{"id":184947,"date":"2025-01-21T19:26:10","date_gmt":"2025-01-21T19:26:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184947"},"modified":"2025-01-21T19:26:13","modified_gmt":"2025-01-21T19:26:13","slug":"acetone-c3h60ch3coch3-lewis-structure-using-dots-to-represent-bonding-electrons","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/21\/acetone-c3h60ch3coch3-lewis-structure-using-dots-to-represent-bonding-electrons\/","title":{"rendered":"Acetone, C3H60(CH3COCH3) Lewis structure using dots to represent bonding electrons"},"content":{"rendered":"\n<ol class=\"wp-block-list\">\n<li>Acetone, C3H60(CH3COCH3) Lewis structure using dots to represent bonding electrons: Lewis structure using lines to represent pairs of bonding electrons: Label the carbon atoms A,B, &amp;C. Carbon atom A: Total number of electron groups: Number of lone pairs: Number of bonding groups\/pairs: Bond angles: Electron geometry: Molecular geometry: Hybridization: Number of hybrid orbitals: Carbon atom B: Total number of electron groups: Number of lone pairs: Number of bonding groups\/pairs: Bond angles: Electron geometry: Molecular geometry: Hybridization: Number of hybrid orbitals: Carbon atom C Total number of electron groups: Number of lone pairs: Number of bonding groups\/pairs: Bond angles: Electron geometry: Molecular geometry: Number of hybrid orbitals: Hybridization: Is this molecule polar? (Y\/N) Can this molecule hydrogen bond with itself? Number of sigma bonds: Number of pi bonds: None 1. Acetone, C3H6O(CH3COCH3) Lewis structure using dots to represent bonding electrons: Lewis structure using lines to represent pairs of bonding electrons: Label the carbon atoms A,B, &amp;C. | Carbon atom A: Total number of electron groups: Number of bonding groups\/pairs: Electron geometry: Molecular geometry: Hybridization: Carbon atom B: Total number of electron groups: Number of bonding groups\/pairs: Electron geometry: Molecular geometry: Number of hybrid orbitals: I Electron geometry: Molecular geometry: Hybridization: Carbon atom C Total number of electron groups: Number of bonding groups\/pairs: Number of lone pairs: Bond angles: Number of hybrid orbitals: Number of lone pairs: Bond angles: Hybridization: Is this molecule polar? (Y\/N) Can this molecule hydrogen bond with itself? Number of sigma bonds: Number of pi bonds: Number of lone pairs: Bond angles: Number of hybrid orbitals: MacBook Pro<\/li>\n<\/ol>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here\u2019s a detailed explanation and the correct answers for each part of the question about acetone (C\u2083H\u2086O):<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Lewis Structure<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Dot Representation<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Central carbon (B) is double-bonded to oxygen and single-bonded to two methyl groups (A and C).<\/li>\n\n\n\n<li>Dots around oxygen represent its lone pairs.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Line Representation<\/strong>:\n<ul class=\"wp-block-list\">\n<li>A double line between carbon (B) and oxygen.<\/li>\n\n\n\n<li>Single lines connecting carbon (B) to carbons (A) and (C).<\/li>\n\n\n\n<li>Carbon (A) and (C) are bonded to three hydrogens each.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Carbon Atom A (CH\u2083 group)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Total electron groups<\/strong>: 4 (3 bonding pairs with hydrogens, 1 bond with carbon B).<\/li>\n\n\n\n<li><strong>Number of lone pairs<\/strong>: 0.<\/li>\n\n\n\n<li><strong>Number of bonding groups\/pairs<\/strong>: 4.<\/li>\n\n\n\n<li><strong>Bond angles<\/strong>: Approximately 109.5\u00b0.<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Tetrahedral.<\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: Tetrahedral.<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3.<\/li>\n\n\n\n<li><strong>Number of hybrid orbitals<\/strong>: 4.<\/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>Carbon Atom B (C=O group)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Total electron groups<\/strong>: 3 (1 double bond with oxygen, 2 single bonds with carbons A and C).<\/li>\n\n\n\n<li><strong>Number of lone pairs<\/strong>: 0.<\/li>\n\n\n\n<li><strong>Number of bonding groups\/pairs<\/strong>: 3.<\/li>\n\n\n\n<li><strong>Bond angles<\/strong>: Approximately 120\u00b0.<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Trigonal planar.<\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: Trigonal planar.<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2.<\/li>\n\n\n\n<li><strong>Number of hybrid orbitals<\/strong>: 3.<\/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>Carbon Atom C (CH\u2083 group)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Total electron groups<\/strong>: 4 (3 bonding pairs with hydrogens, 1 bond with carbon B).<\/li>\n\n\n\n<li><strong>Number of lone pairs<\/strong>: 0.<\/li>\n\n\n\n<li><strong>Number of bonding groups\/pairs<\/strong>: 4.<\/li>\n\n\n\n<li><strong>Bond angles<\/strong>: Approximately 109.5\u00b0.<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Tetrahedral.<\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: Tetrahedral.<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3.<\/li>\n\n\n\n<li><strong>Number of hybrid orbitals<\/strong>: 4.<\/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>Other Characteristics<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Polarity<\/strong>: Yes, acetone is polar because the C=O bond creates a dipole moment.<\/li>\n\n\n\n<li><strong>Hydrogen Bonding with Itself<\/strong>: No, acetone cannot hydrogen bond with itself, as it lacks a hydrogen atom directly bonded to an electronegative atom like oxygen or nitrogen.<\/li>\n\n\n\n<li><strong>Number of Sigma Bonds<\/strong>: 9 (6 C-H bonds, 2 C-C bonds, and 1 C=O sigma bond).<\/li>\n\n\n\n<li><strong>Number of Pi Bonds<\/strong>: 1 (from the C=O double bond).<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (Summary)<\/h3>\n\n\n\n<p>Acetone\u2019s structure is dictated by the hybridization and geometry of each carbon atom. Carbon A and C exhibit sp\u00b3 hybridization, leading to tetrahedral geometry with approximately 109.5\u00b0 bond angles. Carbon B, as part of the C=O group, is sp\u00b2 hybridized, resulting in trigonal planar geometry and 120\u00b0 bond angles. The molecule&#8217;s polar nature arises from the electronegative oxygen, creating a net dipole moment. Acetone does not form hydrogen bonds with itself but has strong dipole-dipole interactions. Its sigma and pi bonds further contribute to its chemical reactivity.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The correct answer and explanation is: Here\u2019s a detailed explanation and the correct answers for each part of the question about acetone (C\u2083H\u2086O): Lewis Structure Carbon Atom A (CH\u2083 group) Carbon Atom B (C=O group) Carbon Atom C (CH\u2083 group) Other Characteristics Explanation (Summary) Acetone\u2019s structure is dictated by the hybridization and geometry of each [&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-184947","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184947","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=184947"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184947\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184947"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184947"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184947"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}