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7.1: Ionic Compounds

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Lesson Objectives

  • Distinguish between inorganic and organic chemistry.
  • Interpret a molecular formula.
  • Explain why an ionic compound is represented by an empirical formula.
  • Be able to determine the charges of monatomic ions formed by the representative elements from the position of each element on the periodic table.
  • Use the Stock system to identify the charge of transition metal ions.
  • Name an ionic compound given its formula.
  • Write the correct formula for an ionic compound given its name.

Lesson Vocabulary

  • binary ionic compound
  • empirical formula
  • inorganic chemistry
  • molecular formula
  • monatomic ion
  • organic chemistry
  • polyatomic ion
  • ternary ionic compound

Check Your Understanding

Recalling Prior Knowledge

  • Which subatomic particles are lost or gained in order to form ions?
  • The elements of which group on the periodic table are particularly stable and unreactive?
  • Which groups contain the representative elements and which groups contain the transition elements?

Chemistry can be broadly divided into two main classes based on the identity of the elements present in the chemical compounds. Organic chemistry is the branch of chemistry that deals with compounds containing carbon. Because of the overwhelmingly large number of organic molecules that exist, the rules for naming organic compounds are complex and will be covered in a later chapter. In this chapter, we will be discussing inorganic chemistry, which is the branch of chemistry dealing with compounds that do not contain carbon. Note that there are a few exceptions of small molecules that contain carbon but are not subject to the rules for naming organic compounds. We will begin our discussion with an examination of two types of chemical formulas, and the remainder of the lesson will focus on converting back and forth between names and formulas for ionic compounds.

Types of Formulas

Recall that a molecule includes two or more atoms that have been chemically combined. A chemical formula that indicates how many of each type of atom are present in a single molecule is referred to as a molecular formula. For example, a molecule of ammonia contains one nitrogen atom and three hydrogen atoms, so it has the following molecular formula:

Another type of chemical formula, the empirical formula, shows the elements in a compound in their lowest whole-number ratio. Glucose is an important simple sugar that cells use as their primary source of energy. Its molecular formula is C6H12O6. Since each of the subscripts is divisible by 6, the empirical formula for glucose is CH2O. When chemists analyze an unknown compound, often the first step is to determine its empirical formula, as we will see in a later chapter. There are a great many compounds whose molecular and empirical formulas are the same. If the molecular formula cannot be simplified into a smaller whole-number ratio, as in the case of H2O or P2O5, then the empirical formula is also the molecular formula.

Ionic compounds are quite different from molecular compounds such as water. Water and other molecular compounds exist as collections of individual molecules (Figure below).

A water molecule consists of one atom of oxygen bonded to two atoms of hydrogen.

Ionic compounds do not exist as discrete molecular units. Instead, an ionic compound consists of a large three-dimensional array of alternating cations and anions. For example, sodium chloride (NaCl) is composed of Na+ and Cl- ions arranged into a structure like the one shown in Figure below.

A crystal of table salt, sodium chloride, is a large array of alternating positive and negative ions. The purple spheres represent the Na+ ions, while the green spheres represent the Cl- ions.

Watch a video about The Structure of Ionic Solids: http://www.youtube.com/watch?v=TLPY9Z6z4Mg (1:35) .

The most straightforward way to describe this structure with a chemical formula is to give the lowest whole-number ratio between the two ions. The formulas for ionic compounds are always empirical formulas. In the case of NaCl, there are equal numbers of sodium ions and chloride ions in the salt crystal. In contrast, a crystal of magnesium chloride has twice as many chloride ions as magnesium ions, so it has a formula of MgCl2.

Monatomic Ions

Many ionic compound consist of a positively charged metal cation and a negatively charged nonmetal anion that are combined in whatever ratio will produce a compound that is electrically neutral. In other words, the total positive charge is exactly canceled out by the total negative charge. In order to accurately write formulas for ionic compounds, it is necessary to know the charges of the ions involved. We will begin our study with monatomic ions, which are ions that consist of a single atom with either a positive or negative charge.

