{"id":223578,"date":"2025-06-02T05:49:24","date_gmt":"2025-06-02T05:49:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223578"},"modified":"2025-06-02T05:49:27","modified_gmt":"2025-06-02T05:49:27","slug":"draw-the-lewis-structure-of-a-cobr2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/draw-the-lewis-structure-of-a-cobr2\/","title":{"rendered":"Draw the Lewis structure of: a) COBr2."},"content":{"rendered":"\n<p>Draw the Lewis structure of: a) COBr2. The Carbon isin the center and bonded to the Oxygen and to the two Bromine atoms. b) Ammonium Ion.<\/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<h3 class=\"wp-block-heading\">a) Lewis Structure of <strong>COBr\u2082<\/strong> (Carbonyl dibromide)<\/h3>\n\n\n\n<p><strong>Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>     ..\nBr \u2013 C = O\n     |\n     Br\n     ..\n<\/code><\/pre>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Carbon is the central atom and forms:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One <strong>double bond<\/strong> with oxygen (C=O)<\/li>\n\n\n\n<li>Two <strong>single bonds<\/strong> with bromine atoms (C\u2013Br)<\/li>\n<\/ul>\n\n\n\n<p><strong>Steps:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count total valence electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 4<\/li>\n\n\n\n<li>Oxygen (O): 6<\/li>\n\n\n\n<li>Bromine (Br): 7 \u00d7 2 = 14<br>\u2192 Total = 4 + 6 + 14 = <strong>24 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Connect atoms with single bonds:<\/strong>\n<ul class=\"wp-block-list\">\n<li>C\u2013O and two C\u2013Br bonds use 6 electrons (3 bonds \u00d7 2 e\u207b)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Complete the octets for O and Br:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Oxygen needs 6 more electrons (3 lone pairs)<\/li>\n\n\n\n<li>Each Br needs 6 more electrons (3 lone pairs)<\/li>\n\n\n\n<li>That\u2019s 6 + 6 + 6 = 18 electrons<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>\u2192 Total used = 6 (bonds) + 18 (lone pairs) = 24 electrons \u2014 matches total.<\/p>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li><strong>Check octets:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Carbon has only 6 electrons so far. To complete its octet, form a <strong>double bond with oxygen<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>\u2192 Final structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C=O double bond<\/li>\n\n\n\n<li>Two single bonds to Br<\/li>\n\n\n\n<li>Oxygen has 2 lone pairs<\/li>\n\n\n\n<li>Each Br has 3 lone pairs<\/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) Lewis Structure of <strong>Ammonium Ion (NH\u2084\u207a)<\/strong><\/h3>\n\n\n\n<p><strong>Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>       H\n       |\nH \u2013 N \u2013 H\n       |\n       H\n<\/code><\/pre>\n\n\n\n<p><em>The entire ion is enclosed in brackets with a \u201c\u207a\u201d charge:<\/em> &amp;nbsp;&amp;nbsp;| &amp;nbsp;&amp;nbsp;H &amp;nbsp;&amp;nbsp;H \\]\u207a **Explanation:** 1. **Count valence electrons:** &#8211; Nitrogen: 5 &#8211; Hydrogen: 1 \u00d7 4 = 4 &#8211; Positive charge = **\u20131 electron** \u2192 Total = 5 + 4 \u2013 1 = **8 electrons** 2. **Form four N\u2013H single bonds:** &#8211; 4 bonds \u00d7 2 electrons = 8 electrons \u2014 all used. 3. **No lone pairs** on nitrogen, as all electrons are in bonding. 4. **Formal charge:** &#8211; Nitrogen: normally 5 valence electrons; now surrounded by 4 bonds (0 lone pairs) \u2192 5 \u2013 4 = **+1 charge** This explains the **positive charge** on the ammonium ion. &#8212; ### Summary: &#8211; **COBr\u2082** has 24 valence electrons; carbon forms a double bond with oxygen and single bonds with two bromines. Oxygen and bromines have lone pairs; carbon satisfies the octet via a double bond. &#8211; **Ammonium ion (NH\u2084\u207a)** forms four N\u2013H bonds, uses 8 electrons, and has a formal positive charge due to the proton donation (H\u207a), leaving nitrogen without lone pairs and with a +1 charge.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-9.jpeg\" alt=\"\" class=\"wp-image-223579\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of: a) COBr2. The Carbon isin the center and bonded to the Oxygen and to the two Bromine atoms. b) Ammonium Ion. The Correct Answer and Explanation is: a) Lewis Structure of COBr\u2082 (Carbonyl dibromide) Structure: Explanation: Carbon is the central atom and forms: Steps: \u2192 Total used = 6 (bonds) [&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-223578","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223578","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=223578"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223578\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223578"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223578"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223578"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}