Practice Exam III
Answer Key

1. Review question really.
For 1s, n=1, l=0   ml= 0
For 4f, n=4  l=3   ml =-3

2.
B:  1s2 2s2 2p1
Mn: [Ar] 4s2 3d5

3. sulfur

4. K or Ca

5.
Cl:  [Ne]  3s2 3p5
Cu: [Ar] 4s1 3d10
Cl-:  [Ne] 3s2 3p6
Cu+:  [Ar] 3d10

6.
Sc:  [Ar]4s2 3d1
Ni2+:  [Ar]3d8
Ag:  [Kr] 5s1 4d10
U:  [Rn] 7s2 6d1 5f3  or [Rn]7s2 5f4
Ba+:  [Xe] 6s2
Cr: [Ar] 4s1 3d5
 

7. Why does the 4s orbital fill before the 3d orbital for Sc?  The 4s level is actually just a bit closer to the nucleus than the 3d level.  That means that the 4s level is lower in energy than the 3d level and according to the Aufbau principle, lower energy levels fill first.

8. How do you know when to use ionic bonding and when to use covalent bonding in Lewis structures?
Metallic atoms bond using ionic bonding.  If there is no metal in the compound, it is bonding with covalent bonds.

9. Rewrite these in order from smallest to largest atom:
Be< Mg < Ca < Sr

N < C < Be

The first series of compounds are in the same column or group.  The biggest difference in size is that Sr is in principle quantum level 5 while Be is in principle quantum level 2.  A higher principle quantum level means that the atom is larger.

The second series of compounds are in the same row or period.  The biggest difference here is the effective nuclear charge (all compounds are in the same principle quantum level of 2).  The effective nuclear charge for N is +5, for C is +4, and for Be is +2.  Nitrogen has 5 valence electrons being pulled in by the strongest effective nuclear charge (of this grouping).  Thus, the electrons are pulled in closer to the nucleus and the atom is smaller.  Be has only 2 electrons being pulled in by the weakest effective nuclear charge (of this grouping of atoms).  The two electrons are not pulled as close to the nucleus so Be is the largest.

10.
Cs < Rb < K < Na
Be < C < N

First ionization energy is the energy to removed one electron from a neutral atom.  In the first series of atoms, Cs is in the highest principle quantum level, making it the largest atom.  As the largest atom, its outermost electron is the furthest away from the nucleus and therefore the easiest to remove.

For the second series, Be has an effective nuclear charge of +2, the weakest effective nuclear charge of this grouping of atom.  The makes Be the biggest atom of the three.  Thus, Be's outermost electron is the furthest from the nucleus and the easiest to remove.

11.
Sr < Ca < Rb < K

C < N < Be

In the first series, the second ionization of Rb and K breaks into the previous energy shell.  The electron configurations for Rb, for example, go from [Kr] 4s1 to [Ar] 3s2 3p5.  This is very unfavorable and requires a lot of energy.  It is harder for K to do this second ionization than for Rb because Rb is larger so the second electron removed is farther from the nucleus.  (Can you write the first and second ionization equations??)

In the second series, Be will have the higest second ionization energy (note that Be's first ionization energy was the smallest) for the same reason.  C will have the smallest energy because it is larger than N and therefore the electron being removed is farther from the nucleus.
 

12.
Cs < In <  P < F

Be < C < N

The most electronegative atom is fluorine and the farther away from F, the less electronegative an atom is.
 
 

13.
H+ < Li+ < He < H-

When you are examining atoms and ions from different rows, it is easiest to look at total number of electrons and protons (rather than effective nuclear charge).  H+ has one proton and no electrons, it is essentially a nucleus.  It is the smallest.  Li+ has three protons to pull in two electrons.  It pulls them in very close to the nucleus.  He has two protons pulling in two electrons.  H- has only one protons to pull in two electrons.  That makes it the biggest.
 
H+ Li+  He H-
total number of protons 1 3 2 1
total number of electrons 0 2 2 2

14. Would you expect K+ to be paramagnetic? What is the electron configuration for K+?  [Ne] 3s2 3p6.  All orbitals have paired electrons, so K+ is diamagnetic.  K [Ar] 4s1 would be paramagnetic because the one valence electron is not paired.
 

15. The effective nuclear charge for both Ga and Ga2+ is three.  But Ga has 3 valence electrons while Ga2+ has only one valence electron.  The nucleus can pull in the one valence electron harder than it can the three valence electrons makeing Ga2+ significantly smaller.