Cations

Cations are positive ions that are formed when a metal atom loses one or more electrons. For the representative elements, cations are generally formed by removing all of the valence electrons from the atom. Since the numbers of valence electrons for the representative elements are constant within a particular group, all we need is the group number of a given element to know its charge when it becomes a cation. Group 1 elements form ions with a 1+ charge, Group 2 metal ions have a 2+ charge, and the ions of Group 13 elements tend to have a 3+ charge. Heavier p-block metals such as tin and lead are special cases and will be discussed with the transition metal ions. Monatomic cations have the same name as their parent element. For example, Na+ is the sodium ion, while Al3+ is an aluminum ion.

Anions

Monatomic anions are negative ions that are formed when a nonmetal atom gains one or more electrons. Nonmetallic atoms typically gain enough electrons to obtain the same electron configuration as the nearest noble gas. All the elements in Group 17 have seven valence electrons, which are arranged into a outer configuration of ns2np5. To achieve a noble gas configuration (ns2np6), each of these elements needs to gain just one electron, resulting in an anion with a 1− charge. Similarly, Group 16 elements can obtain an ns2np6 valence configuration by forming ions with a 2− charge, and the Group 15 nonmetals will form ions with a 3− charge. Naming anions is slightly different than naming cations. The ending of the element’s name is dropped and replaced with the –ide suffix. For example, F- is the fluoride ion, while O2- is the oxide ion. Table below shows the names and charges for common monatomic ions of the representative elements.

Common Monatomic Ions
1+ 2+ 3+ 3- 2- 1-
lithium, Li+ beryllium, Be2+ aluminum, Al3+ nitride, N3- oxide, O2- fluoride, F-
sodium, Na+ magnesium, Mg2+ gallium, Ga3+ phosphide, P3- sulfide, S2- chloride, Cl-
potassium, K+ calcium, Ca2+ arsenide, As3- selenide, Se2- bromide, Br-
rubidium, Rb+ strontium, Sr2+ telluride, Te2- iodide, I-
cesium, Cs+ barium, Ba2+

Transition Metal Ions

Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one type of stable cation. For example, iron sometimes loses two electrons to form the Fe2+ ion, but it is also common for iron to lose three electrons to form the Fe3+ ion. Although they are members of the p block and not the d block, tin and lead also form more than one type of ion. Because the charges of these ions cannot be unambiguously determined by looking at the periodic table, they must have names that also indicate their charge. The Stock system denotes the charge of these ions by using a Roman numeral in parentheses after the name of the ion. For example, the previously mentioned iron ions are named the iron(II) ion and the iron(III) ion. When said out loud, "iron(II) ion" should be read, “iron two ion.” Table below lists the names and formulas of some of the more common transition metal ions.

Common Transition Metal Ions
1+ 2+ 3+ 4+
copper(I), Cu+ cadmium, Cd2+ chromium(III), Cr3+ lead(IV), Pb4+
gold(I), Au+ chromium(II), Cr2+ cobalt(III), Co3+ tin(IV), Sn4+
mercury(I), Hg22+ cobalt(II), Co2+ gold(III), Au3+
silver, Ag+ copper(II), Cu2+ iron(III), Fe3+
iron(II), Fe2+
lead(II), Pb2+
manganese(II), Mn2+
mercury(II), Hg2+
nickel(II), Ni2+
platinum(II), Pt2+
tin(II), Sn2+
zinc, Zn2+

Notice in Table above that there are three cations whose names do not include a Roman numeral. Silver, cadmium, and zinc only form one common type of ion, so the charges on ions of these elements are considered to be implied by the name (1+ for silver, and 2+ for zinc and cadmium). By convention, the Stock system is not used for these elements, and their cations are named in the same way as those of the representative elements. The mercury(I) ion is a special case that consists of a pair of mercury atoms bonded together. The overall ion has a charge of 2+, so each mercury ion can be thought of as carrying a 1+ charge. In some instances, charges greater than 4+ are seen, but because these are less common, they are not included in the table.

There is an older system for naming some of these cations that is still occasionally used. The Latin root of the metal name is written with one of two suffixes: (1) –ic for the ion with a higher charge, and (2) –ous for the ion with a lower charge. For example, the Latin name for iron is ferrum, so the Fe3+ ion is called the ferric ion, and the Fe2+ ion is called the ferrous ion. The primary disadvantage of this system is that the suffixes do not tell you exactly what the charge is for a given ion. For copper, the two most common charges are 1+ and 2+, so Cu2+ is called the cupric ion and Cu+ is the cuprous ion. The Stock system is a much more informative system and will be used as the primary method for naming transition metal compounds throughout this book.

