{"id":197518,"date":"2025-03-07T10:18:10","date_gmt":"2025-03-07T10:18:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197518"},"modified":"2025-03-07T10:18:13","modified_gmt":"2025-03-07T10:18:13","slug":"choose-the-hybridization-for-the-carbon-atom-in-co2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/choose-the-hybridization-for-the-carbon-atom-in-co2\/","title":{"rendered":"Choose the hybridization for the carbon atom in CO2"},"content":{"rendered":"\n

Choose the hybridization for the carbon atom in CO2.<\/p>\n\n\n\n

sp\u00b3d
sp2
sp\u00b3
sp<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

The correct hybridization for the carbon atom in carbon dioxide (CO\u2082) is sp<\/strong>.<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

In the CO\u2082 molecule, carbon is at the center, bonded to two oxygen atoms through double bonds. To determine the hybridization of carbon, we need to consider the following factors:<\/p>\n\n\n\n

    \n
  1. Valence Shell Electron Pair Repulsion (VSEPR) Theory<\/strong>: According to VSEPR theory, the carbon atom in CO\u2082 forms two regions of electron density, each associated with a double bond to an oxygen atom. The electron density is arranged linearly to minimize repulsion, meaning the geometry is linear, with an angle of 180\u00b0 between the bonds. This linear shape indicates that the carbon atom must have two regions of electron density, corresponding to sp hybridization<\/strong>.<\/li>\n\n\n\n
  2. Bonding and Hybridization<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • In sp hybridization, one 2s orbital combines with one 2p orbital to form two sp hybrid orbitals.<\/li>\n\n\n\n
    • These two sp hybrid orbitals are oriented linearly (180\u00b0 apart), which allows for the formation of two \u03c3 (sigma) bonds, one with each oxygen atom.<\/li>\n\n\n\n
    • The remaining unhybridized p orbitals on the carbon atom overlap with unhybridized p orbitals on the oxygen atoms to form two \u03c0 (pi) bonds, which are responsible for the double bonds in CO\u2082.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Electron Configuration<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • The carbon atom in its ground state has the electron configuration 1s\u00b2 2s\u00b2 2p\u00b2. In CO\u2082, it undergoes hybridization where the 2s and one of the 2p orbitals combine to form two sp hybrid orbitals. The other two 2p orbitals remain unhybridized and are used for the formation of the \u03c0 bonds.<\/li>\n<\/ul>\n\n\n\n

          Thus, the carbon atom in CO\u2082 adopts sp hybridization<\/strong>, which corresponds to a linear molecular shape.<\/p>\n\n\n\n

          Image of CO\u2082 Structure with Hybridization:<\/h3>\n\n\n\n

          Let me generate an image to illustrate the structure and hybridization of CO\u2082.<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Here is an illustration of the CO\u2082 molecule with the sp hybridization of the carbon atom, showing the linear structure and bond angles. The carbon atom forms sigma bonds with oxygen using sp hybrid orbitals and pi bonds using unhybridized p orbitals.<\/p>\n","protected":false},"excerpt":{"rendered":"

          Choose the hybridization for the carbon atom in CO2. sp\u00b3dsp2sp\u00b3sp The correct answer and explanation is : The correct hybridization for the carbon atom in carbon dioxide (CO\u2082) is sp. Explanation: In the CO\u2082 molecule, carbon is at the center, bonded to two oxygen atoms through double bonds. To determine the hybridization of carbon, we […]<\/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-197518","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197518","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=197518"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197518\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197518"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197518"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}