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Element Valency PDF

Fluroine Atom

The outer shell of a fluorine atom contains 7 electrons. This means it has one less electron than needed to complete the shell. This gives fluorine a -1 valence.

This element Valency PDF is a downloadable version of the Valences of the Elements table.

Download PDF Download PDF

As in the table, the most common valences are in BOLD text where values in italics are theoretical values based on periodic table trends.

This table requires three sheets of paper to print. If you prefer a more compact version, this information is available in periodic table form. There is a Color Periodic Table of the Elements or a Black and White version. A simpler version listing only the most common valence charges is also available.

This is page 1 of the PDF file. It contains the valences of the first 45 elements.

Element Valences 1-45Elements 46-118 can be found in the PDF.

 

Coulomb Force Example Problem

Coulomb force is the force of either attraction or repulsion between two charged bodies. The force is related to the magnitude and charge on the two bodies and the distance between them by Coulomb’s Law:


where
q1 and q2 is the amount of charge in Coulombs
r is the distance in meters between the charges
k is the Coulomb’s Law constant = 8.99×109 N•m2/C2

The direction of the force depends on the positive or negative charges on the bodies. If the two charges are identical, the force is a repulsive force. If one is positive and the other negative, the force is an attractive force.

This Coulomb force example problem shows how to use this equation to find the number of electrons transferred between two bodies to generate a set amount of force over a short distance.

Example Problem:
Two neutrally charged bodies are separated by 1 cm. Electrons are removed from one body and placed on the second body until a force of 1×10-6 N is generated between them. How many electrons were transferred between the bodies?

Solution:

First, draw a diagram of the problem.

Coulomb Force Example Problem 2

Define the variables:
F = coulomb force = 1×10-6 N
q1 = charge on first body
q2 = charge on second body
e = charge of a single electron = 1.60×10-19 C
k = 8.99×109 N•m2/C2
r = distance between two bodies = 1 cm = 0.01 m

Start with the Coulomb’s Law equation.

As an electron is transferred from body 1 to body 2, body 1 becomes positive and body two becomes negative by the charge of one electron. Once the final desired force is reached, n electrons have been transferred.

q1 = +ne
q2 = -ne

The signs of the charges give the direction of the force, we are more interested in the magnitude of the force. The magnitude of the charges are identical, so we can ignore the negative sign on q2. This simplifies the above equation to:

We want the number of electrons, so solve the equation for n.

Substitute in the known values. Remember to convert 1 cm to 0.01 m to keep the units consistent.

n = 6.59×108

Answer:
6.59×108 electrons were transferred between the two bodies to produce an attractive force of 1×10-6 Newtons.

For another Coulomb force example problem, check out Coulomb’s Law Example Problem.

Coulomb’s Law Example Problem

Coulomb’s Law is a force law between charged bodies. It relates the force to the magnitude and charge on the two bodies and the distance between them by the relationship:


where
q1 and q2 is the amount of charge in Coulombs
r is the distance in meters between the charges
k is the Coulomb’s Law constant = 8.99×109 N•m2/C2

The direction of the force depends on the positive or negative charges on the bodies. If the two charges are identical, the force is a repulsive force. If one is positive and the other negative, the force is an attractive force.

This Coulomb’s Law example problem shows how to use this equation to find the charges necessary to produce a known repulsive force over a set distance.

Example Problem:
The force between two identical charges separated by 1 cm is equal to 90 N. What is the magnitude of the two charges?

Solution:
First, draw a force diagram of the problem.

Setup diagram of Coulomb's Law Example Problem.

Two charges separated by one centimeter experiencing a force of repulsion of 90 N.

Define the variables:
F = 90 N
q1 = charge of first body
q2 = charge of second body
r = 1 cm

Use the Coulomb’s Law equation

The problem says the two charges are identical, so

q1 = q2 = q

Substitute this into the equation

Since we want the charges, solve for q

Enter the values for the variables. Remember to convert 1 cm to 0.01 meters to keep the units consistent.

q = ±1.00×10-6 Coulombs

Since the charges are identical, they are either both positive or both negative. This force will be repulsive.

Answer:
Two identical charges of ±1.00×10-6 Coulombs separated by 1 cm produce a repulsive force of 90 N.

For another Coulomb’s law example problem, check out Coulomb Force Example Problem.

Polyatomic Ions List

Polyatomic ions are ions that contain more than one element. This polyatomic ions list contains many common polyatomic ions grouped by charge. Each entry contains the ion’s name, molecular formula and chemical structure.