Binary Ionic Compounds

A binary ionic compound is a compound made by combining monatomic metal cations with monatomic nonmetal anions.

Naming Binary Ionic Compounds

When given the formula of a chemical compound, you must first decide what kind of compound it is before you can determine how it should be named. A binary ionic compound will contain one metallic element and one nonmetallic element. By convention, the metal is generally written before the nonmetal in the chemical formula for this type of compound. To name a binary ionic compound, simply write the name of the cation followed by the name of the anion. Note that subscripts in the formula do not affect the name. Table below shows three examples.

Examples of Binary Ionic Compounds
Formula Name
KF potassium fluoride
Na3N sodium nitride
Ca3P2 calcium phosphide

Notice that in each of the formulas above, the overall charge of the compound is zero. Since potassium ions (K+) and fluoride ions (F) have charges of equal magnitude, a neutral compound is formed when combining them in a 1:1 ratio, resulting in a formula that contains one of each ion. This would also be the case for a compound such as MgS, which is composed of Mg2+ and S2−. For sodium nitride, the sodium ion (Na+) has a charge of 1+, while the nitride ion (N3−) has a charge of 3-. In order to make a neutral compound, three of the 1+ sodium ions are needed to balance out each 3- nitride ion. This is reflected in the chemical formula by giving Na a subscript of 3. Calcium phosphide is composed of calcium ions (Ca2+) and phosphide ions (P3−). Determining the necessary ratio for ions whose charges both have magnitudes greater than 1 can sometimes be made easier by looking for the least common multiple of the two charges. In this case, the least common multiple of 2 and 3 is 6. To make the compound neutral, three calcium ions (with a total charge of 6+) should be combined with two phosphide ions (with a total charge of 6−). The Ca is given a subscript of 3, while the P is given a subscript of 2.

Practice naming binary ionic compounds with this short crossword puzzle: https://sites.google.com/site/pattihowellsciencespot/uploads/binary%20ionic%20compounds.html?attredirects=0&d=1.

Naming Compounds Using the Stock System

Naming compounds that involve transition metal cations requires the use of the Stock system. Consider the binary ionic compound FeCl3. The name "iron chloride" would be ambiguous, because iron is capable of forming two ions with different charges. The name of any iron-containing compound must reflect which type of iron ion is in the compound. In this case, the subscript in the formula indicates that there are three chloride ions, each with a 1− charge. Therefore, the charge of the single iron ion must be 3+. The correct name of FeCl3 is iron(III) chloride, where the Roman numeral indicates the charge of the cation. Table below includes a few other examples.

Examples of Naming Using the Stock System
Formula Name
Cu2O copper(I) oxide
CuO copper(II) oxide
SnO2 tin(IV) oxide

The first two are both oxides of copper (shown in Figure below). The ratio of copper ions to oxide ions determines the name. Since the oxide ion is O2−, the charges of the copper ion must be 1+ in the first formula and 2+ in the second formula. In the third formula, there is one tin ion for every two oxide ions. This means that the tin must carry a 4+ charge, making the name tin(IV) oxide.

Copper(I) oxide, a red solid, and copper(II) oxide, a black solid, are different compounds because of the charge of the copper ion.

You can practice naming multivalent metal compounds by completing this online crossword puzzle: https://sites.google.com/site/pattihowellsciencespot/uploads/multivalent%20metals%20compounds.html?attredirects=0&d=1.

Writing Formulas for Binary Ionic Compounds

If you know the name of a binary ionic compound, you can write its formula. Start by writing the metal ion and its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many ions of each type are needed in order for the positive and negative charges to cancel each other out. Consider the compound aluminum nitride. The charges on each of these ions can be determined by looking at the groups in which aluminum and nitrogen are found. The ions are:

\text{Al}^{3+} \ \ \ \ \text{N}^{3-}

Since the ions have charges that are equal in magnitude, 1:1 is the lowest ratio of ions that will produce a neutral compound. As a result, the formula of aluminum nitride is AlN. Another compound, lithium oxide, contains the following ions:

\text{Li}^+ \ \ \ \ \text{O}^{2-}

In this case, two lithium ions are required to balance out the charge of each oxide ion. The formula of lithium oxide is Li2O.

