NAME ______________________________ UNIT 7 (1): NOTES: REDOX (PART 1): OXIDATION #'S,
TEXT SUPPORT: PAGES: 631 – 643
I) Redox Reaction: Any reaction in which there is a change in oxidation numbers (LEO says GER)
REDuction / OXidation reaction: Electrons are * lost
of * oxidation states (or numbers)
and * gained
generally in a change
generally (not always) for two different species. 
any chemical entity: a(n) molecule, atom or ion
A) For every oxidation there must be a reduction.
N
GER
LEO
G
B) Oxidation State (a.k.a: * Oxidation Number
): An arbitrarily assigned value which explains
or predicts the number of electrons of a species, involved in making a bond with a species of a
different element.
1) + or – values for oxidation states apply to species of compounds, or of ions in water.
0 is the oxidation state for pure elements. [Thus, the oxidation state(s) of the oxygen atoms in a
molecule of O2 is 0, since the molecule is produced by species of the same (not different) element(s)]
2) The charge on an ion is only ONE category of oxidation states. Every encounter with an
oxidation state (number), is NOT necessarily an encounter with an ion
a) Oxidation states can be applied to species of a molecule as well. They are used to
describe the number *of shared electrons in a bond between the atoms of
a covalent (molecular) substance.
395
3) An oxidation number of a species may be * positive
more strongly to another nucleus
to be gained, relatively speaking.
if its electrons are attracted
or * negative
if the involved electrons tend
a) The metal ions of a compound will most likely be assigned a * positive
number
The nonmetals of a compound may be assigned a negative or a positive number.
There appear to be multiple definitions of oxidation and reduction yet when analyzed,
they all have a common thread of thought …
What is that common thread?
The Oxidized species of a reaction is
the reactant that:
The Reduced species of a reaction is
the reactant that:
becomes more positive (increases) in
oxidation number
becomes more negative (decreases) in
oxidation number
0
0
+1 -2
16Na + S8 → 8Na2S
completely loses e- (as in an ionic bond)
or experiences a shift of e- away from
the parent nucleus in a covalent bond,
thus becoming a more positive species
0
0
+1 -1
H2 + Cl2 → 2HCl
0
0
+1 -2
16Na + S8 → 8Na2S
completely gains e- (as in an ionic bond)
or experiences a shift of e- towards
itself in a covalent bond, thus
becoming more negative species
0
0
Inorganic /
reaction
chemistry
+1 -1
H2 + Cl2 → 2HCl
the reducing agent (or antioxidant)
the oxidizing agent (or oxidant)
loses hydrogen
gains hydrogen
e.g. alcohol oxidized to an aldehyde
CH3CH2OH → CH3COH
gains oxygen bonds with oxygen
e.g. alcohol oxidized to a carboxylic
acid
CH3CH2OH → CH3COOH
loses oxygen
Biological
&/or
Organic
reactions
e.g. Krebs's
cycle
The oxidation state of the C in the COOH
group is MORE positive than it was in the
C-O-H group
By increasing the number of highly
electronegative O, more electrons were drawn
away from that C, making it more positive.
396
4) The assignment of an oxidation state to a species in a compound is closely linked to
the electronegativity values of the bonded species. The electronegativity of a species
reflects the ability of a species to attract bonding electrons, due to the effective nuclear
charge
For instance, let’s take a look at an organic compound, such as propane (C3H8).
We know that the hydrogen atoms will have partial positive charges (δ+) because of a
lesser electronegativity value (2.2) relative to carbon’s value of 2.6.
But, how would that look, in terms of actual oxidation states?
a) For organic compounds, we tend to write the carbon
of the compound first, representing it as the “central atom”
H H H
| | |
H–C–C–C–H
| | |
H H H
b) When dealing with an INORGANIC compound, the species with the positive
oxidation # is often written first in a formula
Translation: *The species with the lower electronegativity value is written
first in a formula
e.g. NaCl, not ClNa
i) There are some exceptions…For 1st year students, NH3 is the most important.
c) Recall: There are 2 situations of serious importance when dealing with oxidation states:
i) We use these + and – oxidation states, when dealing with species of compounds or
ions dissolved in water.
