The carbon atom in the carbonate ion (CO\u2083\u00b2\u207b) is sp\u00b2 hybridized<\/strong>.<\/p>\n\n\n\n Here’s an explanation:<\/p>\n\n\n\n In CO\u2083\u00b2\u207b, the carbon atom forms bonds with three oxygen atoms. These bonds are arranged in a trigonal planar geometry, where the bond angles are approximately 120\u00b0. This suggests that the carbon atom is using three hybrid orbitals to form bonds with the oxygen atoms, indicating an sp\u00b2 hybridization<\/strong>.<\/p>\n\n\n\n Therefore, the carbon atom in CO\u2083\u00b2\u207b is sp\u00b2 hybridized, reflecting its trigonal planar structure and delocalized bonding.<\/p>\n\n\n\n What is state of hybridisation of C in CO3^2 The Correct Answer and Explanation is: The carbon atom in the carbonate ion (CO\u2083\u00b2\u207b) is sp\u00b2 hybridized. Here’s an explanation: In CO\u2083\u00b2\u207b, the carbon atom forms bonds with three oxygen atoms. These bonds are arranged in a trigonal planar geometry, where the bond angles are approximately […]<\/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-261159","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261159","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=261159"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261159\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261159"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261159"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-step reasoning:<\/h3>\n\n\n\n
\n
In CO\u2083\u00b2\u207b, carbon has three bonding regions of electron density (one for each C-O bond). It also has one lone pair of electrons. However, this lone pair is delocalized, so we don’t count it as a separate electron domain in the traditional sense.<\/li>\n\n\n\n
Each of the three C-O bonds is a combination of sigma bonds<\/strong> and pi bonds<\/strong>, with resonance occurring between the bonds, meaning that the double bond character is spread across the three bonds.<\/li>\n\n\n\n
The ideal geometry for sp\u00b2 hybridized atoms is trigonal planar<\/strong>, which is exactly what we observe in CO\u2083\u00b2\u207b, with bond angles close to 120\u00b0 between the oxygen atoms.<\/li>\n\n\n\n
The resonance structures of CO\u2083\u00b2\u207b show that the electron density is spread evenly across the three C-O bonds. This delocalization of electrons further supports the idea of sp\u00b2 hybridization, where one of the orbitals overlaps with the p orbital of oxygen to form a pi bond<\/strong>, and the other two form sigma bonds<\/strong>.<\/li>\n<\/ol>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1113.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"