For compounds in which the ratio of ions is not as obvious, an alternative way to determine the correct formula is to use the crisscross method. In this method, the numerical value of each charge crosses over to become the subscript of the opposite ion. The signs of the charges are dropped. The crisscross method is demonstrated below for aluminum oxide.

The red arrows indicate that the 3 from the 3+ charge will cross over to become the subscript for O, while the 2 from the 2− charge will cross over to become the subscript for Al. The formula for aluminum oxide is Al2O3.

For aluminum oxide, the crisscross method directly produces the correct formula, but in some cases, another step is required. Because ionic compounds are always described by their empirical formulas, they must be written as the lowest whole-number ratio of the ions. In the case of aluminum nitride, the crisscross method would yield a formula of Al3N3, which is not correct. A second step must be performed in which the subscripts are reduced but the ratio is kept the same. Al3N3 can be reduced to AlN, because both formulas describe a 1:1 ratio of aluminum ions to nitride ions. Following the crisscross method to write the formula for lead(IV) oxide would involve the following steps:

The crisscross method first yields Pb2O4 for the formula, but that must be reduced to PbO2, which is the correct formula.

Click here to watch an animation of ionic bonding: http://www.dlt.ncssm.edu/core/Chapter9-Bonding_and_Geometry/Chapter9-Animations/IonicBonding.html.

Sample Problem 7.1: Writing Formulas for Binary Ionic Compounds

Write the correct formulas for the following ionic compounds:

  1. barium chloride
  2. chromium(III) oxide

Step 1: Plan the problem.

In each case, write the metal cation followed by the nonmetal anion. Crisscross the ion charges in order to make each ionic compound neutral. Reduce to the lowest ratio if necessary.

Step 2: Solutions

Step 3: Think about your result.

The formula for barium chloride is BaCl2. Notice that the charge of the chloride ion needs to be written as 1− rather than just – in order to do the crisscross. However, the number 1 is not used as a subscript. The formula for chromium(III) oxide is Cr2O3.

Practice Problems
  1. Write formulas for the binary ionic compounds formed by the following pairs of elements:
    1. cesium and fluorine
    2. calcium and sulfur
    3. aluminum and chlorine
    4. zinc and nitrogen
  2. Write the formula and give the name for the compound formed by each of the following ion pairs:
    1. Fe3+ and O2-
    2. Ni2+ and S2-
    3. Au+ and Cl-
    4. Sn4+ and I-
  3. Give names for the following compounds:
    1. Ag2S
    2. PdO
    3. PtCl4
    4. V2O5

Ternary Ionic Compounds

Not all ionic compounds are composed of only monatomic ions. A ternary ionic compound is an ionic compound composed of three or more elements. In a typical ternary ionic compound, there is still just one type of cation and one type of anion, but the cation or the anion (or sometimes both) is a polyatomic ion.

Polyatomic Ions

A polyatomic ion is an ion composed of more than one atom. The ammonium ion consists of one nitrogen atom and four hydrogen atoms. Together, they comprise a single ion with a 1+ charge and a formula of NH4+. The carbonate ion consists of one carbon atom and three oxygen atoms, and it carries an overall charge of 2−. The formula of the carbonate ion is CO32-. The atoms of a polyatomic ion are tightly bonded together, so the entire ion behaves as a single unit. Figure below shows several models, and Table below lists many of the most common polyatomic ions.

(A) The ammonium ion (NH4+) is a nitrogen atom (blue) bonded to four hydrogen atoms (white). (B) The hydroxide ion (OH-) is an oxygen atom (red) bonded to a hydrogen atom. (C) The carbonate ion (CO32-) is a carbon atom (black) bonded to three oxygen atoms.

