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40        <a href="#quotes" class="w3-bar-item w3-button">Quotes</a>
41        <a href="#welcome" class="w3-bar-item w3-button">Welcome</a>
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43            target="_blank" class="w3-bar-item w3-button">Videos</a>
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46        <a id="expandButton1" onclick="myFunc('topic1')" href="javascript:void(0)"
47            class="w3-bar-item w3-button">Balancing Chemical Reactions (3)</a>
48        <div id="topic1" class="w3-hide w3-animate-left">
49            <a href="#inspectionMethod" class="w3-bar-item w3-button w3-border-bottom">
50                <div class="w3-container">
51                    <h5>Inspection Method</h5>
52                </div>
53            </a>
54            <a href="#algebraicMethod" class="w3-bar-item w3-button w3-border-bottom">
55                <div class="w3-container">
56                    <h5>Algebraic Method</h5>
57                </div>
58            </a>
59            <a href="https://samuelchukwuemeka.github.io/chemical-reactions/chemicalReactions.html" target="_blank"
60                class="w3-bar-item w3-button w3-border-bottom">
61                <div class="w3-container">
62                    <h5>Solved Examples: Solve with Algebraic Method, Check with Inspection Method</h5>
63                </div>
64            </a>
65            <a href="#calculatorChemicalReactions" class="w3-bar-item w3-button w3-border-bottom">
66                <div class="w3-container">
67                    <h5>Calculator: Given the Products of a Chemical Reaction; Balance Chemical Reactions</h5>
68                </div>
69            </a>
70        </div>
71        <a href="#combinationReaction" class="w3-bar-item w3-button">Combination Reaction</a>
72        <a href="#decompositionReaction" class="w3-bar-item w3-button">Decomposition Reaction</a>
73        <a href="#displacementReaction" class="w3-bar-item w3-button">Displacement Reaction</a>
74        <a id="expandButton2" onclick="myFunc('topic2')" href="javascript:void(0)" class="w3-bar-item w3-button">Redox
75            Reactions (6)</a>
76        <div id="topic2" class="w3-hide w3-animate-left">
77            <a href="#oxidationNumber" class="w3-bar-item w3-button w3-border-bottom">
78                <div class="w3-container">
79                    <h5>Oxidation Number</h5>
80                </div>
81            </a>
82            <a href="#redoxOxidationNumber" class="w3-bar-item w3-button w3-border-bottom">
83                <div class="w3-container">
84                    <h5>Redox Reactions by Oxidation Number</h5>
85                </div>
86            </a>
87            <a href="#redoxOxygen" class="w3-bar-item w3-button w3-border-bottom">
88                <div class="w3-container">
89                    <h5>Redox Reactions by Addition of Oxygen</h5>
90                </div>
91            </a>
92            <a href="#redoxHydrogen" class="w3-bar-item w3-button w3-border-bottom">
93                <div class="w3-container">
94                    <h5>Redox Reactions by Removal of Hydrogen</h5>
95                </div>
96            </a>
97            <a href="#redoxBalance" class="w3-bar-item w3-button w3-border-bottom">
98                <div class="w3-container">
99                    <h5>Balance Redox Reactions</h5>
100                </div>
101            </a>
102            <a href="https://samuelchukwuemeka.github.io/chemical-reactions/redoxReactions.html" target="_blank"
103                class="w3-bar-item w3-button w3-border-bottom">
104                <div class="w3-container">
105                    <h5>Solved Examples: Redox Reactions</h5>
106                </div>
107            </a>
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130                        If there is one prayer that you should <b>pray/sing</b> every day and every hour, it is the
131                        <a href="https://www.samdomforpeace.com/Theology/prayer.html" target="_blank">LORD's prayer (Our
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133                        It is the <b>most powerful prayer</b>.
134                        A <b>pure heart</b>, a <b>clean mind</b>, and a <b>clear conscience</b> is necessary for it.
135                    </q> - Samuel Dominic Chukwuemeka
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138                    <q>Love the LORD your GOD with all your heart, with all your soul, with all your mind, and
139                        with all your strength. Love your neighbor as yourself.
140                    </q> - Mark 12: 30 - 31
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142                <p><q>The Joy of the Teacher is the Success of the Students.</q> - Samuel Chukwuemeka</p>
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144        </header>
145
146        <h1>Chemical Reactions</h1>
147
148        <!-- Automatic Slideshow Images -->
149        <div class="mySlides w3-center">
150            <img src="images/chemical-reaction1.png" alt="Chemical Reaction 1"
151                title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970">
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181                <p>
182                    <span id="greetings"></span> <br>
183                    I greet you this day,
184                    
184<script>document.write(Date() + ".");</script>
184 <br>
185                    <b>First:</b> read the notes. <br>
186                    <b>Second:</b> view the videos. <br>
187                    <b>Third:</b> solve the questions/solved examples. <br>
188                    <b>Fourth:</b> check your solutions with my <b>thoroughly-explained</b> solutions. <br>
189                    <b>Fifth:</b> check your answers with the calculators as applicable. Many thanks to the Wolfram
190                    Alpha Widget developers. <br>
191                    Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may
192                    contact me. <br>
193                    If you are my student, please do not contact me here. Contact me via the school's system. Thank you
194                    for visiting.
