{"id":214191,"date":"2025-05-13T18:12:31","date_gmt":"2025-05-13T18:12:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214191"},"modified":"2025-05-13T18:12:35","modified_gmt":"2025-05-13T18:12:35","slug":"draw-the-lewis-structure-of-ch32chch2cho","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/draw-the-lewis-structure-of-ch32chch2cho\/","title":{"rendered":"Draw the Lewis structure of (CH3)2CHCH2CHO"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Draw the Lewis structure of (CH3)2CHCH2CHO<\/p>\n\n\n\n<ol class=\"wp-block-list\"><\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Draw the Lewis structure of [CH3CH(OH2)CH2CH3]+<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Draw the Lewis structure of CH3CH(CN)CH3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Draw the Lewis structure of [(CH3)2O(CH3)]+<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">fe3n2 oxidation number<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\"><\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Sure! Let&#8217;s go through each part of your question in two sections:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>1. Lewis Structures<\/strong><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>A. (CH\u2083)\u2082CHCH\u2082CHO<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This is <strong>isovaleraldehyde<\/strong>, a five-carbon aldehyde. Its structure is:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>        H   H   O\n        |   |  ||\nCH\u2083 \u2014 C \u2014 C \u2014 C \u2014 H\n     |   |   |\n    CH\u2083 H   H\n<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Breakdown:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Two CH\u2083 (methyl) groups attached to a CH (central carbon).<\/li>\n\n\n\n<li>That CH connects to a CH\u2082.<\/li>\n\n\n\n<li>CH\u2082 connects to a CHO (aldehyde).<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Lewis structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each C makes 4 bonds.<\/li>\n\n\n\n<li>O in the aldehyde has a double bond with C and 2 lone pairs.<\/li>\n\n\n\n<li>H makes 1 bond.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>B. [CH\u2083CH(OH\u2082)CH\u2082CH\u2083]\u207a<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This is <strong>a protonated alcohol<\/strong>, specifically a protonated form of butanol.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Structure:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>CH\u2083\u2013CH(OH\u2082\u207a)\u2013CH\u2082\u2013CH\u2083\n<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Notes:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>OH\u2082\u207a means the oxygen has 3 bonds (to H, H, and carbon) and <strong>one lone pair<\/strong>.<\/li>\n\n\n\n<li>The molecule has an overall <strong>+1 charge<\/strong> centered on the oxygen.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>C. CH\u2083CH(CN)CH\u2083<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This is <strong>2-methylpropanenitrile<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Structure:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>    CN\n     |\nCH\u2083\u2013CH\u2013CH\u2083\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>CN is a <strong>nitrile group<\/strong> with a <strong>triple bond<\/strong> between carbon and nitrogen.<\/li>\n\n\n\n<li>N has a <strong>lone pair<\/strong>.<\/li>\n\n\n\n<li>Central carbon is bonded to two CH\u2083 groups and the CN group.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>D. [(CH\u2083)\u2082O(CH\u2083)]\u207a<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This is a <strong>trimethyloxonium ion<\/strong>, also written as (CH\u2083)\u2083O\u207a.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Structure:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>      +\n     O\n   \/ | \\\nCH\u2083 CH\u2083 CH\u2083\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>O has <strong>three single bonds<\/strong> and <strong>no lone pairs<\/strong>, carrying a <strong>+1 charge<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>2. Fe\u2083N\u2082 \u2013 Oxidation Number Explanation<\/strong><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">To find the oxidation number of iron (Fe) in Fe\u2083N\u2082 (iron(II) nitride), follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-step:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Let the oxidation number of Fe be <strong>x<\/strong>.<\/li>\n\n\n\n<li>Nitrogen (N) in nitrides almost always has an oxidation number of <strong>\u20133<\/strong>.<\/li>\n\n\n\n<li>There are 2 nitrogen atoms: <strong>2 \u00d7 (\u20133) = \u20136<\/strong><\/li>\n\n\n\n<li>There are 3 Fe atoms: <strong>3x<\/strong><\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Set up the equation:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>3x + (\u20136) = 0\n=&gt; 3x = +6\n=&gt; x = +2\n<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Answer: Iron (Fe) has an oxidation number of +2 in Fe\u2083N\u2082.<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300+ words):<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Oxidation numbers are used to determine how electrons are transferred or shared in compounds. In ionic compounds like <strong>Fe\u2083N\u2082<\/strong>, we assume that atoms exist as ions with integer oxidation states. Iron (Fe) and nitrogen (N) are both elements that can form multiple oxidation states, but in this case, we are dealing with a nitride compound.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nitrogen is more electronegative than iron and typically forms a <strong>\u20133 oxidation state<\/strong> when bonded to metals, as seen in compounds like <strong>Li\u2083N<\/strong> and <strong>Mg\u2083N\u2082<\/strong>. This is because nitrogen needs three electrons to complete its octet and becomes <strong>N\u00b3\u207b<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The formula <strong>Fe\u2083N\u2082<\/strong> contains:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 iron (Fe) atoms.<\/li>\n\n\n\n<li>2 nitrogen (N) atoms.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">If each nitrogen atom has a \u20133 oxidation state (N\u00b3\u207b), the total negative charge from nitrogen is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">2 \u00d7 (\u20133) = <strong>\u20136<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To balance this charge in a neutral compound, the total positive charge from the iron atoms must be:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>+6<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since there are 3 iron atoms, we divide:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">+6 \/ 3 = <strong>+2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This tells us that each Fe atom has an oxidation number of <strong>+2<\/strong>. Thus, <strong>Fe\u2083N\u2082<\/strong> is made up of three <strong>Fe\u00b2\u207a<\/strong> cations and two <strong>N\u00b3\u207b<\/strong> anions. This compound is therefore called <strong>iron(II) nitride<\/strong>, where the Roman numeral II reflects the +2 oxidation state of iron.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Understanding oxidation numbers is essential in predicting reactivity, identifying redox reactions, and naming compounds systematically.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of (CH3)2CHCH2CHO Draw the Lewis structure of [CH3CH(OH2)CH2CH3]+ Draw the Lewis structure of CH3CH(CN)CH3 Draw the Lewis structure of [(CH3)2O(CH3)]+ fe3n2 oxidation number The Correct Answer and Explanation is: Sure! Let&#8217;s go through each part of your question in two sections: 1. Lewis Structures A. (CH\u2083)\u2082CHCH\u2082CHO This is isovaleraldehyde, a five-carbon [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-214191","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214191","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=214191"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214191\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214191"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214191"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214191"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}