In light of the definition of oxidation state, why do pure elements get an oxidation
state of zero (0) assigned to them? * oxidation states account for the electrons
used to bond species of different elements. Thus, in pure elements, 0 electrons are
used to bond to species of different elements.
ii) When dealing with ions in water
 Ions in aqueous solution are also called *electrolytes or hydrated ions
397
5) Oxidation states can be assigned, using the oxidation states right off of the Periodic Table
OR oxidation states can be calculated.
a) it is vital to know the oxidation #s so we can track the electrons of a reaction...
When fighting crime, we might “follow the money” … In studying chemistry,
we want to “follow the electrons”. So assigning oxidation states is important.
0
+4 -2
0
2Mg(s) + SiO2(s) 
+2 -2
Si(s) + 2 MgO(s)
b) Generally when only ONE value is written on the periodic table, then that is the one
you will use whenever the species is bonded in a compound. Just pluck it right off the
table!
This works great for alkali and alkaline-earth metals, as well as fluorine, aluminum
zinc and silver (See rules 1-6 on the following pages)
+2
Note that there are 2 val. electrons. Will Ca
tend to lose 2 e - or gain 6 e- to achieve a
stable octet?
Do you see from where the +2 comes?
in a compound, every calcium will be a +2
Ca
2-8-8-2 or
[Ar] 4s2
e.g.) CaS,
Ca3(PO4)2
Ca(NO3)2
c) However, for species in groups 14-17 and those transition metals with multiple
oxidation state possibilities, the oxidation number will be calculated
e.g.
[Ar] 3d ↑↓ ↑ ↑ ↑ ↑ 4s↑↓
+2
+3
+2
+3
Fe
2-8-14-2
FeO vs. Fe2O3
i) the calculation is possible, when most oxidation values can be assigned,
using the rules … AND knowing that the sum of all oxidation states must
add up to be a total of 0 ….because compounds are electrically neutral.
398
II) SUMMARY: Assigning Oxidation Numbers:
A) When assigning, use the oxidation value listed on your periodic table, unless there is a special rule.
Rule … Description
Oxid. #
Examples
0 no electrons (0 electrons) used
1
Any element in its *elemental form or any *free element
0
Fe
N20 no electrons (0) used to bond N
to an atom of a different element
(*elemental form or free element refers to atoms or the diatomic molecules)
2
In ALL compounds, the Group 1 metal ions and the Ag ion
ALWAYS have a +1 (This applies to Ag in binary and ternary compounds)
+1
NaCl: sodium is Na+1
AgNO3: silver is Ag+1
3
In ALL compounds, the Group 2 metal ions and the Zn ion
ALWAYS have a +2 (This applies to Zn in binary and ternary compounds)
+2
CaS: calcium is Ca+2
ZnCl2: zinc is Zn+2
4
In ALL compounds, Al ion is ALWAYS +3
+3
Al2O3:each aluminum =Al+3
-1
NaF: fluoride is F-1
OF2: each F is F-1
(This applies to Al in binary and ternary compounds)
5
In ALL compounds F (fluoride ion), is ALWAYS -1
(This applies to binary and to ternary compounds)
As a general rule, when the compound
is binary, pick the first oxid. # listed
In BINARY IONIC compounds:
6
a) GROUP 15 ions = -3
(first one listed on the PT)
-3
Na3N: nitrogen is N-3
b) GROUP 16 ions = -2
(first one listed on the PT)
-2
CaS: sulfide is S-2
c) GROUP 17 ions = -1
(first one listed on the PT)
-1
NaCl: chloride is Cl-1
(Oxidation numbers for all of the above elements in ternary compounds must be
calculated arithmetically)
SOME RULE VARIATIONS EXIST
7
Oxygen is usually a -2,
except when in compounds with fluorine
and in
compounds with the peroxide polyatomic ion (O2)-2
-2 usually NaOH: oxygen is –2
+1 or +2 OF2: oxygen is +2
-1
Na2O2 each oxygen is -1
8
Hydrogen is usually a +1,
except in metal hydrides (a binary ionic compound in which H is bonded
+1 usually H2O: each hydrogen is H+1
-1
NaH: hydrogen is H-1
to a metal but is the 2nd element…with the 1st being an oxidized metal)
9
In any compound, the sum of every species’s oxidation
number must add up to ZERO, because all compounds are
neutral, by their definition (no charges on compounds)
10 For any given simple ion in aqueous solution (monatomic
ion), the oxidation number = the charge of the ion
NaCl: Na+1 plus Cl-1 =
ZnCl2: Zn+2 plus Cl-1
plus Cl-1
0
=0
oxid # = in the ion Br-1 the
ion charge oxidation number = -1
11 In a polyatomic ion, the SUM of every oxidation number of
every particle must equal THE ASSIGNED CHARGE OF of
the polyatomic ion. See Table E of the Reference Charts
399
B) Review and Practice:
1) BINARY COMPOUND:
Any substance made of only two (different) elements
2) TERNARY COMPOUND: Any substance made of three different elements
a) A ternary ionic compound will have a polyatomic ion
b) A ternary covalent compound will not have a polyatomic ion
3)
IONIC COMPOUND:
Any substance made of a metal ion and nonmetal ion or any
compound with a polyatomic ion in it (ternary inorganic compound)
4) Practice: Using your Periodic Table, record the group and the oxidation number of the
underlined species. None of the "exceptions" are used in this practice.
