Chemistry 1000 (Fall 2013)
Problem Set #7: Nonmetals, Acids and Oxidation States
Solutions
Answers to Questions in Silberberg (only those w/out answers at the back of the book)
2.95
(a)
HClO4 = perchloric acid
(b)
HNO3 = nitric acid
(c)
HBrO2 = bromous acid
(d)
HF = hydrofluoric acid
2.110
(a)
(c)
(e)
(g)
(i)
copper(II) sulphate pentahydrate
sulfuric acid
hydrochloric acid
calcium carbonate
sodium hydrogen carbonate
or sodium bicarbonate
(b)
(d)
(f)
(h)
(j)
calcium hydroxide
sodium carbonate
magnesium sulfate heptahydrate
carbon dioxide
sodium hydroxide
2.134
(a)
(b)
(c)
(d)
(e)+(f)
black (C), orange (O), green (Cl), purple (Ar)
brown (Na), blue (Ba)
orange (O2), green (Cl2), purple (Ar)
brown (Na), blue (Ba), black (C)
black and orange (CO, CO2)
black and green (CCl4, C2Cl6, C2Cl4, C2Cl2, etc.)
orange and green (though products will either be radicals or ions)
(g)+(h) brown and green (NaCl)
blue and orange (BaO)
(i)
brown and orange (Na2O)
(j)
blue and green (BaCl2)
(k)
purple (Ar)
13.1
The two core electrons in Li shield close to 2/3 of the positive charge from the three protons, so
the effective nuclear charges for the valence electrons in H and Li are similar. The valence
electron in H is much closer to the nucleus, however, so it requires much more energy to
remove it (compared to the valence electron in Li).
Also, H+ is *much* smaller than Li+ (since H+ is literally a single proton), so the charge density
of H+ is much higher than that of Li+, making it less stable.
13.2
Sorry. No picture.
Ionic hydrides are formed from the Group 1 and 2 metals (first two columns of the periodic
table).
Metallic hydrides are formed from the transition metals (i.e. the d-block).
Covalent hydrides are formed from non-metals (top right corner of periodic table).
13.34
(a)
Boron compounds are covalently bonded. Other Group 13 elements form ionic compounds.
Elemental boron is a very hard metalloid with a very high melting point. The other Group 13
elements are metals, much softer and lower melting than boron.
(b)
Boron hydrides (boranes) contain three-center two-electron bonds (B-H-B sharing a single
electron pair)
Boron is much smaller than the other elements in Group 13. As such, its valence electrons are
held more strongly and it requires too much energy to form B3+ (which would be very small
and have a very high charge density).
13.38
(a)
Ga appears to have an oxidation state of +2
(b)
Ga+ has an oxidation state of +1. Ga in GaCl4- has an oxidation state of +3.
(c)
GaCl4- is tetrahedral with bond angles of ~109.45.
.. .
.. Cl
.
-1
.. .
..
Ga
.
..Cl
Cl
..Cl..
..
..
..
13.47 Allotropes are different forms of the same pure element.
Carbon has many allotropes: diamond, graphite, fullerenes, graphene, nanotubes, etc.
13.59 least acidic Bi2O3, Sb2O3, Sb2O5,
P4O10 most acidic
Generally speaking, acidity of an oxide decreases with the metallic character of the non-O
element in the oxide. Also, acidity of an oxide increases as the oxidation state of the non-O
atoms increase.
13.86
(a)
F2 and Cl2 are gases at STP, Br2 is a liquid at STP, I2 is a solid at STP
(b)
All four halogens are nonpolar so the only intermolecular forces at work are induced dipoleinduced dipole forces (aka London dispersion forces). Because the size of the atoms increases
down the group, the polarizability of the molecules increases down the group. As such, the
induced dipoles become larger and the intermolecular forces become stronger down the group.
13.96
(b)
NaOCl
(c)
HF
(d)
Br2
13.103
(a)
To have an even number of electrons, a xenon fluoride must have an even number of F atoms.
(Xe has 8 valence electrons. F has 7 valence electrons. So, to get an even number of electrons
from F, there must be an even number of F atoms.)
(b)
To have an even number of electrons, XeF3+ and XeF7- must have an odd number of F atoms
because Xe has 8 valence electrons and F has 7 valence electrons.
XeF3+ has one less electron than XeF3 (8 + 3(7) – 1 = 28 valence electrons).
XeF7- has one more electron than XeF7 (8 + 7(7) + 1 = 58 valence electrons).
(c)
In XeF3+, Xe is bonded to three F atoms and has two lone pairs. As such, it has trigonal
bipyramidal electron group geometry and T-shaped molecular geometry.
13.122 The efficiency can be accounted for at any step of the calculation.
In this solution, it’s shown as step 2.
Step 1: Balance the chemical equation
2Ca3 (PO4 )2(s )  6SiO 2  10C (s )  6CaSiO 3(s )  10CO( g )  P4( g )
Step 2: Given that the reaction is only 90.% efficient (i.e. percent yield is 90.%), calculate the
theoretical yield of P4 necessary to give an actual yield of 315 mol.
percent yield 
actual yield
 100%
theoretical yield
theoretical yield 
actual yield
315 mol
 100% 
 100%  350 mol  3.5  102 mol
percent yield
90.%
Step 3: Calculate the number of moles of Ca3(PO4)2 required to produce 350 mol P4
nCa (PO )  350molP4 
3
4 2
2molCa 3(PO4 )2
 700molCa 3(PO4 )2  7.0  102 molCa 3(PO4 )2
1molP4
Step 4: Calculate the mass of Ca3(PO4)2 (M = 310.177 g/mol)
310.177g
1kg
mCa 3(PO4 )2  7.0  102 molCa 3(PO4 )2 

