{"id":257284,"date":"2025-07-16T21:04:02","date_gmt":"2025-07-16T21:04:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257284"},"modified":"2025-07-16T21:04:04","modified_gmt":"2025-07-16T21:04:04","slug":"sketch-the-lewis-dot-structure-for-each-of-the-following-compounds-and-state-the-molecular-geometry","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/sketch-the-lewis-dot-structure-for-each-of-the-following-compounds-and-state-the-molecular-geometry\/","title":{"rendered":"Sketch the Lewis Dot Structure for each of the following compounds and state the molecular geometry."},"content":{"rendered":"\n<p>Sketch the Lewis Dot Structure for each of the following compounds and state the molecular geometry. i. BeF2 ii. BCl3 iii. CCl4 iv. PBr5<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the Lewis structures and molecular geometries for each of the compounds you mentioned:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">i. <strong>BeF2 (Beryllium fluoride)<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Lewis Structure:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Beryllium (Be)<\/strong> is in Group 2 and has 2 valence electrons.<\/li>\n\n\n\n<li><strong>Fluorine (F)<\/strong> is in Group 17 and has 7 valence electrons each.<\/li>\n\n\n\n<li>Be will bond with two fluorine atoms, and each fluorine will form a single bond with Be, using one of their lone pairs. Each fluorine will also have three lone pairs remaining.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure of BeF2 is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">rCopyEdit<code>   F\u2014Be\u2014F\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Beryllium is the central atom, and each fluorine is attached by a single bond.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Geometry:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular geometry is <strong>linear<\/strong>.<\/li>\n\n\n\n<li>Be has two bonding pairs and no lone pairs. According to the VSEPR theory, this results in a <strong>180\u00b0<\/strong> bond angle.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">ii. <strong>BCl3 (Boron trichloride)<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Lewis Structure:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Boron (B)<\/strong> is in Group 13 and has 3 valence electrons.<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong> is in Group 17 and has 7 valence electrons each.<\/li>\n\n\n\n<li>Boron will form three single bonds with three chlorine atoms, and each chlorine will have three lone pairs remaining.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure of BCl3 is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">lessCopyEdit<code>    Cl\n     |\n  Cl\u2014B\u2014Cl\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Boron is the central atom, and each chlorine is bonded to it by a single bond.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Geometry:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular geometry is <strong>trigonal planar<\/strong>.<\/li>\n\n\n\n<li>Boron has three bonding pairs and no lone pairs. This gives a <strong>120\u00b0<\/strong> bond angle between the chlorine atoms.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">iii. <strong>CCl4 (Carbon tetrachloride)<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Lewis Structure:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> is in Group 14 and has 4 valence electrons.<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong> is in Group 17 and has 7 valence electrons each.<\/li>\n\n\n\n<li>Carbon will form four single bonds with four chlorine atoms, and each chlorine will have three lone pairs remaining.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure of CCl4 is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">markdownCopyEdit<code>       Cl\n       |\n  Cl\u2014C\u2014Cl\n       |\n       Cl\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is the central atom, and each chlorine is bonded to it by a single bond.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Geometry:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular geometry is <strong>tetrahedral<\/strong>.<\/li>\n\n\n\n<li>Carbon has four bonding pairs and no lone pairs. This gives a <strong>109.5\u00b0<\/strong> bond angle between the chlorine atoms.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">iv. <strong>PBr5 (Phosphorus pentabromide)<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Lewis Structure:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Phosphorus (P)<\/strong> is in Group 15 and has 5 valence electrons.<\/li>\n\n\n\n<li><strong>Bromine (Br)<\/strong> is in Group 17 and has 7 valence electrons each.<\/li>\n\n\n\n<li>Phosphorus will form five single bonds with five bromine atoms, and each bromine will have three lone pairs remaining.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure of PBr5 is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">lessCopyEdit<code>       Br\n       |\n  Br\u2014P\u2014Br\n       |\n  Br\u2014P\u2014Br\n       |\n       Br\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Phosphorus is the central atom, and each bromine is bonded to it by a single bond.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Geometry:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular geometry is <strong>trigonal bipyramidal<\/strong>.<\/li>\n\n\n\n<li>Phosphorus has five bonding pairs and no lone pairs. The bond angles are <strong>90\u00b0<\/strong> between atoms in the axial positions and <strong>120\u00b0<\/strong> between atoms in the equatorial positions.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Molecular Geometries:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>BeF2<\/strong>: Linear (180\u00b0 bond angle)<\/li>\n\n\n\n<li><strong>BCl3<\/strong>: Trigonal planar (120\u00b0 bond angle)<\/li>\n\n\n\n<li><strong>CCl4<\/strong>: Tetrahedral (109.5\u00b0 bond angle)<\/li>\n\n\n\n<li><strong>PBr5<\/strong>: Trigonal bipyramidal (90\u00b0 and 120\u00b0 bond angles)<\/li>\n<\/ul>\n\n\n\n<p>These geometries are determined by the number of bonding pairs and lone pairs around the central atom, as predicted by VSEPR theory.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-739.jpeg\" alt=\"\" class=\"wp-image-257285\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Sketch the Lewis Dot Structure for each of the following compounds and state the molecular geometry. i. BeF2 ii. BCl3 iii. CCl4 iv. PBr5 The Correct Answer and Explanation is: Here are the Lewis structures and molecular geometries for each of the compounds you mentioned: i. BeF2 (Beryllium fluoride) Lewis Structure: The Lewis structure of [&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-257284","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257284","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=257284"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257284\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}