This exercise is an attempt to get you re-familiarized with the position of
the elements. When you feel weak on the names or symbols, use table S.
Comound
Na2O
Group
Number
1
Oxidation
Number
+1
Per single species
LiF
AlCl3
(binary ionic)
ZnO
SrBr2 (binary ionic)
MgNO3
Ag2SO4
Al2S3
potassium oxide
silver nitrate
lithium hydroxide
boron triiodide
5) Directions: Identify the compound as: binary, ternary, molecular, ionic (use 2 terms/compound)
Compound
KI
CO2
Description: The compound is
*binary ionic
*binary molecular
Na2CO3
* ternary ionic
NH4Br
*ternary ionic
HCl
*binary molecular
C6H12O6
*ternary molecular
400
6) Practice: Assign the oxidation number to each species.
RECAP of Rules
a)
2 AgI(s) + F2(g)  I2(s) + 2 AgF(s)
Pure elements =
Grp 1 and Ag ions in comps =
Grp 2 and Zn ions in comps =
b)
2 H2O(l) 
Al ions in comps are =
2 H2(s) + O2(s)
F ions (fluoride ions) in comps =
Grp 15 ions in binary ionic comps =
16 ions in binary ionic comps =
c)
F2(g) + 2NaI(aq)  2NaF(aq) +
17 ions in binary ionic comps =
I2(g)
Every O in a comp (usually) =
Every H in a comp (usually) =
d)
Mg(s) + 2 HCl(aq) 
e)
Zn(s) + 2HCl(aq)
f)
Hg(l) + Ag2S(s)  HgS(s) + 2 Ag(s)
can you guess at the oxidation state of mercury in this compound?

H2(g) + MgCl2(aq)
ZnCl2(aq) + H2(g)
There is no
rule that applies directly for assigning a value ...
g)
4 Al(s) + 3 O2(s)  2 Al2O3(s)
THINK: CATCH
electron
h)
2 Li(g) + S(s)  Li2S
REACTANTS
i)
N2(g) + 6 Na(s) 
PRODUCTS
2 Na3N
401
D) The Special Rules for oxygen and hydrogen explained
1) Oxygen, has an oxidation number of * -2
, in polyatomic ions or compounds, usually.
Exceptions
a) Each oxygen atom of the peroxide polyatomic ion (O2)-2 is * -1
i) thus, peroxides are compounds in which the oxygen is in the -1 oxidation state:
They are not always easily recognized by first year students, but for us the most
common peroxides are H2O2 , Na2O2 and CaO2
b) When oxygen is combined with fluorine, oxygen will be *
+1
or
+2
Why? * Atoms of fluorine have a greater electronegativity value than atoms of oxygen
and oxygen gets oxidized, while fluorine gets reduced.
a)
CaO(s) + H2O(l)  Ca(OH)2(s)
(Is oxygen bonded to F in any substance or a peroxide?)
b)
4 Cr(s) + 4 K(s) + 7 O2(g)  2 K2Cr2O7(s)
(Is oxygen bonded to F in any substance or peroxide?)
c)
2 H2O2(g)  2 H2O(l) + O2(g)
(Is oxygen bonded to F in any substance or a peroxide?)
d)
O2(g) + F2(g)  2 OF(g)
(Is oxygen bonded to F in any substance or a peroxide?)
e)
O2(g) + 2 F2(g)  2 OF2(g)
(Is oxygen bonded to F in any substance or a peroxide?)