195                </p>
196                <p><strong>Samuel Dominic Chukwuemeka</strong> (SamDom For Peace) <span
197                        style="font-size: medium">B.Eng., A.A.T, M.Ed., M.S</span></p>
198            </div>
199        </div>
200
201
202        <!-- Overview of Chemical Reactions -->
203        <!-- Objectives -->
204        <div class="w3-container" id="objectives">
205            <h2>Objectives</h2>
206            <p>Students will:</p>
207            <p>
208                (1.) Define a chemical reaction. <br>
209                (2.) Write the states of the reactants of a chemical reaction. <br>
210                (3.) Write the states of the products of a chemical reaction. <br>
211                (4.) Predict the products of a chemical reaction, given the reactants of the reaction. <br>
212                (5.) Balance chemical reactions using the Inspection Method. <br>
213                (6.) Balance chemical reactions using the Algebraic Method. <br>
214                (7.) List the types of chemical reactions. <br>
215                (8.) Discuss Combination (or Synthesis) reactions. <br>
216                (9.) Discuss Decomposition reactions. <br>
217                (10.) Discuss Thermal Decomposition reactions. <br>
218                (11.) Discuss Displacement (or Single Displacement or Single Replacement) reactions. <br>
219                (12.) Discuss Double Displacement (or Metathesis) reactions. <br>
220                (13.) Discuss Double Decomposition (or Precipitation) reactions. <br>
221                (14.) Discuss Catalytic reactions. <br>
222                (15.) Discuss Reversible reactions. <br>
223                (16.) Discuss Thermal Dissociation reactions. <br>
224                (17.) Discuss Oxidation-Reduction (or Redox) reactions. <br>
225                (18.) Discuss Combustion reactions.
226            </p>
227        </div>
228        <!-- End Overview of Chemical Reactions -->
229
230
231
232        <!-- Vocabulary Words -->
233        <div class="w3-container" id="vocabularyWords">
234            <h2>Vocabulary Words</h2>
235            <p>
236                Bring it to <b>English:</b> Prefixes and Meanings: <br>
237                mono = 1, di = 2, tri = 3, tetra = 4, penta = 5, hexa = 6, hepta = 7, octa = 8, nona = 9, deca = 10;
238                etc. <br>
239                This implies that $tri-oxo$ means $3$ oxygen atoms; $di-chromate$ means 2 chromium atoms; etc. <br>
240                solid, liquid, gas, aqueous
241            </p>
242        </div>
243        <!-- End Vocabulary Words -->
244
245
246        <!-- Balance Chemical Reactions - Algebraic Method -->
247        <div class="w3-container" id="algebraicMethod">
248            <h2>Algebraic Method for Balancing Chemical Reactions</h2>
249            <p>
250                To balance chemical reactions using the Algebraic Method: <br><br>
251                (1.) Make sure that the reactant(s) and product(s) are correct. <br><br>
252                (2.) Make sure the chemical formulas for the reactant(s) and product(s) are correct. <br><br>
253                (3.) Put variables in front of all the reactant(s). <br><br>
254                (4.) Put variables in front of every product besides one product. <br>
255                If there is one product, do not put a variable. <br>
256                If there are two products, put a variable in front of only one product. You can put a variable in
257                either one of them. It does not matter. <br>
258                If there are three products, put variables in front of any of the two products. <br>
259                If there are four products, put variables in front of any of the three products.
260            </p>
261            <p>
262                Let us do an example. <br>
263                Write a balanced reaction between tetraoxosulphate($VI$)acid with aluminium hydroxide to give water and
264                aluminium tetraoxosulphate($VI$) <br><br>
265                $
266                Valency\:\:of\:\:Al = 3 \\[3ex]
267                Valency\:\:of\:\:SO_4 = 2 \\[3ex]
268                \therefore Al_{2}(SO_4)_3 \\[3ex]
269                Valency\:\:of\:\:Al = 3 \\[3ex]
270                Valency\:\:of\:\:OH = 1 \\[3ex]
271                \therefore Al(OH)_3 \\[3ex]
272                Valency\:\:of\:\:H = 1 \\[3ex]
273                Valency\:\:of\:\:SO_4 = 2 \\[3ex]
274                \therefore H_{2}SO_4 \\[3ex]
275                Valency\:\:of\:\:H = 1 \\[3ex]
276                Valency\:\:of\:\:O = 2 \\[3ex]
277                \therefore H_{2}O \\[3ex]
278                $
279                The unbalanced reaction is: <br>
280                $H_{2}SO_4 + Al(OH)_3 \rightarrow Al_{2}(SO_4)_3 + H_{2}O$ <br><br>