Common Polyatomic Ions
1- 2- 3- 1+ 2+
acetate, CH3COO- carbonate, CO32- arsenate, AsO33- ammonium, NH4+ dimercury, Hg22+
bromate, BrO3- chromate, CrO42- phosphite, PO33-
chlorate, ClO3- dichromate, Cr2O72- phosphate, PO43-
chlorite, ClO2- hydrogen phosphate, HPO42-
cyanide, CN- oxalate, C2O42-
dihydrogen phosphate, H2PO4- peroxide, O22-
hydrogen carbonate, HCO3- silicate, SiO32-
hydrogen sulfate, HSO4- sulfate, SO42-
hydrogen sulfide, HS- sulfite, SO32-
hydroxide, OH-
hypochlorite, ClO-
nitrate, NO3-
nitrite, NO2-
perchlorate, ClO4-
permanganate, MnO4-

The vast majority of polyatomic ions are anions, many of which end in –ate or –ite. Notice that in some cases, such as nitrate (NO3-) and nitrite (NO2-), there are multiple anions that consist of the same two elements. This is particularly common for oxoanions, which are binary anions containing one or more oxygen atoms. A given element may form several oxoanions that all have the same charge but differ in the number of oxygen atoms present. When there are two common oxoanions for a particular element, the one with the greater number of oxygen atoms gets an –ate suffix, while the one with the lower number of oxygen atoms gets an –ite suffix. Some elements form more than two common oxoanions, such as chlorine:

  • ClO-, hypochlorite
  • ClO2-, chlorite
  • ClO3-, chlorate
  • ClO4-, perchlorate

For larger families of oxoanions, the ion with one more oxygen atom than the –ate anion is given a per- prefix, and the ion with one fewer oxygen atom than the –ite anion is given a hypo- prefix.

Naming Ternary Ionic Compounds

As in the case of binary ionic compounds, ternary ionic compounds can be named by writing the name of the cation followed by the name of the anion. Some examples are shown in Table below:

Examples of Ternary Ionic Compounds
Formula Name
NaNO3 sodium nitrate
NH4Cl ammmonium chloride
Fe(OH)3 iron(III) hydroxide

When more than one polyatomic ion is present in a compound, the formula of the entire ion is placed in parentheses, and a subscript outside of the parentheses indicates how many of those ions are in the compound. In the last example above, there is one Fe3+ cation present for every three OH- anions.

Writing Formulas for Ternary Ionic Compounds

Writing a formula for a ternary ionic compound also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Use the crisscross method to ensure that the final formula is neutral. Calcium nitrate is composed of calcium cations and nitrate anions.

The charge is balanced by the presence of two nitrate ions and one calcium ion. Parentheses are used around the nitrate ion because more than one of the polyatomic ion is needed. If only one polyatomic ion is present in a formula, parentheses are not used. For example, the formula for calcium carbonate is CaCO3. The carbonate ion carries a 2− charge, so it exactly balances the 2+ charge of the calcium ion.

Note that there are two polyatomic ions that produce unusual formulas. The Hg22+ ion is called either the dimercury ion or, preferably, the mercury(I) ion. When combined with an anion that has a 1− charge, such as chloride, the formula is Hg2Cl2. Because the cation consists of two Hg atoms bonded together, this formula is not reduced to HgCl. Likewise, the peroxide ion, O22-, is also a unit that must stay together in its formulas. For example, the formula for potassium peroxide is K2O2.

Sample Problem 7.2: Writing Formulas for Ternary Ionic Compounds

Write the correct formulas for the following ionic compounds:

  1. potassium sulfate
  2. zinc phosphate

Step 1: Plan the problem.

In each case, write the metal cation followed by the nonmetal anion. Crisscross the ion charges in order to make the ionic compound neutral. Use parentheses around the polyatomic ion if more than one is present in the final formula. Reduce to the lowest ratio if necessary.

Step 2: Solutions

Step 3: Think about your result.

The formula for potassium sulfate is K2SO4. Two potassium cations with 1+ charges balance out the 2− charge of each sulfate ion. The formula for zinc phosphate is Zn3(PO4)2. Three zinc cations with 2+ charges balance out two phosphate anions with 3− charges.

