{"id":182547,"date":"2025-01-14T07:58:56","date_gmt":"2025-01-14T07:58:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182547"},"modified":"2025-01-14T07:58:58","modified_gmt":"2025-01-14T07:58:58","slug":"use-the-lewis-dot-structure-to-predict-the-electron-domain-geometry-and-molecular-geometry-for-the-following-molecules-or-ions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/14\/use-the-lewis-dot-structure-to-predict-the-electron-domain-geometry-and-molecular-geometry-for-the-following-molecules-or-ions\/","title":{"rendered":"Use the Lewis dot structure to predict the electron domain geometry and molecular geometry for the following molecules or ions"},"content":{"rendered":"\n<p>Use the Lewis dot structure to predict the electron domain geometry and molecular geometry for the following molecules or ions. Determine the molecular polarity by showing the bond dipole moment if the molecules have polar bonds and the direction of the net dipole if any. a. BeCl2 b. XeF2 C. SCI2<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a. <strong>BeCl\u2082 (Beryllium Chloride)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Dot Structure:<\/strong><\/li>\n\n\n\n<li>Beryllium (Be) is in Group 2, so it has two valence electrons.<\/li>\n\n\n\n<li>Chlorine (Cl) is in Group 17, so it has seven valence electrons. Each chlorine atom forms a single bond with beryllium, using one of its electrons.<\/li>\n\n\n\n<li>This results in the following structure: <code>Cl-Be-Cl<\/code><\/li>\n\n\n\n<li><strong>Electron Domain Geometry:<\/strong><br>Beryllium (Be) has two bonding pairs and no lone pairs, so the electron domain geometry is <strong>linear<\/strong>.<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong><br>The molecular geometry is also <strong>linear<\/strong> because there are no lone pairs on the beryllium atom to create any repulsion that would alter the bond angles.<\/li>\n\n\n\n<li><strong>Polarity:<\/strong><br>Each Be-Cl bond is polar because chlorine is more electronegative than beryllium, resulting in a bond dipole pointing toward the chlorine atom. However, since the molecule is linear, the dipoles cancel each other out. Thus, <strong>BeCl\u2082 is nonpolar<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">b. <strong>XeF\u2082 (Xenon Difluoride)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Dot Structure:<\/strong><\/li>\n\n\n\n<li>Xenon (Xe) is in Group 18, so it has eight valence electrons. Fluorine (F) is in Group 17, so it has seven valence electrons.<\/li>\n\n\n\n<li>Two fluorine atoms form single bonds with xenon, using two of xenon&#8217;s electrons, leaving three lone pairs on xenon.<\/li>\n\n\n\n<li>The structure is: <code>F | F - Xe | (lone pairs on Xe)<\/code><\/li>\n\n\n\n<li><strong>Electron Domain Geometry:<\/strong><br>Xenon has five electron domains (two bonding pairs and three lone pairs). The electron domain geometry is <strong>trigonal bipyramidal<\/strong>.<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong><br>Since there are three lone pairs, the molecular geometry is <strong>linear<\/strong>, with the two fluorine atoms positioned opposite each other along the axial positions of the trigonal bipyramidal structure.<\/li>\n\n\n\n<li><strong>Polarity:<\/strong><br>The Xe-F bond is polar, with the bond dipoles pointing toward the fluorine atoms. Since the molecular geometry is linear, the dipoles do not cancel, resulting in a <strong>polar molecule<\/strong> with a net dipole moment pointing from xenon to fluorine.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">c. <strong>SCI\u2082 (Sulfur Dichloride Iodide)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Dot Structure:<\/strong><\/li>\n\n\n\n<li>Sulfur (S) is in Group 16, so it has six valence electrons. Chlorine (Cl) and iodine (I) are both in Group 17, each having seven valence electrons.<\/li>\n\n\n\n<li>The sulfur atom forms two single bonds, one with chlorine and one with iodine, using two of its electrons.<\/li>\n\n\n\n<li>The structure is: <code>Cl | S - I<\/code><\/li>\n\n\n\n<li><strong>Electron Domain Geometry:<\/strong><br>Sulfur has three electron domains (one for each bond and one for the lone pair). The electron domain geometry is <strong>trigonal planar<\/strong>.<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong><br>The molecular geometry is <strong>bent<\/strong> due to the lone pair on sulfur, which pushes the bonding atoms closer together.<\/li>\n\n\n\n<li><strong>Polarity:<\/strong><br>Both the S-Cl and S-I bonds are polar because of the difference in electronegativity between sulfur and the halogens. The bond dipoles point toward chlorine and iodine. Since the molecule is bent, the dipoles do not cancel, resulting in a <strong>polar molecule<\/strong> with a net dipole moment pointing toward the more electronegative atoms (chlorine and iodine).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Use the Lewis dot structure to predict the electron domain geometry and molecular geometry for the following molecules or ions. Determine the molecular polarity by showing the bond dipole moment if the molecules have polar bonds and the direction of the net dipole if any. a. BeCl2 b. XeF2 C. SCI2 The Correct Answer and [&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-182547","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182547","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=182547"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182547\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}