum numbers (n, R , mR , and ms ) for: i. ii. iii. the 25 th electron in selenium (at.no. 34) the 32 nd electron in palladium (at.no. 46) the 77 th electron in plutonium (at.no. 94)
Kindly retain this sheet of exam questions for possible future reference.
Answers 1. A. Samarium (Sm) is the atom with 62 protons. It is an inner-transition metal atom with vse in the 4 f subshell.. Its complete electronic configuration is... 1s2 2s2 3s2 4s2 5s2 6s2 B. 2p6 3p6 4p6 5p6 4f5
3 d 10 4 d 10 5d 1
Thallium (Tl) is the element having at.no. 81. It is a main group metal with 6 s and 6 p type vse. The 6 p-type electron(s) are lost first. 6s 6p0
C.
Nickel is the element having 28 protons. It is a transition metal with 4 s and 3 d vse. The 4 s electrons are lost first. 4s0 3d7
2.
A.
Construct Lewis dot diagram (a). Tin has four vse ( 5 s 2 5 p 2 ). Total # vse = 4 + 4(7) + 2 = 34 # sigma bonds = 4 VSEPR formula is A X 4 E 1 Basis structure is trigonal bipyramid (b).
2F F F F (a) (b) (c) F F Sn 180 Sn 120
F
F
F Sn
290
F F
2-
Sn
F
(d)
i.
reported name of structure w/r to atoms is refer to diagram (d) above. refer to (d) above above YES iii.
TEETER TOTTER
ii. iv.
refer to (c) above
B.
Construct Lewis dot diagram first. Total # vse = 5 + 2(6) + 1 = 18 # sigma bonds = 2 VSEPR formula is A X 2 E 1 Basis structure is plane triangle Two resonance forms are possible and are shown below.
O N O
1-
O N O
1N O GT 120 O
1-
i. ii. iv.
Name of structure is BENT right-most diagram YES
(refer to right-most diagram) iii. right-most diagram
3.
A.
i. ii. i. ii.
14 (is the maximum number of electrons in ANY f-type subshell). n = 6 ( 6 is only possible value) shell THREE subshells in n = 3 shell (
B.
l = 0 ( s ) , 1 ( p ) , and 2 ( d )
Total = NINE orbitals =+2
( s ) subshell only has ONE orbital (p) THREE orbitals (d) FIVE orbitals
C.
i.
m s = + 1/2
ii.
ml
4.
n final = 3
Process is EXOTHERMIC, b/c electron is moving closer to nucleus.
E will have a NEGATIVE
sign.
a.
convert from wavelength to frequency
=
b.
3.0 x 10 17 nm / sec 1283.44 nm
= 2.337468E+14 sec 1-
find absolute value of energy associated with this frequency, and apply negative sign.
E = h
= (6.626E-34 J-sec)(2.337468E+14 sec 1-) = | 1.548806E-19 J | = - 1.548806E-19 J
c.
- 1.548807E-19 J = (-2.1874E-18 J)[
1 1 ] 2 3 n2
1 / n2 = ( 1 / 9 ) - 0.070806 = 0.0403053 n 2 = 24.81 (n and n 2 have to be integers, so 24.81 must be 25)
and n (initial ) = 5 , the electron moved from the FIFTH shell.
5.
Construct Lewis dot diagrams for each linkage shown. Total vse = 1 = 6 + 7 = 14 Number sigma bonds = 2
no pi bonds
H
Cl str A
O
H
O str B
Cl
Formal charges for str A: F.C. (chlorine) = 7 - [ 2 + 4 ] = +1 F.C. (oxygen) = 6 - [ 1 + 6 ] = -1 Formal charges for str B: F.C. (chlorine) = 7 - [ 1 + 6 ] = 0 F.C. (oxygen) = 6 - [ 2 + 4 ] = 0 Better distribution of vse will have Formal Charges closer to ZERO. Consequently, structure B, is a more acceptable distribution than A. 6. A. Draw upon the information from problem 2 b, above. The basis structure is plane triangle. The s and p central atom orbitals that hybridize to bond in triangular structures are ONE s and TWO p-type, or sp 2 hybridization. The third p-type orbital on central N is involved in the pi bonding.
B.
The TWO unbonded electron pairs on oxygen also occupy significant geometrical positions. They and the sigma bond to central N define a sp2 hybrid set, and the remaining p-type orbital on oxygen is available to pi bond with nitrogen.
C.
Sigma bonds are shown in the diagram below as THICK lines which are centered on a line connecting the two nuclear centers. Pi bonds are shown as thin lines which are centered above and below a line connecting the two nuclear centers, i.e., pi bonds do not lie between the two nuclear centers. Pi bonds are weaker than sigma bonds.
1N O O
7. 8.
solution not provided Molecular orbital energy levels for vse of first-row-of-eight diatomic molecules and ions are represented by the following diagram:
sigma * 2 p pi * 2 p 2p - - pi 2 p sigma 2 p sigma * 2s - - - 2p bonding bonding anti-bonding - 2s sigma 2 s b...