+1 Polyatomic Ions

 Ion FormulaStructure
AmmoniumNH4+Ammonium Cation Structure
HydroniumH3O+Hydronium Ion

-1 Polyatomic Ions

 Ion FormulaStructure
AcetateC2H3O2Acetate Anion
BicarbonateHCO3bicarbonate anion structure
BisulfateHSO4Bisulfate Anion Structure
ChlorateCO3Chlorate Anion Structure
ChloriteClO2Chlorite Anion Structure
CyanateOCNCyanate Anion
CyanideCNCyanide Anion Structure
Dihydrogen phosphateH2PO4Dihydrogen Phosphate Anion
Dihydrogen phosphiteH2PO3Dihydrogen Phosphite Anion
HydroxideOHHydroxide Anion
NitrateNO3Nitrate Anion
NitriteNO2Nitrite Anion
PerchlorateClO4Perchlorate Anion
PermanganateMnO4Permanganate Anion
ThiocyanateSCNThiocyanate Anion

-2 Polyatomic Ions

 Ion FormulaStructure
CarbonateCO32-Carbonate Anion
ChromateCrO42-Chromate Anion
DichromateCr2O72-Dichromate Anion
Hydrogen phosphateHPO42-Hydrogen Phosphate Anion
PeroxideO22-Peroxide Anion
SilicateSiO32-Silicate Anion
SulfateSO42-Sulfate Anion
SulfiteSO32-Sulfite Anion
ThiosulfateS2O32-Thiosulfate Anion

-3 Polyatomic Ions

 Ion FormulaStructure
PhosphatePO43-Phosphate Anion
PhosphitePO33-Phosphite Anion

Today In Science History – September 14 – Charles François de Cisternay du Fay

Charles François de Cisternay du Fay

Charles François de Cisternay du Fay (1698-1739)

September 14 is Charles François de Cisternay du Fay’s birthday. Du Fay was a French chemist who made early contributions to the study of electrical charge.

Du Fay discovered there were two types of electric charge. He named these charges after the methods he used to separate them. Vitreous (Latin for glass) electricity is generated when glass is rubbed with fur. Resinous (Latin for resin or amber) electricity is obtained from rubbing amber with silk or paper. He also noticed the different charges would attract each other, but the same kind of charge would repel each other. He developed a theory to explain this behavior where there existed two different fluids. Neutrally charged objects would have equal amounts of both fluids which would neutralize each other. When rubbed, an object would lose one of these fluids and leave an excess of the other.

Today we know these type of electrical ‘fluids’ are actually electrical charges where vitreous is positive charge and resinous is negative charge. These charges are caused when the rubbing forces a transfer of electrons.

Du Fay also noticed he could transfer charges by induction to metals and liquids. He also showed the electrical properties of an object that depended on color were caused by the dye that colored the object and not the color itself. He found glass was an insulator of electrical charge and thread conducts better when wet rather than dry. Most importantly, he discovered two bodies charged with vitreous or two bodies charged with resinous would repel each other and a vitreous charged body would attract a resinous charged body. We know this today as like charges repel, opposites attract.

Today In Science History – August 23 – Charles-Augustin de Coulomb

Coulomb

Charles-Augustin de Coulomb (1736-1806) Pioneer of the study of Electricity

August 23 marks the passing of Charles-Augustin de Coulomb. Coulomb was a French military engineer known for his work with static electricity.

Coulomb is best known for Coulomb’s law. This law is a statement of the force between two electric charges being inversely proportional to the square of the distance between them. and directly proportional to the amount of charge.

where Q1 and Q2 are the charges and r is the distance between them. If the two charges have the same sign (both positive or negative charge), the force is a repulsive force. If they have different signs (one positive and the other negative) the force is attractive.

Coulomb designed a device called a torsion balance to measure very small forces. The torsion balance consists of a bar suspended by a thin wire ribbon. The ribbon acts as a very weak spring that will twist when forces act on the ends of the bar. The greater the twist, the greater the force. Typically, this device is enclosed in an air tight container so air currents won’t move the bar.

Coulomb’s balance was an insulating rod with a metal coated ball on one end suspended by a silk ribbon. He charged the ball with a known amount of static electricity and brought another similarly charged ball close. He then measured the amount the suspended ball would move. He could then calculate the amount of force from the deflection. After multiple trials, he found the relationship that would become Coulomb’s Law.

His military career involved the design and construction of fortifications in France and the West Indies island of Martinique. After the French Revolution, he was appointed to the commission to develop the Metric System.

Coulomb’s contribution to the study of electricity and magnetism was honored in the new Metric System as the unit of electric charge. This measurement is still used today as the SI fundamental measurement of charge. One coulomb is the amount of charge transported by a constant current of one ampere in one second. This amount is approximately equal to the amount of charge on 6.241 x 1018 electrons.

Table of Valences of the Elements

A lithium atom has one outer shell electron, so it's usual valence is +1, but it can lose the electron and have a valence of -1.

A lithium atom has one outer shell electron, so it’s usual valence is +1, but it can lose the electron and have a valence of -1.

This is a table of the valences or oxidation states of the elements. The most common valences are in BOLD. Values in italics are predicted theoretical values.

This information is available on a Color Periodic Table of the Elements or a Black and White version. A simpler version listing only the most common valence charges is also available.