281                <i>
282                    Teacher: So, you have an acid reacting with a base to give a salt and water <br>
283                    What do we call this type of reaction?
284                </i> <br><br>
285                $Acid + Base \rightarrow Salt + Water$ is a <b>Neutralization Reaction</b> <br><br>
286                Let us balance the reaction using the Algebraic method <br>
287                We have two reactants and two products <br>
288                We shall place two different variables in front of the two reactants and one variable in front of one
289                product.
290            </p>
291            <p>
292                $
293                mH_{2}SO_4 + nAl(OH)_3 \rightarrow pAl_{2}(SO_4)_3 + H_{2}O \\[3ex]
294                H = 2m + 3n ~~~~~\:\:\:\:\:\:\: H = 2 \\[3ex]
295                S = m ~~~~~~~~~~~~~~~~~\:\:\:\:\:\:\: S = 3p \\[3ex]
296                O = 4m + 3n ~~~~~\:\:\:\:\:\:\: O = 12p + 1 \\[3ex]
297                Al = n ~~~~~~~~~~~~~~~\:\:\:\:\:\:\: Al = 2p \\[3ex]
298                2m + 3n = 2 ...eqn.(1) \\[3ex]
299                m = 3p ...eqn.(2) \\[3ex]
300                4m + 3n = 12p + 1 ...eqn.(3) \\[3ex]
301                n = 2p ...eqn.(4) \\[3ex]
302                Substitute\:\:eqn.(2)\:\:and\:\:eqn.(4)\:\:in\:\:eqn.(1) \\[3ex]
303                Substitute\:\:3p\:\:for\:\:m\:\:and\:\:2p\:\:for\:\:n\:\:in\:\:eqn.(1) \\[3ex]
304                From\:\:eqn.(1)\:\: 2(3p) + 3(2p) = 2 \\[3ex]
305                6p + 6p = 2 \\[3ex]
306                12p = 2 \\[3ex]
307                p = \dfrac{2}{12} \\[5ex]
308                p = \dfrac{1}{6}
308 \\[5ex]
309                Substitute\:\:the\:\:value\:\:of\:\:p\:\:in\:\:eqn.(2) \\[3ex]
310                From\:\:eqn.(2)\:\: m = 3\left(\dfrac{1}{6}\right) \\[5ex]
311                m = \dfrac{1}{2} \\[5ex]
312                Substitute\:\:the\:\:value\:\:of\:\:p\:\:in\:\:eqn.(4) \\[3ex]
313                From\:\:eqn.(4)\:\: n = 2\left(\dfrac{1}{6}\right) \\[5ex]
314                m = \dfrac{1}{3} \\[5ex]
315                Substitute\:\:the\:\:values\:\:back\:\:in\:\:the\:\:equation \\[3ex]
316                \dfrac{1}{2}H_{2}SO_4 + \dfrac{1}{3}Al(OH)_3 \rightarrow \dfrac{1}{6}Al_{2}(SO_4)_3 + H_{2}O \\[5ex]
317                LCD = 6 \\[3ex]
318                Multiply\:\:each\:\:reactant\:\:and\:\:each\:\:product\:\:by\:\:the\:\:LCD \\[3ex]
319                6 * \dfrac{1}{2}H_{2}SO_4 + 6 * \dfrac{1}{3}Al(OH)_3 \rightarrow 6 * \dfrac{1}{6}Al_{2}(SO_4)_3 + 6 *
320                1H_{2}O \\[5ex]
321                \therefore 3H_{2}SO_4 + 2Al(OH)_3 \rightarrow Al_{2}(SO_4)_3 + 6H_{2}O \\[3ex]
322                $
323            </p>
324            <u>Check by Inspection Method</u> <br>
325            <table>
326                <tr>
327                    <td>
328                        $
329                        \underline{Reactants} \\[3ex]
330                        H = 3 * 2 + 2 * 3 = 6 + 6 = 12 \\[3ex]
331                        S = 3 \\[3ex]
332                        O = 3 * 4 + 2 * 3 = 12 + 6 = 18 \\[3ex]
333                        Al = 2
334                        $
335                    </td>
336                    <td>
337                        $
338                        \underline{Products} \\[3ex]
339                        H = 6 * 2 = 12 \\[3ex]
340                        S = 3 \\[3ex]
341                        O = 4 * 3 + 6 = 12 + 6 = 18 \\[3ex]
342                        Al = 2
343                        $
344                    </td>
345                </tr>
346            </table>
347            <p>
348                For more examples, please review: <a
349                    href="https://samuelchukwuemeka.github.io/chemical-reactions/chemicalReactions.html"
350                    target="_blank">Solved Examples on the Balancing of Chemical Reactions using the Algebraic
351                    Method</a>
352            </p>
353        </div>
354        <!-- End Balance Chemical Reactions - Algebraic Method -->
355
356        <br><br>
357
358        <!-- Calculator for Chemical Reactions -->
