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Wright CHM 1210 - Ionic Compounds Lecture 2 - Forming Ionic Compounds

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Ionic Compounds 1Preview The lectures in this unit cover the formation of ions and ionic compounds, properties of ions and nomenclature of ionic compounds. This lecture covers the formation of ionic compounds, the nature of the ionic bond, and properties of ionic compounds. Ionic Compounds I: Nature of the Ionic Bond Ions do not form in a vacuum. If they gain an electron to form a negatively charged ion they have to gain it from somewhere. And if they lose an electron to form a positively charged ion it has to go somewhere. Because of this it is common to see reactions where metals which like to lose electrons and nonmetals which like to gain electrons react together to form ionic compounds by exchanging electrons. For this reason, every time we see a compound formed with a METAL and a NONMETAL we know it is an ionic compound. Please note: molecules with metals and polyatomic ions (which you should be able to recognize at a glance!) are also ionic compounds. For instance, elemental sodium reacts with elemental chlorine to form the sodium chloride ionic compound by transferring one electron from each sodium atom to each chlorine atom. We write both parts of the reaction as follows: Na  Na+ + 1 e- Cl + e-  Cl- In order to transfer that electron the sodium and chlorine must come in contact with one another. And the moment the electron is transferred they are still in contact with one another. At this moment they are attracted to each other by their newly formed charges because opposite charges attract. They are stuck together! Ionic compounds have a unique kind of a bond in that they are held together only by this mutual attraction. If anything comes by which overcomes the attraction of the individual ions for each other the bond breaks and the ions separate. This is the case when an ionic compound dissolves in water. The water molecules, for reasons we discuss in just a few lectures, are attracted to the individual ions and overcome the mutual ionic attractions. So every time an ionic compound dissolves in water the ions separate. We call this dissociation and express it with the following reaction: NaCl(s) + water  Na+(aq) + Cl-(aq) The little (s) means the NaCl is a solid and the little (aq) means the Na+ and Cl- ions are surrounded by or dissolved in water. See how each of those ions retains its charge? Sodium does not want its electron back, and even if it did chlorine is not about to give it up. They are both more stable as ions than they were as elements.Ionic Compounds 2Solutions with ions dissolved in them are called electrolytes. This name comes from a unique property that solutions with dissolved ions have: these solutions conduct electricity. If we heat up this solution the water boils it off and the salt is left behind. As this happens it re-forms its ionic (attractive) bond to re-create NaCl(s). II. Forming Ionic Compounds Because the nature of the ionic bond depends solely on charge attraction, ions with multiple charges may end up attracting multiple atoms. Whenever an ionic compound forms the overall charge must be neutral, or zero. That is to say the positive charge contribution from the cations must be equal to the negative charge contribution from the anions. To find the formula for an ionic compound, ask the following three questions: 1 – What is the charge on the cation? 2 – What is the charge on the anion? 3 – What is the lowest possible ratio of cations:anions to make a formula with a net charge of 0? Once you have the answer to question #3, make the formula using the symbol for the cation followed by the number of cations required in subscript, and the symbol of the anion followed by the number of anions required in subscript. For example, when we form a compound between Na+ and Cl- the +1 charge from the Na+ is exactly equal to and opposite the -1 charge from the Cl-, so the proper ratio of Na+ ions to Cl- ions is 1:1 in order to have a neutral compound. The correct formula for this compound is NaCl. If instead we form a compound with Na+ ions and O2- ions, we need two Na+ ions to balance out the total -2 charge from the O2- ion. The correct formula for this compound is Na2O. Here weIonic Compounds 3use a subscript 2 to indicate the number of sodium ions in the formula. The subscript of 1 is omitted for oxygen: we do not show a subscript if there is only one ion in the formula. If we make a compound between Mg2+ and Cl-, we need two Cl- ions to balance out the total +2 charges from the Mg2+ ion. The correct formula for this compound is MgCl2. Please note that in our formulas for ionic compounds we do not ever show the charge of the ion: only the number of each ion present in the total formula. One quick method for determining the proper ratio of cations to anions is to take the charge value of each ion and apply that number as a subscript to the other ion. This method helps us with more complicated charge ratios, although the final formula must contain the lowest ratio of cations to anions. Consider forming compounds between Ca2+ and N3-; or between Ti4+ and O2-: Following this process always gives a charge balance: 3 Ca2+ ions have a total charge contribution of 6+ while 2 N3- ions have a total charge contribution of 6-. However, care must be taken that the final formula has the lowest whole number ratio of ions. Using this method with Ti4+ and O2- gives a formula which does not have the lowest whole number ratio of ions. Reducing the formula gives the proper answer: TiO2. Notice that the charges are still balanced: 1 Ti4+ ion has a total charge contribution of 4+, while 2 O2- ions have a total charge contribution of 4-. It is balanced! Determining the correct number and type of each ion in the balanced formula of an ionic compound can be done with more complex ionic compounds too. Ionic compounds formed with polyatomic ions follow the same rules. Just remember: the atoms of a polyatomic ion come in a bunch with the charge attached. If you want more than one of that ion you need to make sure it is clear that you have more than one of the whole ion. In these cases we use parenthesesIonic Compounds 4to help us keep track of how many atoms and how many ions are in the ionic formula. The following are a few examples of forming ionic compound with polyatomic ions. To make a formula for the compound formed between the Mg2+ ion and the sulfate polyatomic ion (SO42-) we


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