The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows:<\/p>\n\n\n\n
Step 1: Count the valence electrons.<\/h3>\n\n\n\n\n- Carbon (C) has 4 valence electrons.<\/li>\n\n\n\n
- Hydrogen (H) has 1 valence electron (each hydrogen atom).<\/li>\n\n\n\n
- Oxygen (O) has 6 valence electrons (each oxygen atom).<\/li>\n\n\n\n
- The ion has a negative charge (\u22121-1\u22121), which adds 1 more electron.<\/li>\n<\/ul>\n\n\n\n
Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) + 3 \\times 1 \\ (\\text{H}) + 2 \\times 6 \\ (\\text{O}) + 1 \\ (\\text{negative charge}) = 12 \\ \\text{electrons}.4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.<\/p>\n\n\n\n
Step 2: Draw the skeleton structure.<\/h3>\n\n\n\n\n- Place the carbon in the center, with two oxygens attached to it.<\/li>\n\n\n\n
- One oxygen is double-bonded to the carbon, and the other oxygen is single-bonded to the carbon and carries a negative charge.<\/li>\n<\/ul>\n\n\n\n
Step 3: Distribute the electrons.<\/h3>\n\n\n\n\n- The double-bonded oxygen gets 4 electrons (2 for the bond, 2 for lone pairs).<\/li>\n\n\n\n
- The single-bonded oxygen gets 2 electrons for the bond and 3 lone pairs (6 electrons total), plus 1 additional electron due to the negative charge.<\/li>\n<\/ul>\n\n\n\n
Step 4: Check for octets and minimize charges.<\/h3>\n\n\n\n\n- The carbon has 4 bonds, so it satisfies the octet rule.<\/li>\n\n\n\n
- The double-bonded oxygen has 2 bonds and 2 lone pairs, fulfilling the octet rule.<\/li>\n\n\n\n
- The single-bonded oxygen has 1 bond, 3 lone pairs, and 1 negative charge, also fulfilling the octet rule.<\/li>\n<\/ul>\n\n\n\n
Thus, the structure is:CH3C=O\u2212\\text{CH}_3\\text{C} = \\text{O}^- CH3\u200bC=O\u2212 \u2223\\ \\ \\ \\ \\ \\ | \u2223 O\\ \\ \\ \\ \\text{O} O<\/p>\n\n\n\n
Electron Geometry:<\/h3>\n\n\n\n
The electron geometry around the central carbon is trigonal planar<\/strong> because the carbon is connected to three regions of electron density (one double bond and two single bonds).<\/p>\n\n\n\nMolecular Geometry:<\/h3>\n\n\n\n
The molecular geometry is trigonal planar<\/strong> as well because there are no lone pairs on the carbon, and all regions of electron density are bonds.<\/p>\n\n\n\nPolarity:<\/h3>\n\n\n\n\n- The molecule is polar<\/strong>. The C\u2013O bond has a dipole moment due to the difference in electronegativity between carbon and oxygen. While the acetate ion has symmetry, the negative charge on the single-bonded oxygen causes a net dipole moment. Therefore, the ion is polar.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate the molecular geometry diagram for the acetate ion.<\/p>\n\n\n\n![\"Generated](\"https:\/\/sdmntprnorthcentralus.oaiusercontent.com\/files\/00000000-2928-622f-b48a-b2c8055cbede\/raw?se=2025-07-04T20%3A03%3A14Z&sp=r&sv=2024-08-04&sr=b&scid=a88becc5-cfc6-50a2-b064-9964cfbe5d4d&skoid=bbd22fc4-f881-4ea4-b2f3-c12033cf6a8b&sktid=a48cca56-e6da-484e-a814-9c849652bcb3&skt=2025-07-04T06%3A40%3A31Z&ske=2025-07-05T06%3A40%3A31Z&sks=b&skv=2024-08-04&sig=jJj5eKaN\/25khfmnypOv6Rk0rlwmvhEji4HLOlJXgRU%3D\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-384.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the lewis structure for CH3CO2- ? What is the electron and molecular geometry? Draw the molecular geometry. Is it polar or non- polar? The Correct Answer and Explanation is: The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows: Step 1: Count the valence electrons. Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) […]<\/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-244112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112","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=244112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) + 3 \\times 1 \\ (\\text{H}) + 2 \\times 6 \\ (\\text{O}) + 1 \\ (\\text{negative charge}) = 12 \\ \\text{electrons}.4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.<\/p>\n\n\n\n
Step 2: Draw the skeleton structure.<\/h3>\n\n\n\n\n- Place the carbon in the center, with two oxygens attached to it.<\/li>\n\n\n\n