16. The electronic configuration for Si is [Ne]3s2 3p2 and for Cl is [Ne] 3s2 3p5.  The effective nuclear charge of Si is 4 and for Cl is 7.  The number of valence electrons for Si is 4 and for Cl is 7.  Cl is a smaller atom because of the higher effective nuclear charge pulling on its seven valence electrons.  Smaller atoms have their valence electrons closer to the nucleus.  Electrons closer to the nucleus require more energy to remove resulting in a higher ionization energy.  Thus, it is harder to ionize Cl than it is to ionize Si.

17.  When you remove the first electron from, say, Mg, you have Mg+ left.  Mg+ is smaller than Mg because the effective nuclear charge, 2, did not change when you removed an electron, but the number of valence electrons it is pulling on was reduced to one. Smaller atoms have their valence electrons closer to the nucleus.  Electrons closer to the nucleus require more energy to remove resulting in a higher ionization energy.  Thus, it is harder to ionize a second electron from Mg+ than it is to ionize the first electron from Mg.

18.  The electronic configuration for Si is [Ne]3s2 3p2 and for P is [Ne] 3s2 3p3.  The effective nuclear charge of Si is 4 and for P is 5.  The number of valence electrons for Si is 4 and for P is 5.  P is a smaller atom because of the higher effective nuclear charge pulling on its five valence electrons.

19.
Na+ has no dots around it
Na has one dot around it
Ca has two dots around it, not paired up but separate
P has five dots around it, one set paired up and three dots alone.

20.
Sr has a 2+ charge and no dots.  Oxygen has 8 dots, brackets and a 2- written as a superscript.

Mg has a 2+ charge and no dots.  Two Bromines each have 8 dots, brackets and a 1- written as a superscript.

21 & 22. Give the Lewis structure for the following covalently bonded compounds.
 
 
Compound EG MG Angle
AsCl3  tetrahedral trigonal pyramidal < 109.5
H2CO  double bond between C and O trigonal planar trigonal planar  120
NOCl  N in the middle 
double bond between N and O
trigonal planar bent or angular <120
XeF4  two lone pairs on Xe octahedral square planar 90 and 180
F2CCF2
double bond between two carbons 
geometries given are for either carbon
trigonal planar trigonal planar  120
AlCl4-
donít forget brackets
tetrahedral  tetrahedral  109.5
KrO2 
double bond between Kr and O
tetrahedral  bent or angular < 109.5
CS
Carbon-Sulfur double bonds
linear linear 180
IF4- 
two lone pairs on iodine
octahedral square planar 90 and 180
SeO2 
Se-O double bonds 
trigonal planar bent or angular <120
F2CO 
carbon-oxygen double bond
trigonal planar  trigonal planar  120
CH3+  l
less than an octet around carbon 
no double bonds!
trigonal planar trigonal planar 120
SCl4 
lone pair on S
trigonal bipyramidal  seesaw 120, 90, 180
SCl2  two lone pairs on S
   EG= tetrahedral MG=bent or angular  angle = < 109.5
tetrahedral bent or angular < 109.5

23. Draw the resonance structures for SO42-and the resonance hybrid for this molecule. What molecular geometry will it have?

Two oxygens are doubly bonded to sulfur and two oxygens are singly bonded to sulfur.  I found six equivalent resonance structures. (It helps to see them if you label the oxygens 1-4 and then look for pairs of doubly bonded oxygens:  oxygens 1 & 2, 1 & 3, 1 & 4, etc.)  The geometry is tetrahedral.

24.
Aufbau: Mg = 1s2  2s2  2p6  3s1  3p1  (Didn't fill lowest energy level first!)
Pauli:  a box and arrow diagram with two arrows in one blank both pointing in the same direction
Hund:  a box and arrow diagram with two 2p electrons arranged so that two are in one blank and the other two blanks are empty
OTHER answers are possible!
 

25. Give the formal charge on NH2 -, CH3 + and CO.

for N:   5 - 4 - (1/2 x 4) = -1  H:   1 - 0 - (1/2 x 2) = 0
for C:   4 - 0 - (1/2 x 6) = 1    H:   1 - 0 - (1/2 x 2) = 0
for C:   4 - 2 - (1/2 x 6) = -1  O:   6 - 2 - (1/2 x 6) = 1