Correct Answer:<\/strong> Explanation:<\/strong><\/p>\n\n\n\n The carbon atom in formaldehyde (CH\u2082O) forms one double bond with oxygen and two single bonds with hydrogen. This results in three regions of electron density around carbon, giving it a trigonal planar geometry<\/strong>. To accommodate this shape, carbon undergoes sp\u00b2 hybridization<\/strong>, where one s orbital and two p orbitals mix to form three sp\u00b2 orbitals. These hybrid orbitals form three sigma (\u03c3) bonds<\/strong>: two with hydrogen and one with oxygen. The second bond in the carbon\u2013oxygen double bond is a pi (\u03c0) bond, which does not involve hybrid orbitals.<\/p>\n\n\n\n The ideal bond angle in trigonal planar geometry is 120\u00b0<\/strong>. However, due to the presence of a double bond (which has more electron density and repulsion), the H\u2013C\u2013H angle becomes slightly less than 120\u00b0<\/strong>, and the H\u2013C=O angles become slightly more than 120\u00b0<\/strong>.<\/p>\n\n\n\n The formal charges in the best Lewis structure of CH\u2082O are all zero<\/strong>. Carbon has four bonds (zero formal charge), oxygen has two bonds and two lone pairs (zero formal charge), and each hydrogen has one bond (zero formal charge).<\/p>\n\n\n\n Therefore, all the given statements about CH\u2082O are true.<\/p>\n\n\n\n In the ICl\u2082\u207b ion, iodine is the central atom, surrounded by two bonding pairs and three lone pairs<\/strong>, making a total of five electron domains. This corresponds to a trigonal bipyramidal electron geometry<\/strong>, where the three lone pairs occupy the equatorial positions<\/strong> to minimize repulsion. The molecule itself has a linear shape<\/strong>, with the two bonding pairs 180\u00b0 apart.<\/p>\n\n\n\n The hybridization of iodine in ICl\u2082\u207b is sp\u00b3d<\/strong>, not sp\u00b2. Also, the angle between lone pairs is 120\u00b0<\/strong> in the equatorial plane, not 109.5\u00b0<\/strong>, which is associated with sp\u00b3<\/strong> hybridization.<\/p>\n\n\n\n Thus, the statement about ICl\u2082\u207b is false<\/strong>.<\/p>\n\n\n\n What is true about the CH2O molecule? Carbon utilizes sp2 hybridization. The H-C-H bond angle is expected to be slightly less than the H-C-O bond angle. The molecule contains 3 sigma bonds. The formal charge of all atoms equals zero in the best Lewis structure. All of the responses above are true. QUESTION 25 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-238458","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238458","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=238458"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238458\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238458"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238458"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238458"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
All of the responses above are true<\/strong> (for CH\u2082O).
For the ICl\u2082\u207b ion statement: It is false<\/strong> \u2013 the lone pairs occupy sp\u00b3d<\/strong> hybrid orbitals and are oriented 120\u00b0 and 180\u00b0<\/strong>, not 109.5\u00b0 from each other.<\/p>\n\n\n\n
\n\n\n\nPart 1: CH\u2082O Molecule (Formaldehyde)<\/strong><\/h3>\n\n\n\n
\n\n\n\nPart 2: ICl\u2082\u207b Ion<\/strong><\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-742.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"