402
2) Hydrogen is most often assigned an oxidation number of * +1
compounds.
in polyatomic ions and
Exceptions: When hydrogen is combined with a metal ion, it will be assigned an
oxidation number of -1. Compounds in which H is the second element, and is bonded to a
metal are called: ionic hydrides e.g. NaH, LiH, CaH2 The ionic bond dominates.
Please note : There is a difference between ionic hydrides such as NaH and the ternary
ionic compounds called Arrhenius bases: e.g. sodium hydroxide (NaOH.)
Why? Hydrogen is a nonmetal which often likes to act as a metal. When reacted with an actual metal
though, hydrogen is forced to act as the nonmetal it really is. Thus it is the H atom which
becomes negative by default because the metal must be oxidized and thus be positive.
Remember …it has to do with electronegativity values and thus effective nuclear charge ….
Hydrogen has a lesser electronegativity value than virtually any other nonmetal … BUT, its
value of 2.2 is greater than any metal. Thus, H when reacted directly with a metal, is the
reduced species …and takes on a -1 oxidation state.
a) 2 Li(s) + H2(g) 
2 LiH(s)
b) Ca(s) + H2(g) 
CaH2(s)
Take a look at the reaction of lithium hydride with aluminum chloride, producing
lithium aluminum hydride and lithium chloride.
Can you assign the oxidation states to each species?
4 LiH + AlCl3 → LiAlH4 + 3 LiCl
II) Calculating an oxidation number:
The rules only go so far. Rules 1-5 (for our course) work very well, and (for our course)
will not have variants. However, rule 6 is ONLY reasonable for binary ionic compounds.
But the elements of Groups 15, 16, and 17 … and not to mention ions of transition metals can
be found in compounds that are not necessarily ionic or binary.
Chemists do have a means of accounting for the electronic status of each species in most other
inorganic compounds (beyond the binary ionic compounds). So, how do we get the oxidation
state of nitrogen in NaNO3 , the chromium in MgCrO4 , or the sulfur in (SO4)-2 ? We
do a little algebra …
403
A) The oxidation number reflects the number of electrons being used to bond, and thus, it makes sense
to imagine that the oxidation state of a particular species changes, as the species to which it is
bonded, changes.
1) This is especially true for most of the Transition metals and the all of the nonmetals.
These oxidation states should NOT be assumed, but rather, they should be algebraically
calculated
a) This is especially true for nonmetals in TERNARY INORGANIC compounds
b) PROCESS: Assign known oxidation numbers using the rules and assign the
unknown species, as "x". Then, algebraically solve for "x"
RECAP of Rules
Pure elements =
Fact: You cannot use the rules for Grp 15, 16, 17 elements in the following
compounds.
Grp 1 and Ag ions in comps =
Grp 2 and Zn ions in comps =
Al ions in comps are =
What is the oxidation number of S in Li2SO3?
F ions (fluoride ions) in comps =
Grp 15 ions in binary ionic comps =
16 ions in binary ionic comps =
17 ions in binary ionic comps =
Every O in a comp (usually) =
Every H in a comp (usually) =
What is the oxidation number of Fe in Fe2O3?
What is the oxidation number of Mn in Ca(MnO4)2
404
PRACTICE: ASSIGNING OXIDATIONS NUMBERS IN TERNARY COMPOUNDS OR THOSE WITH TRANSITION METALS
PROCEDURE :
1) Assign "x" to the element you are investigating.
2) Use the rules and periodic table values to assign known oxidation numbers.
3) Multiply every oxidation number by the appropriate subscript to create an equation and set the
equation equal to zero (for compounds) and equal to the charge (for PAI).
4) Arithmetically solve for "x" by multiplying subscripts by oxidation # and set it all = to 0
Calculate and assign the oxidation number of :
(answers are below….)
PRACTICE 1: Calculate and assign the oxidation number
Answers are on the next page…
Review: 1) Assign "x" to the element you are investigating and known oxidation states to all other species.
2) Multiply every oxidation number by the appropriate subscript to create an equation and set the equation equal to zero (for compounds)
OR equal to the charge for polyatomic ions.