359        <div id="calculatorChemicalReactions" class="w3-container">
360            <h2>Calculator for Chemical Reactions</h2>
361            <p>
362                <u>This calculator will:</u> <br>
363                (1.) Give the <b>product(s)</b> of an <b>unbalanced</b> chemical reaction <b>when given the
364                    reactant(s)</b> <br>
365                (2.) <b>Balance</b> a chemical reaction <b>when given the reactant(s) and the product(s)</b> <br><br>
366
367
368                <u>To use the calculator:</u> <br>
369                (1.) If you want to determine the products of a chemical reaction: <br>
370                <b>1st:</b> Type <b>only the reactants</b> and the <b>arrow</b> in the big textbox <br>
371                The <b>arrow</b> is the <b>hyphen</b> <u>and</u> the <b>greater than</b> symbol on your keyboard:
372                <b>-></b> <br>
373                <b>2nd:</b> Copy and paste it in the small textbox (actual calculator) <br>
374                <i>
375                    You may also type directly in the small textbox if you wish. The big textbox is just for you to see
376                    everything
377                    you typed, before you copy and paste it into the small textbox. <br>
378                    The small textbox is the actual calculator.
379                </i> <br>
380                <b>3rd:</b> Click the <b>Submit</b> button <br>
381                <b>4th:</b> Review the <b>Sample reactions</b> in the output. <br>
382                You may need to click the <b>Show formulas</b> link on the RHS (Right Hand Side) to see the chemical
383                reaction(s). <br>
384                Those are the <b>likely products</b> of the chemical reaction. <br>
385                They are not balanced. So, you will need to copy any one you want to balance and read on to Number (2.)
386                <br><br>
387
388                (2.) If you want to balance any chemical reaction: <br>
389                <b>1st:</b> Type the <b>reactants</b>, then the <b>arrow</b>, then the <b>products</b> in the big
390                textbox <br>
391                The <b>arrow</b> is the <b>hyphen</b> <u>and</u> the <b>greater than</b> symbol on your keyboard:
392                <b>-></b> <br>
393                <b>2nd:</b> Copy and paste it in the small textbox (actual calculator) <br>
394                <i>
395                    You may also type directly in the small textbox if you wish. The big textbox is just for you to see
396                    everything
397                    you typed, before you copy and paste it into the small textbox. <br>
398                    The small textbox is the actual calculator.
399                </i> <br>
400                <b>3rd:</b> Click the <b>Submit</b> button <br>
401                <b>4th:</b> Review the balanced chemical reaction in the output. <br><br>
402
403                <b>Please review examples below</b>
404            </p>
405
406            <ul class="w3-ul">
407                <li class="yellow w3-hover-black w3-xlarge w3-padding-32 w3-center">See Examples (based on these solved
408                    examples: <a href="https://samuelchukwuemeka.github.io/chemical-reactions/chemicalReactions.html"
409                        target="_blank">Balance Chemical Reactions</a>)</li>
410                <li class="w3-padding-16">To determine the products of a reaction, Type: $SO_2 + O_2 \rightarrow$
411                    <u>as</u> SO2 + O2 ->
412                </li>
413                <li class="w3-padding-16">To balance the reaction, Type: $SO_2 + O_2 \rightarrow SO_3$ <u>as</u> SO2 +
414                    O2 -> SO3</li>
415                <li class="w3-padding-16">To determine, Type: $C_4H_{10} + O_2 \rightarrow$ <u>as</u> C4H10 + O2 -></li>
416                <li class="w3-padding-16">To balance, Type: $C_4H_{10} + O_2 \rightarrow CO_2 + H_2O$ <u>as</u> C4H10 +
417                    O2 -> CO2 + H2O</li>
418                <li class="w3-padding-16">To determine, Type: $Cu + HNO_3 \rightarrow$ <u>as</u> Cu + HNO3 -></li>
