{"id":146389,"date":"2024-09-30T19:38:00","date_gmt":"2024-09-30T19:38:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=146389"},"modified":"2024-09-30T19:38:03","modified_gmt":"2024-09-30T19:38:03","slug":"draw-the-lewis-structure-of-carbon-dioxide-co%e2%82%82-and-then-determine-if-the-molecule-is-polar-ornonpolar","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/09\/30\/draw-the-lewis-structure-of-carbon-dioxide-co%e2%82%82-and-then-determine-if-the-molecule-is-polar-ornonpolar\/","title":{"rendered":"Draw the Lewis structure of carbon dioxide (CO\u2082) and then determine if the molecule is polar ornonpolar"},"content":{"rendered":"\n<p>Draw the Lewis structure of carbon dioxide (CO\u2082) and then determine if the molecule is polar or<br>nonpolar.<br>+<br>A) Nonpolar<br>B) Polar<\/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\">Lewis Structure of Carbon Dioxide (CO\u2082)<\/h3>\n\n\n\n<p>To draw the Lewis structure of carbon dioxide (CO\u2082), follow these steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the Total Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) has 4 valence electrons.<\/li>\n\n\n\n<li>Each Oxygen (O) has 6 valence electrons, and there are two O atoms.<\/li>\n\n\n\n<li>Total: (4 + (6 \\times 2) = 16) valence electrons.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the Central Atom:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is less electronegative than oxygen, so it will be the central atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Place Oxygen Atoms Around the Central Atom:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Arrange the two oxygen atoms on either side of the carbon atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Distribute Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place two electrons (one pair) between the carbon and each oxygen to form bonds.<\/li>\n\n\n\n<li>After forming two single bonds, 12 electrons remain (16 total &#8211; 4 used in bonds = 12).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Complete the Octets of the Oxygen Atoms:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each oxygen needs 8 electrons to complete its octet. Place six electrons (three pairs) around each oxygen.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Form Double Bonds:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each oxygen has 8 electrons now, but carbon only has 4. To satisfy the octet rule for carbon, convert one lone pair from each oxygen into a double bond.<\/li>\n\n\n\n<li>The final Lewis structure shows carbon double-bonded to each oxygen:<br>[<br>O=C=O<br>]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Polarity of CO\u2082<\/h3>\n\n\n\n<p><strong>Answer: A) Nonpolar<\/strong><\/p>\n\n\n\n<p>Carbon dioxide (CO\u2082) is a linear molecule with a symmetrical arrangement of its bonds. The molecular geometry can be described as linear because the angle between the oxygen-carbon-oxygen (O-C-O) atoms is 180 degrees.<\/p>\n\n\n\n<p>Despite the presence of polar covalent bonds (since oxygen is more electronegative than carbon), the symmetry of the CO\u2082 molecule causes the dipoles to cancel each other out. Each C=O bond has a dipole moment pointing towards the oxygen atoms, but because the molecule is linear, these dipoles are equal in magnitude and opposite in direction.<\/p>\n\n\n\n<p>As a result, the net dipole moment of the molecule is zero, making CO\u2082 nonpolar. This lack of polarity has implications for its behavior in various chemical contexts. For instance, CO\u2082 does not interact significantly with polar solvents, like water, and is considered a nonpolar gas. This nonpolarity is also why CO\u2082 is relatively insoluble in water compared to more polar substances.<\/p>\n\n\n\n<p>In conclusion, CO\u2082 is a nonpolar molecule due to its symmetrical linear structure, leading to the cancellation of the individual bond dipoles.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of carbon dioxide (CO\u2082) and then determine if the molecule is polar ornonpolar.+A) NonpolarB) Polar The Correct Answer and Explanation is : Lewis Structure of Carbon Dioxide (CO\u2082) To draw the Lewis structure of carbon dioxide (CO\u2082), follow these steps: Polarity of CO\u2082 Answer: A) Nonpolar Carbon dioxide (CO\u2082) is a [&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-146389","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/146389","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=146389"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/146389\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=146389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=146389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=146389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}