 217kg
1mol
1000g
13.123 Picture A shows that the oxide of E (EO3) has trigonal planar molecular geometry. As such, it
must also have trigonal planar electron group geometry.
Picture B shows that the chloride of E (ECl4) has see-saw molecular geometry. As such, it
must have trigonal bipyramidal electron group geometry.
(a)
Since F and Cl are both halogens, expect that EF4 has the same geometry as ECl4. This see-saw
geometry corresponds to four bonded atoms and one lone pair. Element E has contributed 6
valence electrons to the Lewis structure (since total valence electrons = 1(6) + 4(7) = 34).
EF5- has one additional bonded atom, but must have the same number of lone pairs as EF4.
(Total valence electrons = 1(6) + 5(7) + 1 = 42)
.. ..F ..
.. .
.
.. E ..F..
..
.. F .. ..F
..
(c)
13.124
(a)+(b)
.. ..F .. ..
..
.. ..F . E-1 F..
.. . .. .
.
.. F .. F..
..
Therefore, EF5- has octahedral electron group geometry and square pyramidal molecular
geometry.
Since all elements are less electronegative than F, and we have established that a neutral atom
of E has 6 valence electrons, the oxidation state of E in both of these compounds must be +4.
-1
..
C
+1
.
O.
(c)
The electronegativity difference between C and O means that, within the bond, electrons are
pulled toward the oxygen atom (which would normally make O partially negative and C
partially positive). The formal charges show that electrons are also pulled in the opposite
direction – toward the carbon (which has the negative formal charge) in order for carbon to
have a complete octet.
If the C-O bond was a double bond instead of a triple bond and the oxygen had two lone pairs,
there would be no formal charges on either atom. The problem with that structure would be
that C would only have six electrons. To give it the octet, oxygen must share one of those pairs
of electrons, giving a triple bond instead.
16.5
A conjugate acid-base pair is a pair of compounds that differ by H+.
When H+ is removed from the conjugate acid, the product is the conjugate base.
When H+ is added to the conjugate base, the product is the conjugate acid.
16.115
(a)
NaCl(aq)
[Ni(OH2)6]2+ is a stronger acid than [Na(OH2)6]+ so the NiCl2(aq) solution will be more acidic
(lower pH). Therefore, the NaCl(aq) solution will be less acidic (higher pH).
16.132 To be a Lewis base, a molecule or ion have an electron pair available to donate. This is often
(though not always) a lone pair. Furthermore, the Lewis base cannot become unstable as a
result of sharing this electron pair.
e.g. F2 has six lone pairs, but it is not a Lewis base because that would result in formation of a
product containing a fluorine atom with a formal charge of +1. Fluorine atoms are too
electronegative to be stable with a +1 formal charge.
