Determine the hybridization about carbon in H2CO.
a) sp
b) sp2
c) sp3
d) sp3d<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is b) sp2<\/strong>.<\/p>\n\n\n\n In formaldehyde (H\u2082CO), the central carbon atom is bonded to two hydrogen atoms and one oxygen atom (via a double bond). To determine the hybridization of the carbon atom in H\u2082CO, we need to look at its bonding and the electron geometry.<\/p>\n\n\n\n Since the electron geometry is trigonal planar, the carbon in formaldehyde undergoes sp\u00b2 hybridization.<\/p>\n\n\n\n Thus, the correct hybridization for the carbon atom in H\u2082CO is sp\u00b2<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Determine the hybridization about carbon in H2CO.a) spb) sp2c) sp3d) sp3d The Correct Answer and Explanation is : The correct answer is b) sp2. Explanation: In formaldehyde (H\u2082CO), the central carbon atom is bonded to two hydrogen atoms and one oxygen atom (via a double bond). To determine the hybridization of the carbon atom in […]<\/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-189093","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189093","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=189093"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189093\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189093"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189093"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189093"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
The carbon atom in H\u2082CO is bonded to three atoms: two hydrogens and one oxygen. The carbon forms:<\/li>\n<\/ol>\n\n\n\n\n
Therefore, the carbon atom has three regions of electron density (two from the hydrogens and one from the oxygen double bond).<\/li>\n<\/ul>\n\n\n\n\n
With three regions of electron density, the electron geometry around the carbon atom is trigonal planar<\/strong>. This means that the angle between the bonds is 120\u00b0, which is characteristic of sp\u00b2 hybridization.<\/li>\n\n\n\n
To form three regions of electron density (three bonds), the carbon atom mixes one s orbital and two p orbitals. This creates three sp\u00b2 hybrid orbitals. These hybrid orbitals overlap with the orbitals of hydrogen and oxygen to form sigma bonds. The remaining unhybridized p orbital on carbon forms the pi bond in the double bond with oxygen.<\/li>\n<\/ol>\n\n\n\n