{"id":203913,"date":"2025-03-21T02:51:44","date_gmt":"2025-03-21T02:51:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=203913"},"modified":"2025-03-21T02:51:47","modified_gmt":"2025-03-21T02:51:47","slug":"question-7-what-is-the-molecular-geometry-of-the-sbr6-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/21\/question-7-what-is-the-molecular-geometry-of-the-sbr6-molecule\/","title":{"rendered":"QUESTION 7 What Is The Molecular Geometry Of The SBr6 Molecule"},"content":{"rendered":"\n<p>QUESTION 7 What Is The Molecular Geometry Of The SBr6 Molecule? Octahedral Tetrahedral O Trigonal Pyramidal O Trigonal Bipyramidal Bent<\/p>\n\n\n\n<p>QUESTION 8 Which Of The Following Has A Net Dipole Moment? OBF3 O SO2 CO2 O SBr6 O CHA<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>QUESTION 7: Molecular Geometry of SBr\u2086<\/strong><br><strong>Correct Answer: Octahedral<\/strong><\/p>\n\n\n\n<p>The sulfur hexabromide (SBr\u2086) molecule follows the octahedral molecular geometry. This is because sulfur (S) is the central atom, surrounded by six bromine (Br) atoms. The molecular geometry is determined by the number of bonding and lone pairs around the central atom.<\/p>\n\n\n\n<p>To determine the geometry, we use the VSEPR (Valence Shell Electron Pair Repulsion) theory:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sulfur belongs to Group 16 and has six valence electrons.<\/li>\n\n\n\n<li>Each bromine atom contributes one electron through bonding, forming six single bonds.<\/li>\n\n\n\n<li>Since sulfur forms six sigma bonds and has no lone pairs, the electron pair arrangement is <strong>octahedral<\/strong>, meaning the bond angles are approximately <strong>90\u00b0<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the correct answer is <strong>octahedral<\/strong>, as all six bromine atoms are symmetrically arranged around sulfur.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>QUESTION 8: Molecule with a Net Dipole Moment<\/strong><br><strong>Correct Answer: SO\u2082<\/strong><\/p>\n\n\n\n<p>A molecule has a net dipole moment when it has <strong>polar bonds<\/strong> and an <strong>asymmetrical shape<\/strong>, leading to an uneven distribution of charge. Let\u2019s analyze the given molecules:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>BF\u2083 (Boron Trifluoride):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Geometry:<\/strong> Trigonal planar<\/li>\n\n\n\n<li><strong>Dipole Moment:<\/strong> <strong>No<\/strong>, because the dipoles cancel out due to symmetry.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>SO\u2082 (Sulfur Dioxide):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Geometry:<\/strong> Bent<\/li>\n\n\n\n<li><strong>Dipole Moment:<\/strong> <strong>Yes<\/strong>, because the two polar S=O bonds do not cancel out.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>CO\u2082 (Carbon Dioxide):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Geometry:<\/strong> Linear<\/li>\n\n\n\n<li><strong>Dipole Moment:<\/strong> <strong>No<\/strong>, since the bond dipoles cancel.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>SBr\u2086 (Sulfur Hexabromide):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Geometry:<\/strong> Octahedral<\/li>\n\n\n\n<li><strong>Dipole Moment:<\/strong> <strong>No<\/strong>, as the symmetry cancels dipoles.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>CH\u2084 (Methane):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Geometry:<\/strong> Tetrahedral<\/li>\n\n\n\n<li><strong>Dipole Moment:<\/strong> <strong>No<\/strong>, as all bonds are identical.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Thus, <strong>SO\u2082 is the only molecule with a net dipole moment.<\/strong><\/p>\n\n\n\n<p>Now, generating an image for <strong>SBr\u2086&#8217;s octahedral shape<\/strong>:<\/p>\n\n\n\n<p>Here is the 3D molecular model of sulfur hexabromide (SBr\u2086) with its octahedral geometry. Let me know if you need any modifications!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-1131.png\" alt=\"\" class=\"wp-image-203914\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>QUESTION 7 What Is The Molecular Geometry Of The SBr6 Molecule? Octahedral Tetrahedral O Trigonal Pyramidal O Trigonal Bipyramidal Bent QUESTION 8 Which Of The Following Has A Net Dipole Moment? OBF3 O SO2 CO2 O SBr6 O CHA The correct answer and explanation is: QUESTION 7: Molecular Geometry of SBr\u2086Correct Answer: Octahedral The sulfur [&hellip;]<\/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-203913","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203913","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=203913"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203913\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203913"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203913"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203913"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}