{"id":236583,"date":"2025-06-16T10:41:29","date_gmt":"2025-06-16T10:41:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=236583"},"modified":"2025-06-16T10:41:31","modified_gmt":"2025-06-16T10:41:31","slug":"draw-the-lewis-structure-of-acetone-ch32co-c-in-the-middle-with-two-cs-and-an-o-bonded-to-it","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/draw-the-lewis-structure-of-acetone-ch32co-c-in-the-middle-with-two-cs-and-an-o-bonded-to-it\/","title":{"rendered":"Draw the Lewis structure of acetone, (CH3)2CO (C in the middle, with two Cs and an O bonded to it)"},"content":{"rendered":"\n

Draw the Lewis structure of acetone, (CH3)2CO (C in the middle, with two Cs and an O bonded to it). What is the hybridization on each carbon atom in the molecule? b. On the Lewis structure, identify every bond as a sigma or pi bond. (\u00cf\u0192 or \u00cf\u20ac) c. For the Carbon-Hydrogen bond, what are the orbitals that overlap to make that bond?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here is the diagram of the Lewis structure of acetone ((CH\u2083)\u2082CO):<\/p>\n\n\n\n

mathematicaCopyEdit       H   H\n        \\  \/\n    H \u2014 C \u2014 C == O\n        \/ \\\n       H   C \u2014 H\n            |\n            H\n<\/code><\/pre>\n\n\n\n
Or more neatly, in a structural way:<\/p>\n\n\n\n
mathematicaCopyEdit     H   H\n      \\  \/\nH \u2014 C \u2014 C == O\n      \/ \\\n     H   C \u2014 H\n          |\n          H\n<\/code><\/pre>\n\n\n\n
Explanation:<\/strong><\/p>\n\n\n\n
Acetone is a ketone with the formula (CH\u2083)\u2082CO. The structure consists of a central carbonyl carbon that bonds to two methyl groups and an oxygen atom. Each methyl group has three hydrogen atoms attached to its carbon. The carbonyl carbon forms a double bond with oxygen, which consists of one sigma bond and one pi bond.<\/p>\n\n\n\n
In this molecule, the hybridization of each atom determines how the atoms are arranged. The carbonyl carbon uses sp\u00b2 hybridization because it forms three sigma bonds: one with each methyl carbon and one with oxygen. The remaining unhybridized p orbital on this carbon overlaps with a p orbital on oxygen to form the pi bond. This hybridization leads to a trigonal planar geometry around the carbonyl carbon, with bond angles close to 120 degrees.<\/p>\n\n\n\n
The methyl carbons are sp\u00b3 hybridized because they form four sigma bonds: three with hydrogen atoms and one with the central carbon. This hybridization results in a tetrahedral geometry around each methyl carbon, with bond angles close to 109.5 degrees.<\/p>\n\n\n\n
Each C-H bond results from the overlap of an sp\u00b3 orbital from the carbon with an s orbital from the hydrogen, forming a stable sigma bond. The C-C single bonds are sigma bonds formed by the overlap of sp\u00b3 and sp\u00b2 orbitals. The C=O double bond includes a sigma bond formed by the overlap of sp\u00b2 orbitals and a pi bond from the side-by-side overlap of unhybridized p orbitals.<\/p>\n\n\n\n
This combination of hybridizations gives acetone its overall shape and bonding characteristics.<\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
Draw the Lewis structure of acetone, (CH3)2CO (C in the middle, with two Cs and an O bonded to it). What is the hybridization on each carbon atom in the molecule? b. On the Lewis structure, identify every bond as a sigma or pi bond. (\u00cf\u0192 or \u00cf\u20ac) c. For the Carbon-Hydrogen bond, what are […]<\/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-236583","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236583","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=236583"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236583\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=236583"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=236583"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=236583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}