e.g. NH3 only has one lone pair, but it is a good Lewis base. N can adopt a +1 formal charge.
To be a Lewis acid, a molecule or ion must contain an atom that has low electron density
(positive charge, partial positive charge or incomplete octet). Furthermore, it must have the
capacity to form an additional bond to the Lewis base. Look for:
 metal cations
 molecules containing atoms with incomplete octets (e.g. BF3, AlCl3, BeCl2, BH3)
 molecules containing polar multiple bonds (e.g. CO2, SO3)
16.135
(a)
Water can be the product of an acid-base reaction according to any of the three definitions.
(H+ is the best Lewis acid and OH- is a good Lewis base.)
(b)
Water is always a product of an Arrhenius acid-base reaction.
16.136
(a)
NH3 is a Brønsted base and a Lewis base, but it is not an Arrhenius base (as it does not contain
OH- ions).
(b)
BF3 is a Lewis acid but not an Arrhenius acid or a Brønsted acid (as it does not contain any
acidic hydrogen atoms).
Note that the “acidic” part of this answer is important. BH3 would also qualify as a Lewis acid
but not an Arrhenius acid or a Brønsted acid. It contains hydrogen atoms, but none of them
are acidic. Cations of any metal would also be good answers to this question (though the
Group 1 cations are such weak acids that we consider them to be neutral. Recall that acidity of
a metal cation increases with the charge of the cation and the electronegativity of the metal.)
16.155 In the first step, the carboxylic acid (RCO2H) is the Lewis base and H+ is the Lewis acid.
In the second step, the alcohol (R’OH) is the Lewis base and the cation is the Lewis acid.
16.161 First, a series of Lewis acid-base reactions occur between Fe3+ and water:
Fe 3  6H 2O(l )  Fe OH2 6 (aq )
3
Then, a Brønsted acid-base reaction occurs between the aqua ion and water:
Fe OH2 6 3(aq )  H 2O(l )  Fe OH2 5 OH 2(aq )  H 3O(aq )
23.54
(a)
If they are not kept separate, the Cl2 and NaOH produced in the chlor-alkali process will react
to give NaOCl, NaCl and H2O.
Additional Practice Problems
1.
For each of the following pairs of ions, indicate which you would expect to be a stronger Lewis
acid and justify your answer.
(a)
Al3+ (53 pm) vs. Fe3+ (65 pm)
Al3+
The two ions have the same charge, but Al3+ has a smaller radius so its charge density is
greater. As such, Lewis bases will be more strongly attracted to Al3+, making it the stronger
Lewis acid.
(b)
Mg2+ (72 pm) vs. Sc3+ (75 pm)
Sc3+
The two ions have similar radii, but Sc3+ has a greater positive charge so its charge density is
greater. As such, Lewis bases will be more strongly attracted to Sc3+, making it the stronger
Lewis acid.
(c)
Be2+ (27 pm) vs. Fe3+ (65 pm)
Be2+
Be2+ is smaller, but Fe3+ has the greater positive charge. When comparing the acidity of aqua
complexes of various cations, we calculated pKa using the following formula:
 z2

