What is the O-S-O bond angle in the molecule of SO2? <\/p>\n\n\n\n
A. 180 <\/p>\n\n\n\n
B. 130 <\/p>\n\n\n\n
C. 119 <\/p>\n\n\n\n
D. 150<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is C. 119\u00b0<\/strong>.<\/p>\n\n\n\n The bond angle in sulfur dioxide (SO\u2082) is approximately 119\u00b0<\/strong>. This value is primarily determined by the molecule\u2019s shape and electron pair geometry, which can be understood using Valence Shell Electron Pair Repulsion (VSEPR) theory<\/strong>.<\/p>\n\n\n\n Thus, the bond angle in SO\u2082 is close to 119\u00b0, which makes option C<\/strong> the correct choice.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the O-S-O bond angle in the molecule of SO2? A. 180 B. 130 C. 119 D. 150 The Correct Answer and Explanation is : The correct answer is C. 119\u00b0. Explanation: The bond angle in sulfur dioxide (SO\u2082) is approximately 119\u00b0. This value is primarily determined by the molecule\u2019s shape and electron pair […]<\/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-152321","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152321","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=152321"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152321\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=152321"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=152321"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=152321"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
SO\u2082 has a bent (or V-shaped) molecular geometry. The sulfur atom (S) is the central atom, and it forms two double bonds with two oxygen atoms (O). The sulfur atom also has a lone pair of electrons, which influences the molecular geometry.<\/li>\n\n\n\n
In VSEPR theory, the molecular shape is determined by the repulsion between electron pairs around the central atom. The sulfur atom in SO\u2082 has three regions of electron density: two bonding pairs (the S=O bonds) and one lone pair. This gives a total of three electron domains, which corresponds to a trigonal planar electron domain geometry<\/strong>.<\/li>\n\n\n\n
In an ideal trigonal planar geometry (like in a molecule such as BF\u2083), the bond angles between the three electron regions would be 120\u00b0. However, lone pairs exert greater repulsion than bonding pairs because lone pairs are closer to the nucleus and spread out more. This additional repulsion from the lone pair pushes the bonding pairs slightly closer together, reducing the bond angle from the ideal 120\u00b0 to about 119\u00b0.<\/li>\n\n\n\n
As a result, SO\u2082 adopts a bent molecular geometry with an O-S-O bond angle of approximately 119\u00b0<\/strong>. This angle is slightly less than 120\u00b0 due to the lone pair-bond pair repulsion, which compresses the angle between the two S=O bonds.<\/li>\n<\/ol>\n\n\n\n