3) Arithmetically solve for "x" by multiplying subscripts by oxidation # and set it all = to 0
a)
S in
Na2SO4
b)
Cl in LiClO4
c)
S in CaS2O3
d)
N in
e)
Mn in KMnO4
f)
Cr in MgCrO4
g)
O in OF2 (exception)
h)
H in
i)
Cl in LiClO3
j)
Fe in FeCl2
k)
N in Ca(NO3)2
KNO3
CaH2 (exception)
405
l)
S in Al2(SO3)3
m)
S in Al2(SO4)3
n)
Cu in CuO
o)
p)
q)
q)
Answers:
a) +6
k) +5
Cr in CrCl3
Cr in Na2Cr2O7
O in H2O2
(exception)
b) +7
l) +4
c) +2
m) +6
d) +5
n) +2
e) +7
o) +3
f) +6
p) +6
g) +2
q) -1
h) -1
i) +5
Fe in FeCl2
j) +2
PRACTICE : Solve for the element in question. The sum, in this case is NOT equal to zero, but to the
charge of the polyatomic ion itself.
THIS IS AN APPLICATION OF RULE 10
a) N in
b) S in
e.g.:
-1
(NO3)
x -2
(PO4)-3
x – 8 = -3 (not 0, because it is an ION)
-2
(SO4)
x = +5
-2
c) S in (S2O3)
-1
d) Mn in (MnO4)
ans: a) +5, b) +6
e) C
in
c) +2
d) +7 e) +3
(C2O4)-2
Note: The oxidation number of a transition metal ion found in a polyatomic ion is a constant for that polyatomic ion
+1
Proof:
x
-2
Na2Cr2O7
-3 +1
CaCr2O7
Al2(Cr2O7)3
Make your life easier:
x
-2
(NH4)2Cr2O7
The N of (NO3)-1
is a * +5
The S of (SO4)-2
is a * +6
The P of (PO4)-3
is a * +5
406
Check Out:
The Tollens’ Test: https://www.youtube.com/watch?v=y-4qqcCxD6g
Electron Transport Chain: 1 https://www.youtube.com/watch?v=f82O5qvok5I
2
https://tinyurl.com/McGraw-Hill-one
3
https://tinyurl.com/McGraw-Hill-2
text only: http://staff.jccc.net/pdecell/cellresp/etstext.html#top
Bozeman Science: The importance of oxygen: https://www.youtube.com/watch?v=o61dTk4J9AU
407
NAME ______________________________
GRADED: TEXT ASSIGNMENT: REDOX pages 630 -633 & p. 635
FIRST: A brief review of what is meant by an oxidation number (oxidation state) from the first quarter, is in order.
An oxidation number is a positive or negative number assigned to a species. It is assigned to help understand
the number of electrons involved in bonding to a species of a different element, and to indicate the degree of
oxidation or reduction.
+1
-1
Given: Na + Cl2  2 NaCl
0
0
For instance, the sodium ion of NaCl is given an oxidation state of +1, compared to the original Na0. This +1
value indicates that 1 electron was involved in the bonding process to the Cl, (that's the "1") and that the electron
has been lost (that's the "+") to the chloride. Thus, that simple symbol of the +1 oxidation state, indicates two
pieces of important information.
The chloride conversely, is assigned a -1, indicating that the chloride species is a reduced species (compared to
Cl0 it is more negative), due to a gain of 1 electron. The “gained” electron is the same electron, as that lost by
the sodium. Thus, 1 electron was lost… 1 was gained… illustrating the Law of the Conservation of Charge.
***********************************************************************************************
NOW: Read pages 630-633 & p. 635 of your textbook (skip page 634). Take notes. You should use those notes to help
the questions, on the next page. While reading the text, read for the following goals.
Read to: grasp / identify / list / comprehend / apply the
 ....various definitions of what is meant by oxidation and reduction. Familiarize yourself with each definition, but focus
on those dealing with the loss/gain or shift of electrons
 ...difference of electron change during a redox reaction dealing with ions and the electron change during a redox
reaction dealing with covalent bond formation.
********************************************
Be sure to have completed the textbook reading, in light of the goals. THEN, read this passage and answer the questions
found on the next page.
The citric acid cycle is the common pathway by which organic fuel molecules of the cell are
oxidized during cellular respiration. These fuel molecules, glucose, fatty acids, and amino acids, are
broken down and fed into the Krebs cycle, becoming oxidized to acetyl coenzyme A (acetyl CoA)
before entering the cycle. The Krebs cycle is one of the processes of eukaryotic cellular respiration.
Compared to the anaerobic process of fermentation, much more energy is released during aerobic
cellular respiration, when glucose, is fully degraded by the Krebs cycle, where it is completely
oxidized to H2O and.CO2. In the glucose molecule, the carbons exist with various oxidation states,
such as -1, 0 and +1. The carbon of CO2 exists in the +4 oxidation state.
