{"id":43183,"date":"2025-06-29T14:34:10","date_gmt":"2025-06-29T14:34:10","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43183"},"modified":"2025-06-29T14:34:11","modified_gmt":"2025-06-29T14:34:11","slug":"determine-the-electron-pair-geometry-and-molecular-shape-of-cbrusing-lewis-structure","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-electron-pair-geometry-and-molecular-shape-of-cbrusing-lewis-structure\/","title":{"rendered":"Determine the electron pair geometry and molecular shape of CBrusing Lewis structure."},"content":{"rendered":"\n<p>Determine the electron pair geometry and molecular shape of CBr<br>using Lewis structure. Are the bonds in this molecule polar or nonpolar? Is the overall molecule polar or nonpolar? Explain your answer.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s break this down step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Lewis Structure of CBr<\/h3>\n\n\n\n<p>Carbon (C) is in Group 14, and bromine (Br) is in Group 17 of the periodic table. For the molecule <strong>CBr<\/strong>, we assume a simple C\u2013Br single bond.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Valence electrons<\/strong>:<ul><li>Carbon has 4 valence electrons.<\/li><li>Bromine has 7 valence electrons.<\/li><\/ul>So, in total, the molecule has <strong>4 + 7 = 11 valence electrons<\/strong>.<\/li>\n\n\n\n<li><strong>Drawing the structure<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Carbon will form a single bond with bromine. Each bond involves two electrons, so that&#8217;s 2 electrons used.<\/li>\n\n\n\n<li>After bonding, carbon has 2 electrons left to complete its valence shell.<\/li>\n\n\n\n<li>Bromine has 5 remaining electrons, so it will form 3 lone pairs.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>So, the structure is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">rubyCopyEdit<code>:Br - C:\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon has 4 valence electrons, 2 of which are used in the bond with bromine, leaving 2 electrons as a lone pair.<\/li>\n\n\n\n<li>Bromine has 7 valence electrons, 2 used for bonding, leaving 5 electrons as 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Electron Pair Geometry and Molecular Shape<\/h3>\n\n\n\n<p>Now, let&#8217;s determine the geometry of the molecule.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron pairs around carbon<\/strong>: There are <strong>two electron pairs<\/strong> around the carbon atom (one bonding pair with bromine and one lone pair).<\/li>\n\n\n\n<li>Since we have two regions of electron density, the <strong>electron pair geometry<\/strong> will be <strong>linear<\/strong>.<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: When considering the molecular shape (which only accounts for the positions of the atoms), the shape will be <strong>linear<\/strong> as well. The lone pair on carbon doesn\u2019t influence the shape in this case because it\u2019s just one lone pair.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Bond Polarity<\/h3>\n\n\n\n<p>Next, let&#8217;s examine the bond polarity. The <strong>C\u2013Br bond<\/strong> is <strong>polar<\/strong> because bromine is more electronegative than carbon. This creates a dipole where the electron density is shifted towards the bromine atom.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Overall Molecular Polarity<\/h3>\n\n\n\n<p>To determine whether the overall molecule is polar or nonpolar, we consider the vector sum of the bond dipoles.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since the molecule is linear and there is a polar bond (C\u2013Br), the overall molecular dipole will be in the direction of the C\u2013Br bond, making the molecule <strong>polar<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron pair geometry<\/strong>: Linear<\/li>\n\n\n\n<li><strong>Molecular shape<\/strong>: Linear<\/li>\n\n\n\n<li><strong>Bond polarity<\/strong>: The C\u2013Br bond is polar.<\/li>\n\n\n\n<li><strong>Overall molecular polarity<\/strong>: The molecule is <strong>polar<\/strong> because it has a polar bond and there is no symmetry to cancel out the dipole.<\/li>\n<\/ul>\n\n\n\n<p>This explanation shows that <strong>CBr<\/strong> (presumably meant to represent a molecule like <strong>CBr4<\/strong>) has polar bonds, and the molecular polarity depends on its shape and symmetry.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-433.jpeg\" alt=\"\" class=\"wp-image-43184\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-433.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-433-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-433-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-433-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the electron pair geometry and molecular shape of CBrusing Lewis structure. Are the bonds in this molecule polar or nonpolar? Is the overall molecule polar or nonpolar? Explain your answer. The Correct Answer and Explanation is: Let&#8217;s break this down step by step: Step 1: Lewis Structure of CBr Carbon (C) is in Group [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-43183","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43183","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=43183"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43183\/revisions"}],"predecessor-version":[{"id":43185,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43183\/revisions\/43185"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43183"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43183"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}