419                <li class="w3-padding-16">To balance, Type: $Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO + H_2O$ <u>as</u> Cu
420                    + HNO3 -> Cu(NO3)2 + NO + H2O</li>
421                <li class="w3-padding-16">To determine, Type: $Fe(OH)_3 + H_2SO_4 \rightarrow$ <u>as</u> Fe(OH)3 + H2SO4
422                    -></li>
423                <li class="w3-padding-16">To balance, Type: $Fe(OH)_3 + H_2SO_4 \rightarrow Fe_2(SO_4)_3 + H_2O$
424                    <u>as</u> Fe(OH)3 + H2SO4 -> Fe2(SO4)3 + H2O
425                </li>
426                <li></li>
427            </ul>
428
429            <p class="w3-center">
430                Type: <label for="calculatorChemicalEquations"></label><input type="text" class="solve"
431                    name="calculatorChemicalEquations" id="calculatorChemicalEquations" autocomplete="off" required>
432            </p>
433            
433<script type="text/javascript" id="WolframAlphaScriptb5f30e9491f4acaf9c0ae7274da803a8"
434                src="//www.wolframalpha.com/widget/widget.jsp?id=b5f30e9491f4acaf9c0ae7274da803a8&theme=red&width=700&height=500"></script>
434
435        </div>
436        <!-- End Calculator for Chemical Reactions -->
437
438        <br><br><br><br>
439
440        <!-- Oxidation Number -->
441        <div id="oxidationNumber" class="w3-container">
442            <h2>Oxidation Number</h2>
443            <p> <b>Oxidation number</b> is also known as <b>Oxidation state</b>. <br>
444                The oxidation number of an element is the number of electrons the element gains or loses in a chemical
445                combination with another element or compound. <br>
446                Generally; metals have positive oxidation numbers (because they donate electrons), and non-metals have
447                negative oxidation numbers (because they accept electrons). <br>
448                Be it as it may, some elements have several oxidation numbers. It just depends on the chemical formula.
449                <br>
450                <u>Recall:</u> When elements react, there is a transfer of electrons (electrovalent bonding), or sharing
451                of electrons (covalent bonding). <br>
452                When an element donates (loses) an electron or receives (gains) an electron, it becomes an ion. <br>
453                Metals are electron donors, while non-metals are electron acceptors. <br>
454                The oxidation number can be positive, zero, or negative. <br>
455                A substance can be an atom, element, compound, molecule, or an ion.
456            </p>
457
458            <h3>Why Study Oxidation Numbers?</h3>
459            <p>
460                Oxidation numbers helps us to: <br>
461                <mark>(1.)</mark> Name compounds according to the IUPAC (International Union of Pure and Applied
462                Chemistry) nomenclature. <br>
463                <mark>(2.)</mark> Balance reduction-oxidation (redox) reactions.
464            </p>
465
466            <h3>Rules for Assigning Oxidation Numbers</h3>
467            <p>
468                <mark>(1.)</mark> The oxidation number of any element in its free state (when it is not combined with
469                any substance) is 0. <br>
470                <u>Recall:</u> Some elements are monoatomic or polyatomic in their free states. <br>
471                <u>Example:</u> <br>
472                The oxidation numbers of monoatomic $Na$ (sodium), diatomic $O_2$ (oxygen),
473                triatomic $O_3$ (ozone), tetra-atomic $P_4$ (phosphorus), and octa-atomic $S_8$ (sulphur)
474                among others is 0.
475            </p>
476            <p>
477                <mark>(2.)</mark> The oxidation number of any compound is 0. <br>
478                This applies to binary, ternary, and tetranary compounds among others. <br>
479                <u>Recall:</u> A binary compound is a compound made up of two elements. A ternary compound is a compound
480                made up of three elements. A tetranary or quaternary compound is a compound made up of four elements.