The Krebs cycle was first postulated in 1937 by Hans Krebs, and represents an efficient way for cells
to produce energy during the degradation of energy-rich molecules. The electrons removed from
intermediate metabolic products during the Krebs cycle are used to reduce coenzyme molecules
nicotinamide adenine dinucleotide [NAD+] and flavin mononucleotide [FAD]) to NADH and
FADH2, respectively. ....
Citation: Krebs Cycle - Coa, Reaction, Acetyl, and Molecules - JRank Articles http://science.jrank.org/pages/3776/KrebsCycle.html#ixzz1mxAVB9nn (edited from the original passage)
408
Question 1: According to the passage, NAD+ is reduced to NADH, while FAD is reduced to FADH2. According
to the text reading, what definition of redox is being rather overtly, used to state that there has been a reduction?
_______________________________________________________________________________________
Question 2: According to the passage, carbon species of the glucose molecule are oxidized and used to produce
carbon dioxide. Identify at least 2 reasons as to why it is appropriate, according to the various definitions of
redox, to say that carbon species of glucose have been oxidized, in this situation.
1)_____________________________________________________________________________________
2)______________________________________________________________________________________
Question 3: The following chemical reaction is a redox reaction. You have been given the oxidation states of
every species as they existed as reactants, and as they exist as products.
0
0
+1 -2
16 Ag + S8  8 Ag2S
a) Each atom of silver and of sulfur is assigned a 0 for an oxidation state. Based upon the definition
of the term oxidation state, explain the assignment of "0" to the elements of silver and sulfur.
________________________________________________________________________________________
________________________________________________________________________________________
b) Based upon the reading and your analysis of the oxidation states, is Ag+1 the oxidized form or the reduced
form of the element, silver? Defend your answer.
________________________________________________________________________________________
_________________________________________________________________________________________
0
0
+2 -2
Question 4: Given the balanced chemical reaction: Mg + O2  2 MgO
Based on concepts in the
0
reading, in theroy, the Mg becomes more stable by losing 2 electrons, and the atoms of dioxygen become more
stable by gaining 2 electrons, each. Using ideas from quantum mechanics, suggest a reason chemists might cite,
for each ion species being more stable than the atoms.
When comparing the electron configurations of the reactant species to the product species ________________
_________________________ _______________________________________________________________
409
Question 5: The alcohol, ethanol, CH3CH2OH, can be partially oxidized to the aldehyde ethanal, CH3COH
Ethanol CH3CH2OH can also be oxidized further to ethanoic acid, CH3COOH, a carboxylic acid.
In both cases, a second reactant (other than the ethanol) is reduced.
___a) Using the definitions of redox, dealing with oxygen and/ or hydrogen identify why it is accurate to claim
that ethanol's conversion to ethanal is a form of oxidation
I) ethanol gains hydrogen(s)
II) ethanol loses hydrogen(s)
III) ethanol gains oxygen(s)
1)
2)
3)
4)
only I is correct
only II is correct
both I and II are correct
both II and III are correct
____b) Using the definitions of redox, dealing with oxygen and/ or hydrogen identify why it is accurate to
claim that ethanol's conversion to ethanoic acid is a more extensive form of oxidation
I) ethanol gains hydrogen(s)
II) ethanol loses hydrogen(s)
III) ethanol gains oxygen(s)
1)
2)
3)
4)
only I is correct
only II is correct
both I and II are correct
both II and III are correct
___c) In reality it is quite accurate to believe that the oxidation state of the second carbon of CH3CH2OH
has been changed in each of these oxidative processes.
From the point of view of bonding electrons, during these "oxidations", to ethanal or ethanoic acie, that
second carbon of the ethanol reactant has probably had
1) electrons shifted away from it, and thus become a more positive oxidation state
2) electrons shifted towards it, and thus become a more negative oxidation state
3) electrons shifted away from it, and thus become a more negative oxidation state
Question 6: In the reading there is a passage: "When a metal and a nonmetal react and form ions, it is easy to
identify complete transfer of electrons. But some reactions involve covalent compounds that is compounds in
which complete electron transfer does not occur..."
The ideas surrounding which of the following concepts can be directly linked to predicting whether electrons
will be transferred completely or not transferred completely? (circle one)
(1) atomic radius
(2) density
(3) conservation of mass
(4) electronegativity
410