pK a  15.14  88.16 pm  0.0960 Pauling  1.50 *You may have used the version without
 r

the “fudge factor”. The conclusions drawn are the same with either version of the formula.*
To compare the acidity of Be2+ and Fe3+, it should therefore be reasonable to compare z2/r for
the two ions. The cation with the larger value for z2/r should be more acidic.
Be2+ has z2/r = +0.148 pm-1
Fe3+ has z2/r = +0.138 pm-1
Therefore, Be2+ is the stronger Lewis acid.
2.
Write a balanced chemical equation for a Brønsted acid-base reaction between a hydride ion
and ammonia. Is this reaction likely to proceed in water? Why or why not?
H   NH 3  H 2  NH 2

This reaction is not likely to proceed in water because water is a stronger acid than ammonia.
Instead of the reaction above, the reaction below will occur:
H   H 2O  H 2  OH 
(Hydride is a very strong base. It cannot be an acid because if H- gives up H+, we are left with
just a pair of electrons!)
3.
For each of the following pairs of reagents:
i.
draw a Lewis structure for each reagent
(each CH3 is attached to the central atom by a bond to C)
ii.
identify the Lewis acid and the Lewis base
ii.
draw a Lewis structure of the product
(a)
SbCl5
+ (CH3)2S 
H
H
C
..S..
H
C
..
.. Cl .. ..
.
Cl
..
.. .
..Cl
.. Sb .. .
.. .
. Cl
..Cl
.. .
H
+
H
..
.. Cl
..
..
H
H
C
+1S
H
H
C
H
Lewis acid
(accepting electrons)
+ (CH3)3P 
(CH3)3Al
H
H
C
H
.
P.
C
H
+
H
H
H
H
C
H
H
H
H
H
H
Lewis base
(donating electrons)
(b)
..
.. Cl .. ..
.
+1 Cl
.. .
Sb ..
.
.. .
. Cl
.
..Cl
..
H
C
Al
H
H
C
H
H
H
C
H
H
-1
+1
H
H
H
Lewis base
(donating electrons)
H
H
H
C
C
H
C
P
Al
C
H
H
H
H
C
H
H
H
C
H
H
Lewis acid
(accepting electrons)
Bond angles in the pictures above are not drawn accurately. Atoms with 4 electron groups are
tetrahedral – not square!
4.
(a)
For each of the following pairs of acids, indicate which is the stronger Brønsted acid and justify
your answer. Where possible, calculate the pKa values to confirm your answer.
[Fe(OH2)6]3+ vs. [Fe(OH2)6]2+
[Fe(OH2)6]3+
 32

 0.0960 pm 1 1.8  1.50  0.39
Fe3+: pK a  15.14  88.16 pm
 65 pm

(2.93 without “fudge factor”)
 22

 0.0960 pm 1 1.8  1.50  8.02
Fe2+: pK a  15.14  88.16 pm
 77 pm

(10.56 without “fudge factor”)
3+
The two ions have similar radii, but Fe has a greater positive charge so its charge density is
greater. Thus, the Fe-O bond will be stronger/more covalent and electrons in the O-H bonds
will be more strongly pulled toward the center of the molecule. This will weaken the O-H
bonds, making it easier for a base to remove H+ from [Fe(OH2)6]3+ than [Fe(OH2)6]2+
(b)
[Al(OH2)6]3+ vs. [Ga(OH2)6]3+
[Al(OH2)6]3+
 32