481                <br>
482                <u>Example:</u> <br>
483                The oxidation numbers of the binary compound $CuCl_2$ (copper (II) chloride), ternary compound $KMnO_4$
484                (potassium tetra-oxo manganate (VII) - also known as potassium permanganate),
485                and the tetranary compound $NaHCO_3$ (sodium hydrogen tri-oxo carbonate (IV) - also known as baking
486                soda) is 0. <br>
487                <i>Do you wonder how I got the roman numerals: II, VII, IV? Do you know what they mean?</i> <br>
488                The oxidation number of <b>copper</b> in <b>copper (II)</b> chloride is <b>2</b>. <br>
489                The oxidation number of <b>manganese</b> in potassium tetra-oxo <b>manganate (VII)</b> is <b>7</b>. <br>
490                The oxidation number of <b>carbon</b> in sodium hydrogen tri-oxo <b>carbonate (IV)</b> is <b>4</b>. <br>
491                <i>Is it making a little sense?</i>
492                <i>As you can see, the oxidation number helps us in naming compounds according to the IUPAC naming
493                    system.</i>
494            </p>
495            <p>
496                <mark>(3.)</mark> The sum of the oxidation numbers of each individual element in a compound is equal to
497                the oxidation number of the compound. <br>
498                Because the oxidation number of any compound is 0, this implies that the sum of the oxidation numbers of
499                each element in a compound is 0. <br>
500                <u>Example:</u> <br>
501                The oxidation number of $CuCl_2 = 0$ <br>
502                This means that the sum of the oxidation numbers of copper and chlorine is zero. <br>
503                We already know from the IUPAC nomenclature that the oxidation number is +2. <br>
504                With that knowledge, we can find the oxidation number of chlorine in copper (II) chloride. <br>
505                Let the oxidation number of $Cl$ in $CuCl_2 = x$ <br>
506                $Cu = 2$ <br>
507                $Cl_2 = 2 * x = 2x$ <br>
508                $CuCl_2 = 0$ <br>
509                $\therefore 2 + 2x = 0$ <br>
510                $2x = 0 - 2$ <br>
511                $2x = -2$ <br>
512                $x = -1$ <br>
513                This means that the oxidation number of chlorine in copper (II) chloride is $-1$. <br>
514                <i>Do you see the usefulness of this $3^{rd}$ rule?</i>
515            </p>
516
517            <p>
518                <mark>(4.)</mark> The oxidation number of a monoatomic ion is the charge on the ion. <br>
519                <u>Recall:</u> A monoatomic ion is an ion formed from one atom. <br>
520                <u>Examples:</u> <br>
521                The oxidation number of aluminium, $Al$ in the aluminium ion, $Al^{3+}$ is $+3$. <br>
522                The oxidation number of selenium, $Se$ in the selenium ion (known as selenide), $Se^{2-}$ is $-2$.
523            </p>
524
525            <p>
526                <mark>(5.)</mark> The oxidation number of a polyatomic ion is the charge on the ion. <br>
527                <u>Recall:</u> A polyatomic ion is an ion formed from more than one atom. <br>
528                This implies that the sum of the oxidation numbers of each atom in the polyatomic ion is equal to the
529                charge on the ion.
530                <u>Example:</u> <br>
531                In this polyatomic ion, $Cr_2O_7^{2-}$ (Can you name the ion? &#x263A;&#x263A;&#x263A;); . <br>
532                The sum of the oxidation numbers of $Cr$ and $O$ is $-2$ <br>
533                Say the oxidation number of oxygen is $-2$, we can find the oxidation number of chromium <br>
534                Let the oxidation number of chromium = $x$ <br>
535                $Cr = 2 * x = 2x$ <br>
536                $O = 7 * -2 = -14$ <br>
537                $Cr_2O_7^{2-} = -2$ <br>
538                $\therefore 2x + (-14) = -2$ <br>
539                $2x - 14 = -2$ <br>
540                $2x = -2 + 14$ <br>
541                $2x = 12$ <br>
542                $x = +6$ <br>
543                This means that the oxidation number of chromium in the ion is $+6$ <br>
544                <i>Ask students to try and name the ion according to IUPAC nomenclature</i> <br>
545                The ion is the hepta-oxo dichromate(VI) ion.
546            </p>
547
548            <p>
549                <mark>(6.)</mark> The oxidation number of an alkali metal in any compound is always $+1$. <br>
550                <u>Recall:</u>
550 Alkali metals are the elements in Group 1A of the periodic table. <br>
551                They include: lithium $Li$, sodium $Na$, potassium $K$, rubidium $Rb$, cesium $Cs$, and Francium $Fr$.
552                <u>Example:</u> <br>
553                The oxidation number of sodium in any compound of sodium is always $+1$. <br>
554                So, the oxidation number of $Na$ in $NaHCO_3$ is $+1$.
555            </p>
556
557            <p>
558                <mark>(7.)</mark> The oxidation number of an alkaline earth metal in any compound is always $+2$. <br>
559                <u>Recall:</u> Alkaline earth metals are the elements in Group 2A of the periodic table. <br>
560                They include: beryllium $Be$, magnesium $Mg$, calcium $Ca$, strontium $Sr$, barium $Ba$, and Radium
561                $Ra$.
