1<!DOCTYPE html> 2<html lang="en"> 3 4<head> 5 6 7 <title>Chemical Reactions</title> 8 <meta charset="UTF-8"> 9 <meta name="description" 10 content="Samuel Dominic Chukwuemeka (SamDom For Peace) gives all the credit to our LORD GOD and Anointed Savior, JESUS CHRIST. We are experts in chemical reactions." /> 11 <meta name="keywords" content="samuel, dominic, chukwuemeka, samuel dominic chukwuemeka, chemistry, 12 reactions, chemical reactions, inorganic chemistry, physical chemistry, 13 organic chemistry, reactants, products, chemical change, chemical combinations, 14 combination, decomposition, dsiplacement, double decomposition, catalytic, 15 catalysts, reversible, irreversible, thermal dissociation, redox, oxidation-reduction, 16 oxidation, reduction, electrons, hydrogen, oxygen, electron transfer, oxidising agents, 17 reducing agents, electrolysis, solid, liquid, gas, aqueous" /> 18 <link href="styles/chemical-reaction.css" rel="stylesheet" type="text/css" media="all"> 19 <link rel="shortcut icon" type="image/x-icon" href="images/favicon.ico"> 20 <meta name="viewport" content="width=device-width, initial-scale=1.0"> 21</head> 22 23 24<body> 25 26 <!-- Side Navigation --> 27 <nav class="w3-sidebar w3-bar-block w3-collapse w3-animate-left" style="z-index:3;width:320px;" id="mySidebar"> 28 <div class="w3-center"> 29 <img src="images/samuelchukwuemeka.jpg" alt="Samuel Dominic Chukwuemeka aka SamDom For Peace" 30 title="Samuel Dominic Chukwuemeka" style="text-align: center"> 31 </div> 32 <div id="google_translate_element" class="w3-center"></div> 33
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39 40 <a href="#quotes" class="w3-bar-item w3-button">Quotes</a> 41 <a href="#welcome" class="w3-bar-item w3-button">Welcome</a> 42 <a href="https://www.youtube.com/playlist?list=PLA9Cap2CEoniViSjnY-XCoetpo14rDIPw&disable_polymer=true" 43 target="_blank" class="w3-bar-item w3-button">Videos</a> 44 <a href="#objectives" class="w3-bar-item w3-button">Objectives</a> 45 <a href="#vocabularyWords" class="w3-bar-item w3-button">Vocabulary Words</a> 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> 108 </div> 109 <a href="#references" class="w3-bar-item w3-button">References</a> 110 <a href="https://chukwuemekasamuel.weebly.com/" target="_blank" class="w3-bar-item w3-button">Contact</a> 111 <a href="https://chukwuemekasamuel.com/portfolio.html" target="_blank" 112 class="w3-bar-item w3-button">Portfolio</a> 113 </nav> 114 115 116 <!-- Overlay effect when opening the side navigation on small screens --> 117 <div class="w3-overlay w3-hide-large w3-animate-opacity" onclick="w3_close()" style="cursor:pointer" 118 title="Close Sidemenu" id="myOverlay"></div> 119 120 <!-- Page content --> 121 <div class="w3-main" style="margin-left:320px;"> 122 <a href="javascript:void(0)" class="w3-hide-large w3-red
122w3-button w3-right w3-margin-top w3-margin-right" 123 onclick="document.getElementById('id01').style.display = 'block'"></a> 124 125 <!-- Top header --> 126 <header class="w3-container green" id="quotes"> 127 <div class="w3-content"> 128 <p> 129 <q> 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 132 FATHER in Heaven prayer)</a> <br> 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 136 </p> 137 <p id="messages1"> 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 - 31 141 </p> 142 <p><q>The Joy of the Teacher is the Success of the Students.</q> - Samuel Chukwuemeka</p> 143 </div> 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"> 152 </div> 153 <div class="mySlides w3-center"> 154 <img src="images/chemical-reaction2.png" alt="Chemical Reaction 2" 155 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 156 </div> 157 <div class="mySlides w3-center"> 158 <img src="images/chemical-reaction3.png" alt="Chemical Reaction 3" 159 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 160 </div> 161 <div class="mySlides w3-center"> 162 <img src="images/chemical-reaction4.png" alt="Chemical Reaction 4" 163 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 164 </div> 165 <div class="mySlides w3-center"> 166 <img src="images/chemical-reaction5.png" alt="Chemical Reaction 5" 167 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 168 </div> 169 <div class="mySlides w3-center"> 170 <img src="images/chemical-reaction6.png" alt="Chemical Reaction 6" 171 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 172 </div> 173 <div class="mySlides w3-center"> 174 <img src="images/chemical-reaction7.png" alt="Chemical Reaction 7" 175 title="Samuel Chukwuemeka aka Samdom For Peace" height="400" width="970"> 176 </div> 177 178 <!-- Welcome Greeting --> 179 <div class="w3-container" id="welcome"> 180 <div class="w3-content"> 181 <p> 182 <span id="greetings"></span> <br> 183 I greet you this day, 184
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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? ☺☺☺); . <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> 735 </div> 736 <!-- End References --> 737 738 <!-- Footer --> 739 <footer class="green w3-center w3-padding-24"> 740 <a href="https://www.youtube.com/user/samuelchukwuemeka" title="Subscribe to my YouTube channel" 741 target="_blank"><img src="images/youtube.png" alt="Subscribe to my YouTube channel" /></a> 742 <a href="https://twitter.com/samdom4peace" title="Follow me on twitter" target="_blank"><img 743 src="images/twitter.png" alt="Follow me on twitter" /></a> 744 <a href="http://samueldominicchukwuemeka.blogspot.com/?view=flipcard" title="Subscribe and follow my blog" 745 target="_blank"><img src="images/blogger.jpg" alt="Subscribe and follow my blog" /></a> 746 <p id="left-align">© 2018 <a href="https://www.samuelchukwuemeka.com/" 747 target="_blank">Samdom4Peace</a> Designs.</p> 748 749 </footer> 750 <!-- End Footer --> 751 752 </div> 753 <!-- End Page content --> 754 755 756
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