{"id":236130,"date":"2025-06-16T05:10:49","date_gmt":"2025-06-16T05:10:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=236130"},"modified":"2025-06-16T05:10:52","modified_gmt":"2025-06-16T05:10:52","slug":"in-the-carbon-skeleton-structure-below-identify-the-carbon-atoms-with-sp2-hybridization","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/in-the-carbon-skeleton-structure-below-identify-the-carbon-atoms-with-sp2-hybridization\/","title":{"rendered":"In the carbon skeleton structure below, identify the carbon atom(s) with sp2 hybridization"},"content":{"rendered":"\n

In the carbon skeleton structure below, identify the carbon atom(s) with sp2 hybridization. 4th attempt Jd See Periodic Table See Hint TIP: Select each sp2 hybridized carbon: Selected atoms will be highlighted in green:<\/p>\n\n\n\n

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

To determine which carbon atoms in a carbon skeleton are sp\u00b2 hybridized<\/strong>, it’s necessary to examine their bonding environment. sp\u00b2 hybridization<\/strong> occurs when a carbon atom forms three sigma (\u03c3) bonds<\/strong> and one pi (\u03c0) bond<\/strong>, typically seen in double bonds<\/strong> or in aromatic rings<\/strong> like benzene.<\/p>\n\n\n\n

\n\n\n\n

Correct Answer<\/strong><\/h3>\n\n\n\n

The sp\u00b2 hybridized carbon atoms<\/strong> in a structure include:<\/p>\n\n\n\n

    \n
  • Each carbon involved in a C=C double bond.<\/strong><\/li>\n\n\n\n
  • Each carbon in a benzene ring or any conjugated planar ring structure.<\/strong><\/li>\n<\/ul>\n\n\n\n

    These carbon atoms form three sigma bonds (including bonds to other carbon or hydrogen atoms) and have an unhybridized p orbital that overlaps to form a pi bond.<\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/strong><\/h3>\n\n\n\n

    Hybridization is a concept in valence bond theory that explains the shapes of molecules based on the mixing of atomic orbitals. For carbon, three major hybridization states are possible: sp\u00b3<\/strong>, sp\u00b2<\/strong>, and sp<\/strong>.<\/p>\n\n\n\n

    The sp\u00b2 hybridization<\/strong> involves one s orbital and two p orbitals combining to form three sp\u00b2 hybrid orbitals. This configuration leads to a trigonal planar geometry<\/strong> with 120\u00b0 bond angles<\/strong>. The remaining unhybridized p orbital<\/strong> participates in the formation of a pi bond<\/strong>, which is the defining feature of a double bond<\/strong>.<\/p>\n\n\n\n

    In carbon skeleton structures, sp\u00b2 carbon atoms are easily recognized by their involvement in double bonds<\/strong> or aromatic rings<\/strong>. In double bonds, one bond is a sigma bond (from the overlap of sp\u00b2 orbitals), and the other is a pi bond (from the side-to-side overlap of p orbitals). In aromatic systems, such as benzene, each carbon is bonded to two adjacent carbon atoms and one hydrogen, with a delocalized pi system spread across the ring.<\/p>\n\n\n\n

    By visually analyzing the carbon structure, carbon atoms participating in double bonds or aromatic rings should be identified. These atoms have precisely three regions of electron density\u2014consistent with sp\u00b2 hybridization.<\/p>\n\n\n\n

    Therefore, all carbon atoms in double bonds or aromatic rings must be selected, as they conform to the bonding pattern and geometry characteristic of sp\u00b2 hybridization.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    In the carbon skeleton structure below, identify the carbon atom(s) with sp2 hybridization. 4th attempt Jd See Periodic Table See Hint TIP: Select each sp2 hybridized carbon: Selected atoms will be highlighted in green: The Correct Answer and Explanation is: To determine which carbon atoms in a carbon skeleton are sp\u00b2 hybridized, it’s necessary to […]<\/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-236130","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236130","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=236130"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236130\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=236130"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=236130"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=236130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}