562                <u>Example:</u> <br>
563                The oxidation number of calcium in any compound of calcium is always $+2$. <br>
564                So, the oxidation number of $Ca$ in $Ca_3(PO_4)_2$ is $+2$.
565            </p>
566
567            <p>
568                <mark>(8.)</mark> The oxidation number of Fluorine in any compound is always $-1$. <br>
569                <u>Examples:</u> <br>
570                The oxidation number of $F$ in Bismuth Fluoride, $BiF_3$ is $-1$. <br>
571                The oxidation number of $F$ in Silicon Fluoride (Silicon tetrafluoride), $SiF_4$ is $-1$.
572            </p>
573
574
575
576            <p>
577                <mark>(9.)</mark> The oxidation number of hydrogen in any compound is usually $+1$, except in binary
578                metallic hydrides. <br>
579                The oxidation number of hydrogen in binary metallic hydrides is $-1$. <br>
580                <u>Recall:</u> A binary metallic hydride is a compound of a metal and hydrogen. <br>
581                <u>Examples:</u> <br>
582                The oxidation number of hydrogen in Ammonia, $NH_3$ is $+1$. Nitrogen is not a metal. So, $NH_3$ is not
583                a metallic hydride. <br>
584                But, the oxidation number of hydrogen in Lithium Hydride, $LiH$ is $-1$. Lithium is a metal. So, $LiH$
585                is a metallic hydride.
586            </p>
587
588            <p>
589                <mark>(10.)</mark> The oxidation number of oxygen in any compound is usually $-2$; except in peroxides,
590                or if it is bonded with fluorine. <br>
591                The oxidation number of oxygen in peroxides is $-1$. <br>
592                The oxidation number of oxygen in compounds of oxygen and fluorine (oxygen fluorides) varies. <br>
593                However, because we know the oxidation number of fluorine ($-1$), we can calculate the oxidation number
594                of oxygen. <br>
595                <u>Recall:</u> A <u>peroxide</u> is a compound that contains <u>two oxygen atoms</u> joined together by
596                a <u>single bond.</u> <br>
597                <u>Examples:</u> <br>
598                The oxidation number of oxygen in Hydrogen Peroxide, $H_2O_2$ is $-1$. <br><br>
599                For the oxygen fluorides, let us do some examples. <br>
600                What is the oxidation number of oxygen in Oxygen difluoride, $OF_2$? <br>
601                Let the oxidation state of oxygen be $x$ <br>
602                We already know from Rule (8.) to be $-1$ <br>
603                $x + 2(-1) = 0$ <br>
604                $x - 2 = 0$ <br>
605                $x = +2$ <br><br>
606
607                What is the oxidation number of oxygen in dioxygen difluoride, $O_2F_2$? <br>
608                Let the oxidation state of oxygen be $x$ <br>
609                We already know from Rule (8.) to be $-1$ <br>
610                $2x + 2(-1) = 0$ <br>
611                $2x - 2 = 0$ <br>
612                $2x = 2$ <br>
613                $x = +1$ <br>
614                As you see, oxygen has various oxidation states when it is bonded with fluorine.
615            </p>
616
617            <p>
618                <mark>(11.)</mark> The oxidation number of chlorine $Cl$, bromine $Br$, and iodine $I$ in any compound
619                is usually $-1$, except in combination with oxygen or fluorine. <br>
620                When either of these elements combine with oxygen or fluorine, we can find their oxidation numbers
621                because we know the oxidation state of oxygen, $-2$ and fluorine, $-1$. <br>
622                <u>Examples:</u> <br>
623                What is the oxidation number of chlorine in Sodium chloride (also known as your common salt), $NaCl$?
624                <br>
625                Let the oxidation state of chlorine be $x$ <br>
626                We already know from Rule (6.) that the oxidation number of sodium is $+1$ <br>
627                $1 + x = 0$ <br>
628                $x = -1$ <br><br>
629
630                What is the oxidation number of chlorine in Dichlorine heptoxide, $Cl_2O_7$? <br>
631                Let the oxidation state of chlorine be $x$ <br>
632                $2x + 7(-2) = 0$ <br>
633                $2x - 14 = 0$ <br>
634                $2x = 14$ <br>
635                $x = +7$ <br>
636                <i>Ask students to rename the compound according to the IUPAC nomenclature.</i> <br>
637                The old name is Dichlorine heptoxide <br>
638                The IUPAC name is Chlorine (VII) oxide
639            </p>
640
641            <p>
642                <mark>(12.)</mark> For all other elements not mentioned in Rules (1.) through (9.): <br>
643                If you were not asked to calculate their oxidation numbers, then assume their oxidation numbers to be
644                their valencies (as determined by their electronic configurations in shells, not orbitals). <br>
645                <u>Recall:</u> The <u>valency</u> of an element is the combining power of the element.