Al3+: pK a  15.14  88.16 pm
 0.0960 pm 1 1.5  1.50  0.17
 53 pm

(0.17 without “fudge factor”)
 32

Ga3+: pK a  15.14  88.16 pm
 0.0960 pm 1 1.6  1.50  1.50
 62 pm

(2.34 without “fudge factor”)
3+
The two ions have the same charge, but Al has a smaller radius so its charge density is
greater. Thus, the Al-O bond will be stronger/more covalent and electrons in the O-H bonds
will be more strongly pulled toward the center of the molecule. This will weaken the O-H
bonds, making it easier for a base to remove H+ from [Al(OH2)6]3+ than [Ga(OH2)6]3+
(c)
HClO3 vs. HClO4
pK a  8  5(2)  2 for HClO3 [i.e. O2Cl(OH)]
pK a  8  5(3)  7 for HClO4 [i.e. O3Cl(OH)]
HClO4
HClO4 has an additional oxygen atom pulling electron density away from the chlorine atom
thereby leaving chlorine with a more positive oxidation state. This means that the electrons in
the O-H bond are more strongly pulled toward the chlorine atom, making it easier for a base to
remove H+ from HClO4.
(d)
H3PO4 vs. H2SO4
pK a  8  5(1)  3
H2SO4
for H3PO4 [i.e. OP(OH)3]
pK a  8  5(2)  2 for H2SO4 [i.e. O2S(OH)2]
The sulfur atom in H2SO4 has a more positive oxidation state than the phosphorus atom in
H3PO4. This means that the electrons in the O-H bonds of H2SO4 are more strongly pulled
toward the central atom, making it easier for a base to remove H+ from H2SO4.
5.
(a)
(b)
Arrange the oxides below in order from the most acidic through amphoteric to the most basic:
Cl2O7
SO3
CO2
Al2O3
BaO
acid
acid
acid amphoteric base
Assign oxidation states to the non-oxygen element in each oxide.
In each of these molecules, O has an oxidation state of -2. Since the sum of the oxidation states
must equal the overall charge of the molecule, the remaining oxidation states can be calculated:
Cl has an oxidation state of +7 in Cl2O7
S has an oxidation state of +6 in SO3
C has an oxidation state of +4 in CO2
Al has an oxidation state of +3 in Al2O3
Ba has an oxidation state of +2 in BaO
(c)
6.
(a)
What is the relationship between the oxidation state of the central atom(s) and the acidity of the
oxide?
Oxides in which the central atom(s) has/have a more positive oxidation state are more strongly
acidic than those which in which the central atom(a) has/have a less positive oxidation state.
Draw the aqua ions of Ag+ (r = 115 pm; χ= 1.9) and Na+ (r = 99 pm; χ= 0.9).
H
H
H O
H O
H
O
O H
O H
H
H
H
Ag
H
(b)
+
O H
H
H
H O
H O
H
+
H
O H
Na
O
H
O H
H
O H
H
Use the formula below to calculate the pKa of these two aqua ions.
 12

pK a ( Ag  complex )  15.14  88.16 pm
 0.0960(1.9  1.50) pm 1   10.99
 115 pm

 12

pK a ( Na  complex )  15.14  88.16 pm
 0.0960(0.9  1.50) pm 1   19.33 *
 99 pm

*The measured value for [Na(OH2)6] + is not this high – actual pKa = 14.2. The “fudge factor”
doesn’t work very well for alkali metals; however, the relative acidity of the Ag+ and Na+
complexes is correct.
(c)
Which of these two aqua ions is more acidic? Explain why this might be. (Your explanation
should not just be “because the pKa values say so”; consider the properties of the two ions.)
The hexaaquasilver(I) ion is more acidic. This means that electron density is pulled more
strongly toward Ag(I) in hexaaquasilver(I) than it’s pulled toward sodium in the sodium aqua
complex. Since the radii of the two ions are similar, this is due primarily to the greater
electronegativity of silver compared to sodium.
7.
(a)
(b)
(c)
Give formulas for the following common acids: nitric acid, sulfuric acid, hydrobromic acid,
perchloric acid, carbonic acid.
HNO3
H2SO4
HBr
HClO4
H2CO3
What is the oxidation state of the central atom in each of these compounds?
N is +5
S is +6
n/a
Cl is +7
C is +4
How does the oxidation state of each common acid relate to the periodic group number of the
element?
In the “common oxoacids”, the oxidation state of the central atom is equal to the group number
minus ten.
8.
9.
(a)
Does nitrogen exhibit allotropy? Does phosphorus? Describe the allotropes formed.
Nitrogen does not. Phosphorus does.
See your notes/text for details of the phosphorus allotropes.
Phosphorus reacts readily with halogens. If 6.1 grams of white phosphorus reacts with chlorine
gas to produce 41 grams of a phosphorus halide, what is the molecular formula of the product?
P4(s)
+
2x Cl2(g)
→
4 PClx
6.1 g
???
41 g
123.895 g/mol
70.9054 g/mol
???
nP4  6.1g 
1mol
 0.049mol
123.895 g
n product  n P4  moleratio  0.049molP4 
M product 
m product
n product

4molproduct
 0.20molproduct
1molP4
41g
g
 2.1 10 2
0.20mol
mol
The molar mass of the product is ~208 g/mol (w/ only 2 sig. fig.). The product could
reasonably be either PCl3 (M = 137 g/mol) or PCl5 (M = 208 g/mol). Therefore, the product is
PCl5.
(b)
Draw a Lewis structure of the product and indicate its molecular geometry.
The product is trigonal bipyramidal:
..
..Cl
..
.. .
..Cl
. ..
.
Cl
.... .
P
.
Cl
.. .
..Cl ..
..