- One oxygen is double-bonded to the carbon, and the other oxygen is single-bonded to the carbon and carries a negative charge.<\/li>\n<\/ul>\n\n\n\n
Step 3: Distribute the electrons.<\/h3>\n\n\n\n\n- The double-bonded oxygen gets 4 electrons (2 for the bond, 2 for lone pairs).<\/li>\n\n\n\n
- The single-bonded oxygen gets 2 electrons for the bond and 3 lone pairs (6 electrons total), plus 1 additional electron due to the negative charge.<\/li>\n<\/ul>\n\n\n\n
Step 4: Check for octets and minimize charges.<\/h3>\n\n\n\n\n- The carbon has 4 bonds, so it satisfies the octet rule.<\/li>\n\n\n\n
- The double-bonded oxygen has 2 bonds and 2 lone pairs, fulfilling the octet rule.<\/li>\n\n\n\n
- The single-bonded oxygen has 1 bond, 3 lone pairs, and 1 negative charge, also fulfilling the octet rule.<\/li>\n<\/ul>\n\n\n\n
Thus, the structure is:CH3C=O\u2212\\text{CH}_3\\text{C} = \\text{O}^- CH3\u200bC=O\u2212 \u2223\\ \\ \\ \\ \\ \\ | \u2223 O\\ \\ \\ \\ \\text{O} O<\/p>\n\n\n\n
Electron Geometry:<\/h3>\n\n\n\n
The electron geometry around the central carbon is trigonal planar<\/strong> because the carbon is connected to three regions of electron density (one double bond and two single bonds).<\/p>\n\n\n\nMolecular Geometry:<\/h3>\n\n\n\n
The molecular geometry is trigonal planar<\/strong> as well because there are no lone pairs on the carbon, and all regions of electron density are bonds.<\/p>\n\n\n\nPolarity:<\/h3>\n\n\n\n\n- The molecule is polar<\/strong>. The C\u2013O bond has a dipole moment due to the difference in electronegativity between carbon and oxygen. While the acetate ion has symmetry, the negative charge on the single-bonded oxygen causes a net dipole moment. Therefore, the ion is polar.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate the molecular geometry diagram for the acetate ion.<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the lewis structure for CH3CO2- ? What is the electron and molecular geometry? Draw the molecular geometry. Is it polar or non- polar? The Correct Answer and Explanation is: The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows: Step 1: Count the valence electrons. Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) […]<\/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-244112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112","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=244112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 3: Distribute the electrons.<\/h3>\n\n\n\n\n- The double-bonded oxygen gets 4 electrons (2 for the bond, 2 for lone pairs).<\/li>\n\n\n\n
- The single-bonded oxygen gets 2 electrons for the bond and 3 lone pairs (6 electrons total), plus 1 additional electron due to the negative charge.<\/li>\n<\/ul>\n\n\n\n
Step 4: Check for octets and minimize charges.<\/h3>\n\n\n\n\n- The carbon has 4 bonds, so it satisfies the octet rule.<\/li>\n\n\n\n
- The double-bonded oxygen has 2 bonds and 2 lone pairs, fulfilling the octet rule.<\/li>\n\n\n\n
- The single-bonded oxygen has 1 bond, 3 lone pairs, and 1 negative charge, also fulfilling the octet rule.<\/li>\n<\/ul>\n\n\n\n
Thus, the structure is:CH3C=O\u2212\\text{CH}_3\\text{C} = \\text{O}^- CH3\u200bC=O\u2212 \u2223\\ \\ \\ \\ \\ \\ | \u2223 O\\ \\ \\ \\ \\text{O} O<\/p>\n\n\n\n
Electron Geometry:<\/h3>\n\n\n\n
The electron geometry around the central carbon is trigonal planar<\/strong> because the carbon is connected to three regions of electron density (one double bond and two single bonds).<\/p>\n\n\n\nMolecular Geometry:<\/h3>\n\n\n\n