646            </p>
647
648
649            <hr>
650
651            <h3>Can we have fractional oxidation numbers?</h3>
652            <i>Ask students to name this compound, $Fe_3O_4$ the common way - with prefixes.</i> <br>
653            <i>Then, ask them to calculate the oxidation number of $Fe$ in $Fe_3O_4$.</i> <br>
654            <p>
655                The common name for $Fe_3O_4$ is Triiron tetraoxide <br>
656                We know from Rule (10.) that the oxidation number of oxygen is $-2$ <br>
657                Let the oxidation number of iron be $x$ <br>
658                $3(x) + 4(-2) = 0$ <br>
659                $3x - 8 = 0$ <br>
660                $3x = 8$ <br>
661                $x = {8\over3}$ <br>
662                Hmmm...what happended? This is a fraction! Let me go back to the definition of an oxidation number. <br>
663                The number of electrons cannot be a fraction. It has to be an integer! <br>
664                <i>Your students are correct if they make this argument.</i>
665            </p>
666
667            <p>
668                <u>Here is the explanation</u>: <br>
669                When we say the oxidation number of $Ca$ in $Ca_3(PO_4)_2$ is $+2$; <br>
670                We mean that the oxidation number of <u>a calcium atom</u> is $+2$ <br>
671                We do realize that we have $3$ calcium atoms in the compound, $Ca_3(PO_4)_2$ <br>
672                But, we found the oxidation number of $1$ calcium atom. <br>
673                If we have to find the oxidation number of 3 calcium atoms, then it would be $3(+2) = +6$ <br>
674                So, we have been finding the oxidation numbers of single ($1$) atoms of elements all along. <br>
675                <mark>Usually, atoms of the same element in a compound have the same oxidation number. <br> </mark>
676                As you can see, the $3$ calcium atoms each have the same oxidation number of $+2$ <br>
677                <mark>However, it is possible for atoms of the same element in a compound to have different oxidation
678                    numbers. <br> </mark>
679                So, in the case of $Fe_3O_4$; <br>
680                We have $3$ atoms of Iron <br>
681                <u>Each</u> of the two of the iron atoms has an oxidation number of $+3$ while the other iron atom has
682                an oxidation number of $+2$ <br>
683                This is known as <u>Fragmenting</u>. <br>
684                The oxidation number of the 3 iron atoms is $+8$ <br>
685                From our calculation, we got the oxidation number of an iron atom to be ${8\over3}$ <br>
686                As you can notice, this is an average. <br>
687                So, you can say that the oxidation numbers we have been calculating are averages. <br>
688                In essence; oxidation numbers cannot be fractions, because the number of electrons donated or accepted
689                cannot be fractions. <br>
690                So, if you get fractions and you are sure you followed the rules;
690 then you may need to apply
691                <u>fragmenting</u> to know the oxidation states of each atom. <br>
692                <i>Some students may ask you how you fragment - how do you know that each of the two iron atoms had
693                    oxidation number of $+3$ while the other iron atom had an oxidation number of $+2$ </i> <br>
694                <i>Please be prepared to answer. Just tell them it is beyond the scope of this topic.</i> <br>
695                <i>If they ask whether it is acceptable to write the oxidation number of $Fe$ in $Fe_3O_4$ as
696                    ${8\over3}$, let them know they can do so.</i> <br>
697                <i>What if they ask you to name the compound according to IUPAC nomenclature?</i> <br>
698                $Fe_3O_4$ is Iron (II, III) oxide. <br>
699                It is an ionic compound whose atoms have different oxidation numbers. <br>
700                One atom of iron (II), $Fe^{2+}$ has an oxidation number of $+2$ <br>
701                Two atoms of iron (III), $Fe^{3+}$ each has an oxidation number of $+3$ <br>
702                Hence, the Iron (II, III) oxide <br>
703                <i>The very curious ones may ask more questions. Please be prepared to answer.</i>
704            </p>
705            <p>
706                For more examples, please review: <a
707                    href="https://samuelchukwuemeka.github.io/chemical-reactions/redoxReactions.html"
708                    target="_blank">Solved Examples on Redox Reactions</a>
709            </p>
710        </div>
711        <!-- End Oxidation Number -->
712
713        <br><br><br><br>
714
715
716
717
718        <!-- References -->
719        <div class="w3-container w3-padding-32" id="references">
720            <h2 class="w3-padding-16">References</h2>
721            <hr>
722            <p>
723                Chukwuemeka, S.D (2016, April 30). <i>Samuel Chukwuemeka Tutorials - Math, Science, and Technology.</i>
724                Retrieved from <a href="https://www.samuelchukwuemeka.com/"
725                    target="_blank">https://www.samuelchukwuemeka.com</a>
726                <br><br>
727                Chemistry. (n.d.) West African Examinations Council (WAEC). Retrieved May 30, 2020, from
728                https://waeconline.org.ng/e-learning/Chemistry/chemmain.html
729                <br><br>
730                Free Jamb Past Questions And Answer For All Subject 2020. (2020, January 31). Vastlearners.
731                https://www.vastlearners.com/free-jamb-past-questions/
732                <br><br>
733
734            </p>
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