{"id":261417,"date":"2025-07-20T06:18:00","date_gmt":"2025-07-20T06:18:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261417"},"modified":"2025-07-20T06:18:03","modified_gmt":"2025-07-20T06:18:03","slug":"sketch-the-lewis-dot-structure-for-each-of-the-following-compounds-and-state-the-molecular-geometry-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/sketch-the-lewis-dot-structure-for-each-of-the-following-compounds-and-state-the-molecular-geometry-2\/","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-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let\u2019s walk through the Lewis structures and molecular geometries for each of the compounds:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">i. <strong>BeF2 (Beryllium Fluoride)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Beryllium (Be) is in Group 2 and has 2 valence electrons. Fluorine (F) is in Group 17 and has 7 valence electrons.<\/li>\n\n\n\n<li>Be will form single bonds with two fluorine atoms. Each F will share one electron to form a bond with Be.<\/li>\n\n\n\n<li>The remaining electrons will be placed as lone pairs on the fluorine atoms.<\/li>\n<\/ol>\n\n\n\n<p><strong>Lewis Structure<\/strong>:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">t<code>F\u2014Be\u2014F<br><\/code><\/pre>\n\n\n\n<p>Each F has 3 lone pairs, and Be has no lone pairs.<\/p>\n\n\n\n<p><strong>Molecular Geometry<\/strong>: The electron pairs around Be are in a linear arrangement, with a bond angle of 180\u00b0.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">ii. <strong>BCl3 (Boron Trichloride)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Boron (B) is in Group 13 and has 3 valence electrons. Chlorine (Cl) is in Group 17 and has 7 valence electrons.<\/li>\n\n\n\n<li>Boron forms three single bonds with three chlorine atoms. Each Cl shares one electron with B, and the remaining electrons are placed as lone pairs on Cl.<\/li>\n<\/ol>\n\n\n\n<p><strong>Lewis Structure<\/strong>:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><code>  Cl<br>   |<br>Cl\u2014B\u2014Cl<br><\/code><\/pre>\n\n\n\n<p>Each Cl has 3 lone pairs, and B has no lone pairs.<\/p>\n\n\n\n<p><strong>Molecular Geometry<\/strong>: The electron pairs around B form a trigonal planar geometry, with bond angles of 120\u00b0.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">iii. <strong>CCl4 (Carbon Tetrachloride)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Carbon (C) is in Group 14 and has 4 valence electrons. Chlorine (Cl) has 7 valence electrons.<\/li>\n\n\n\n<li>Carbon forms four single bonds with four chlorine atoms. Each Cl shares one electron with C, and the remaining electrons are placed as lone pairs on Cl.<\/li>\n<\/ol>\n\n\n\n<p><strong>Lewis Structure<\/strong>:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><code>   Cl<br>    |<br>Cl\u2014C\u2014Cl<br>    |<br>   Cl<br><\/code><\/pre>\n\n\n\n<p>Each Cl has 3 lone pairs, and C has no lone pairs.<\/p>\n\n\n\n<p><strong>Molecular Geometry<\/strong>: The electron pairs around C form a tetrahedral geometry, with bond angles of 109.5\u00b0.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">iv. <strong>PBr5 (Phosphorus Pentabromide)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Phosphorus (P) is in Group 15 and has 5 valence electrons. Bromine (Br) is in Group 17 and has 7 valence electrons.<\/li>\n\n\n\n<li>Phosphorus forms five single bonds with five bromine atoms. Each Br shares one electron with P, and the remaining electrons are placed as lone pairs on Br.<\/li>\n<\/ol>\n\n\n\n<p><strong>Lewis Structure<\/strong>:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><code>    Br<br>    |<br>Br\u2014P\u2014Br<br>    |<br>    Br<br>    |<br>   Br<br><\/code><\/pre>\n\n\n\n<p>Each Br has 3 lone pairs, and P has no lone pairs.<\/p>\n\n\n\n<p><strong>Molecular Geometry<\/strong>: The electron pairs around P form a trigonal bipyramidal geometry, with bond angles of 90\u00b0, 120\u00b0, and 180\u00b0.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Summary of Molecular Geometries:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>BeF2<\/strong>: Linear (180\u00b0)<\/li>\n\n\n\n<li><strong>BCl3<\/strong>: Trigonal Planar (120\u00b0)<\/li>\n\n\n\n<li><strong>CCl4<\/strong>: Tetrahedral (109.5\u00b0)<\/li>\n\n\n\n<li><strong>PBr5<\/strong>: Trigonal Bipyramidal (90\u00b0, 120\u00b0, 180\u00b0)<\/li>\n<\/ul>\n\n\n\n<p>These geometries are determined by the number of bonding electron pairs and lone pairs around the central atom. The repulsion between these electron pairs arranges them to minimize the overall energy of the molecule.<\/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-banner6-1147.jpeg\" alt=\"\" class=\"wp-image-261418\"\/><\/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: Let\u2019s walk through the Lewis structures and molecular geometries for each of the compounds: i. BeF2 (Beryllium Fluoride) Lewis Structure: Lewis Structure: tF\u2014Be\u2014F Each F [&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-261417","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261417","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=261417"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261417\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261417"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261417"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261417"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}