 NUMBER SYMBOLELEMENTVALENCE
1HHydrogen1, 0, -1
2HeHelium0
3LiLithium1, -1
4BeBeryllium2
5BBoron3, 2, 1
6CCarbon4, 3, 2, 1, -1, -2, -4
7NNitrogen5, 4, 3, 2, 1, 0, -1, -2, -3
8OOxygen2, 1, 0, -1, -2
9FFluorine0, -1
10NeNeon0
11NaSodium1, -1
12MgMagnesium2
13AlAluminum3, 1
14SiSilicon4, 3, 2, 1, -1, -2, -4
15PPhosphorus5, 4, 3, 2, 1, 0, -1, -2, -3
16SSulfur6, 5, 4, 3, 2, 1, 0, -1, -2
17ClChlorine6, 5, 4, 3, 2, 1, 0, -1, -2
18ArArgon0
19KPotassium1, -1
20CaCalcium2
21ScScandium3, 2, 1
22TiTitanium4, 3, 2, 0, -1, -2
23VVanadium5, 4, 3, 2, 1, 0, -1, -2
24CrChromium6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4
25MnManganese7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3
26FeIron6, 5, 4, 3, 2, 1, 0, -1, -2
27CoCobalt5, 4, 3, 2, 1, 0, -1
28NiNickel6, 4, 3, 2, 1, 0, -1
29CuCopper4, 3, 2, 1, 0
30ZnZinc2, 1, 0
31GaGallium3, 2, 1
32GeGermanium4, 3, 2, 1
33AsArsenic5, 3, 2, -3
34SeSelenium6, 4, 2, 1, -2
35BrBromine7, 5, 4, 3, 1, 0, -1
36KrKrypton2, 0
37RbRubidium1, -1
38SrStrontium2
39YYttrium3, 2
40ZrZirconium4, 3, 2, 1, 0, -2
41NbNiobium5, 4, 3, 2, 1, 0, -1, -3
42MoMolybdenum6, 5, 4, 3, 2, 1, 0, -1, -2
43TcTechnetium7, 6, 5, 4, 3, 2, 1, 0, -1, -3
44RuRuthenium8, 7, 6, 5, 4, 3, 2, 1, 0, -2
45RhRhodium6, 5, 4, 3, 2, 1, 0, -1
46PdPalladium4, 2, 0
47AgSilver3, 2, 1, 0
48CdCadmium2, 1
49InIndium3, 2, 1
50SnTin4, 2, -4
51SbAntimony5, 3, -3
52TeTellurium6, 5, 4, 2, 1, -2
53IIodine7, 5, 3, 1, 0, -1
54XeXenon8, 6, 4, 3, 2, 0
55CsCesium1, -1
56BaBarium2
57LaLanthanum3, 2
58CeCerium4, 3, 2
59PrPraseodymium4, 3, 2
60NdNeodymium4, 3, 2
61PmPromethium3
62SmSamarium3, 2
63EuEuropium3, 2
64GdGadolinium3, 2, 1
65TbTerbium4, 3, 1
66DyDysprosium4, 3, 2
67HoHolmium3, 2
68ErErbium3
69TmThulium3, 2
70YbYtterbium3, 2
71LuLutetium3
72HfHafnium4, 3, 2, 1
73TaTantalum5, 4, 3, 2, 1, -1, -3
74WTungsten6, 5, 4, 3, 2, 1, 0, -1, -2, -4
75ReRhenium7, 6, 5, 4, 3, 2, 1, 0, -1, -3
76OsOsmium8, 7, 6, 5, 4, 3, 2, 1, 0, -2
77IrIridium6, 5, 4, 3, 2, 1, 0, -1
78PtPlatinum6, 5, 4, 2, 0
79AuGold7, 5, 3, 2, 1, 0, -1
80HgMercury2, 1
81TlThallium3, 1
82PbLead4, 2
83BiBismuth5, 3, 1, -3
84PoPolonium6, 4, 2, -2
85AtAstatine7, 5, 3, 1, -1
86RnRadon2, 0
87FrFrancium1
88RaRadium2
89AcActinium3
90ThThorium4, 3, 2
91PaProtactinium5, 4, 3
92UUranium6, 5, 4, 3, 2
93NpNeptunium7, 6, 5, 4, 3, 2
94PuPlutonium7, 6, 5, 4, 3, 2
95AmAmericium7, 6, 5, 4, 3, 2
96CmCurium6, 5, 4, 3, 2
97BkBerkelium4, 3, 2
98CfCalifornium5, 4, 3, 2
99EsEinsteinium4, 3, 2
100FmFermium4, 3, 2
101MdMendelevium3, 2, 1
102NoNobelium3, 2
103LrLawrencium3, 2
104RfRutherfordium4, 3
105DbDubnium5, 4
106SgSeaborgium6, 5, 4
107BhBohrium7, 6, 5, 4, 3
108HsHassium8, 7, 4, 3, 2
109MtMeitnerium6, 5, 4, 3, 2, 1
110DsDarmstadtium6, 5, 4, 3, 2, 1
111RgRoentgenium3, -1
112CnCopernicium2, 1
113NhNihonium1
114FlFlerovium2
115McMoscovium3, 1
116LvLivermorium4, 2
117TsTennessineunknown
118OgOganesson8, 6, 4, 2