The molecular geometry is trigonal planar<\/strong> as well because there are no lone pairs on the carbon, and all regions of electron density are bonds.<\/p>\n\n\n\nPolarity:<\/h3>\n\n\n\n\n- The molecule is polar<\/strong>. The C\u2013O bond has a dipole moment due to the difference in electronegativity between carbon and oxygen. While the acetate ion has symmetry, the negative charge on the single-bonded oxygen causes a net dipole moment. Therefore, the ion is polar.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate the molecular geometry diagram for the acetate ion.<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the lewis structure for CH3CO2- ? What is the electron and molecular geometry? Draw the molecular geometry. Is it polar or non- polar? The Correct Answer and Explanation is: The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows: Step 1: Count the valence electrons. Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) […]<\/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-244112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112","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=244112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 4: Check for octets and minimize charges.<\/h3>\n\n\n\n\n- The carbon has 4 bonds, so it satisfies the octet rule.<\/li>\n\n\n\n
- The double-bonded oxygen has 2 bonds and 2 lone pairs, fulfilling the octet rule.<\/li>\n\n\n\n
- The single-bonded oxygen has 1 bond, 3 lone pairs, and 1 negative charge, also fulfilling the octet rule.<\/li>\n<\/ul>\n\n\n\n
Thus, the structure is:CH3C=O\u2212\\text{CH}_3\\text{C} = \\text{O}^- CH3\u200bC=O\u2212 \u2223\\ \\ \\ \\ \\ \\ | \u2223 O\\ \\ \\ \\ \\text{O} O<\/p>\n\n\n\n
Electron Geometry:<\/h3>\n\n\n\n
The electron geometry around the central carbon is trigonal planar<\/strong> because the carbon is connected to three regions of electron density (one double bond and two single bonds).<\/p>\n\n\n\nMolecular Geometry:<\/h3>\n\n\n\n
The molecular geometry is trigonal planar<\/strong> as well because there are no lone pairs on the carbon, and all regions of electron density are bonds.<\/p>\n\n\n\nPolarity:<\/h3>\n\n\n\n\n- The molecule is polar<\/strong>. The C\u2013O bond has a dipole moment due to the difference in electronegativity between carbon and oxygen. While the acetate ion has symmetry, the negative charge on the single-bonded oxygen causes a net dipole moment. Therefore, the ion is polar.<\/li>\n<\/ul>\n\n\n\n
Now, I’ll generate the molecular geometry diagram for the acetate ion.<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the lewis structure for CH3CO2- ? What is the electron and molecular geometry? Draw the molecular geometry. Is it polar or non- polar? The Correct Answer and Explanation is: The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows: Step 1: Count the valence electrons. Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) […]<\/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-244112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112","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=244112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Thus, the structure is:CH3C=O\u2212\\text{CH}_3\\text{C} = \\text{O}^- CH3\u200bC=O\u2212 \u2223\\ \\ \\ \\ \\ \\ | \u2223 O\\ \\ \\ \\ \\text{O} O<\/p>\n\n\n\n
Electron Geometry:<\/h3>\n\n\n\n
The electron geometry around the central carbon is trigonal planar<\/strong> because the carbon is connected to three regions of electron density (one double bond and two single bonds).<\/p>\n\n\n\n The molecular geometry is trigonal planar<\/strong> as well because there are no lone pairs on the carbon, and all regions of electron density are bonds.<\/p>\n\n\n\n Now, I’ll generate the molecular geometry diagram for the acetate ion.<\/p>\n\n\n\n What is the lewis structure for CH3CO2- ? What is the electron and molecular geometry? Draw the molecular geometry. Is it polar or non- polar? The Correct Answer and Explanation is: The Lewis structure for the acetate ion CH3CO2\u2212\\text{CH}_3\\text{CO}_2^-CH3\u200bCO2\u2212\u200b (or CH3COO\u2212\\text{CH}_3\\text{COO}^-CH3\u200bCOO\u2212) is as follows: Step 1: Count the valence electrons. Total valence electrons:4 (C)+3\u00d71 (H)+2\u00d76 (O)+1 (negative charge)=12 electrons.4 \\ (\\text{C}) […]<\/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-244112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112","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=244112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Molecular Geometry:<\/h3>\n\n\n\n
Polarity:<\/h3>\n\n\n\n
\n
<\/figure>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"