The difference in bond angles between formaldehyde (COH\u2082)<\/strong> and carbonyl fluoride (COF\u2082)<\/strong> arises from variations in electronegativity<\/strong> and electron repulsion<\/strong> around the central carbon atom.<\/p>\n\n\n\n The H\u2013C\u2013H bond angle in COH\u2082 is approximately 117\u2013118\u00b0<\/strong>, which is slightly less than the ideal 120\u00b0<\/strong> for a trigonal planar geometry. In contrast, the F\u2013C\u2013F bond angle in COF\u2082 is around 108\u00b0<\/strong>, a larger deviation from 120\u00b0. This difference is due to the greater lone-pair and bond-pair repulsion in COF\u2082<\/strong> caused by the highly electronegative fluorine atoms pulling bonding electrons closer to themselves, reducing electron density between carbon and fluorine and increasing repulsion between the C=O and C\u2013F bonds.<\/p>\n\n\n\n Both formaldehyde (COH\u2082)<\/strong> and carbonyl fluoride (COF\u2082)<\/strong> have a trigonal planar structure<\/strong> around the central carbon atom, which is sp\u00b2 hybridized. In an ideal trigonal planar molecule, bond angles would be 120\u00b0<\/strong>. However, actual bond angles deviate from this due to differences in the atoms attached to the central atom and their electron-withdrawing effects.<\/p>\n\n\n\n In COH\u2082<\/strong>, the H\u2013C\u2013H angle is around 117\u2013118\u00b0<\/strong>. Hydrogen atoms are small and have low electronegativity, which means they do not strongly attract bonding electrons. This results in relatively weak repulsion between the C\u2013H bonds. The slightly smaller angle than 120\u00b0 is mainly due to the electron-withdrawing effect of the carbonyl (C=O) group, which pulls electron density away from the bonding region, causing a slight compression of the H\u2013C\u2013H angle.<\/p>\n\n\n\n In COF\u2082<\/strong>, the F\u2013C\u2013F bond angle is about 108\u00b0<\/strong>, a more significant reduction from the ideal 120\u00b0. Fluorine is highly electronegative, and it pulls electron density strongly toward itself, reducing bonding electron density near carbon. This makes the bonding electron clouds<\/strong> more localized around fluorine, which increases repulsion between the F\u2013C and C=O bonds<\/strong>. Additionally, because fluorine atoms have lone pairs<\/strong>, these contribute to greater lone-pair\u2013bond-pair repulsion<\/strong>, further compressing the F\u2013C\u2013F angle.<\/p>\n\n\n\n Thus, the larger deviation in COF\u2082 compared to COH\u2082 is primarily due to the higher electronegativity and lone pairs of fluorine<\/strong>, which create greater electron repulsion and result in a smaller bond angle.<\/p>\n\n\n\n The H-C-H bond angle in COH2 is about 117-118\u00c2\u00b0, while the F-C-F bond angle in COF2 is 108\u00c2\u00b0. Explain the difference in the deviations from the expected 120\u00c2\u00b0 and explain the source of the difference in the two molecules. The Correct Answer and Explanation is: The difference in bond angles between formaldehyde (COH\u2082) and carbonyl […]<\/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-227319","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227319","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=227319"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227319\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227319"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227319"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227319"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nCorrect Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"