Practice Problems
  1. Name the following compounds:
    1. NH4NO3
    2. Na2Cr2O7
    3. PbCO3
    4. Mg(CH3COO)2
  2. Write formulas for the following compounds:
    1. potassium hydrogen sulfate
    2. iron(III) oxalate
    3. sodium peroxide
    4. tin(IV) chromate

Watch a humorous video lecture about Naming Ionic Compounds from BossChemDude: http://www.youtube.com/watch?v=q2s8hQ5NIpE (8:30)

Lesson Summary

  • Inorganic chemistry is the study of chemical compounds that do not contain carbon.
  • Molecular formulas show the type and number of atoms that occur in a molecule. Empirical formulas show the atoms or ions that make up a compound in their lowest whole-number ratio. Because ionic compounds have an extended three-dimensional structure, only empirical formulas should be used to describe them.
  • Atoms of the representative elements generally form monatomic ions by losing or gaining enough electrons to attain the electron configuration of the nearest noble gas.
  • Most transition metals are capable of forming multiple cations with different charges. When the charge of an ion cannot be determined by the identity of the element, Roman numerals are used in the name of the ion. Specifying the charge of an ion in this way is referred to as the Stock system.
  • Ionic compounds are named by writing the name of the cation followed by the name of the anion.
  • Ionic compounds are electrically neutral, so the total amount of positive charge must balance the total amount of negative charge. The crisscross method can be used to determine the correct formulas for ionic compounds.
  • Some ionic compounds contain polyatomic ions. A set of parentheses is used when more than one polyatomic ion is present in a formula.

Lesson Review Questions

Reviewing Concepts

  1. What element is present in all organic compounds?
  2. Write the molecular formula of a compound whose molecules contain one atom of nitrogen and three atoms of fluorine.
  3. A Roman numeral is not used in the names of compounds containing which three transition elements?
  4. Give an example of each of the following: (1) a monatomic cation, (2) a monatomic anion, (3) a polyatomic cation, and (4) a polyatomic anion.
  5. What is the overall charge of an ionic compound?
  6. When are parentheses used in the formula of an ionic compound?

Problems

  1. Write the empirical formula for each of the following compounds:
    1. C2H6
    2. Hg2Cl2
    3. Sb2O5
    4. C8H16O2
  2. Using only the periodic table, write the symbol of the common ion formed by each of the following elements:
    1. Sr
    2. I
    3. Se
    4. Ba
    5. P
    6. Rb
    7. Al
    8. Br
  3. Name the following monatomic ions:
    1. O2-
    2. Li+
    3. W3+
    4. Cu2+
    5. Ga3+
    6. F-
  4. Write formulas for binary ionic compounds formed from each of the following pairs of elements:
    1. potassium and sulfur
    2. silver and chlorine
    3. calcium and oxygen
    4. aluminum and iodine
    5. barium and nitrogen
    6. sodium and selenium
  5. Name the following polyatomic ions:
    1. SO32-
    2. MnO4-
    3. CO32-
    4. ClO-
    5. CH3COO-
    6. H2PO4-
  6. Write the symbol and charge for the following polyatomic ions:
    1. hydrogen sulfide ion
    2. nitrate ion
    3. perchlorate ion
    4. chromate ion
    5. phosphate ion
    6. oxalate ion
  7. Name the following ionic compounds:
    1. KClO3
    2. Cd(NO3)2
    3. CuCl
    4. Ca3(PO4)2
    5. NaCN
    6. (NH4)2SO3
    7. Li2O2
    8. PbS2
    9. Rb2SO4
    10. Mn3P2
    11. NiCO3
    12. Co(OH)3
    13. Hg2Br2
    14. Zn(NO2)2
  8. Write correct formulas for the following ionic compounds:
    1. copper(II) bromide
    2. aluminum hydroxide
    3. silver sulfide
    4. barium acetate
    5. mercury(II) nitrate
    6. lead(II) chromate
    7. potassium permanganate
    8. sodium hydrogen carbonate
    9. calcium silicate
    10. tin(IV) sulfate
    11. ammonium phosphate
    12. gold(III) fluoride
    13. magnesium bromate
    14. chromium(VI) oxide

Further Reading / Supplemental Links

Points to Consider

The large class of inorganic chemical compounds can be sub-classified into ionic compounds and molecular compounds.

  • How is the structure of a molecular compound different from the structure of an ionic compound?
  • Relate molecular compounds to the Law of Multiple Proportions.
  • What is the nomenclature for molecular compounds?

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