{"id":226995,"date":"2025-06-05T17:55:45","date_gmt":"2025-06-05T17:55:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226995"},"modified":"2025-06-05T17:56:09","modified_gmt":"2025-06-05T17:56:09","slug":"draw-the-ocn-lewis-structure-one-way-of-writing-the-lewis-structure-of-the-cyanate-ion-ocn-places-one-double-bond-between-the-carbon-atom-and-the-oxygen-atom-and-another-double-bond-between-the-car","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/draw-the-ocn-lewis-structure-one-way-of-writing-the-lewis-structure-of-the-cyanate-ion-ocn-places-one-double-bond-between-the-carbon-atom-and-the-oxygen-atom-and-another-double-bond-between-the-car\/","title":{"rendered":"Draw the OCN- Lewis structure: One way of writing the Lewis structure of the cyanate ion OCN- places one double bond between the carbon atom and the oxygen atom and another double bond between the carbon atom and the nitrogen atom"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Draw the OCN- Lewis structure: One way of writing the Lewis structure of the cyanate ion OCN- places one double bond between the carbon atom and the oxygen atom and another double bond between the carbon atom and the nitrogen atom; Don&#8217;t forget the lone pair of electrons. What are the formal charges on the oxygen, carbon, and nitrogen atoms, respectively, for this structure? Select one:<br>a. 0, 0, -1<br>b. -1, 0,0<br>c. -2, 1, 0<br>d. none of these<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Count the total valence electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Oxygen (O):<\/strong> 6 valence electrons<\/li>\n\n\n\n<li><strong>Carbon (C):<\/strong> 4 valence electrons<\/li>\n\n\n\n<li><strong>Nitrogen (N):<\/strong> 5 valence electrons<\/li>\n\n\n\n<li><strong>Charge (\u207b1):<\/strong> Add 1 electron<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Total = 6 + 4 + 5 + 1 = 16 valence electrons<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Draw the structure<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We\u2019re told to consider a structure with:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A <strong>double bond between carbon and oxygen<\/strong> (C=O)<\/li>\n\n\n\n<li>A <strong>double bond between carbon and nitrogen<\/strong> (C=N)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So the skeletal structure is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>O = C = N\u207b\n<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Now, distribute the remaining electrons to complete octets and calculate formal charges.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Assign lone pairs<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In this structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Oxygen<\/strong> has two bonds \u2192 4 electrons in bonds \u2192 needs 4 more (2 lone pairs)<\/li>\n\n\n\n<li><strong>Carbon<\/strong> has 4 bonds \u2192 complete octet \u2192 no lone pairs<\/li>\n\n\n\n<li><strong>Nitrogen<\/strong> has two bonds \u2192 4 electrons in bonds \u2192 needs 4 more (2 lone pairs)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So now:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>O<\/strong>: 2 lone pairs (4 electrons)<\/li>\n\n\n\n<li><strong>C<\/strong>: 0 lone pairs<\/li>\n\n\n\n<li><strong>N<\/strong>: 2 lone pairs (4 electrons)<\/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\"><strong>Step 4: Calculate formal charges<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Formal charge = Valence electrons &#8211; (Lone pair electrons + \u00bd Bonding electrons)<\/strong><\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Oxygen:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Valence: 6<\/li>\n\n\n\n<li>Lone pairs: 4 electrons<\/li>\n\n\n\n<li>Bonding: 4 electrons (2 bonds)<\/li>\n\n\n\n<li>FC = 6 &#8211; (4 + 2) = <strong>0<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Carbon:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Valence: 4<\/li>\n\n\n\n<li>Lone pairs: 0<\/li>\n\n\n\n<li>Bonding: 8 electrons (4 bonds)<\/li>\n\n\n\n<li>FC = 4 &#8211; (0 + 4) = <strong>0<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Nitrogen:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Valence: 5<\/li>\n\n\n\n<li>Lone pairs: 4<\/li>\n\n\n\n<li>Bonding: 4<\/li>\n\n\n\n<li>FC = 5 &#8211; (4 + 2) = <strong>-1<\/strong><\/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\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>O: 0, C: 0, N: -1<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>Correct choice: a. 0, 0, -1<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>300-Word Explanation:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The Lewis structure of the cyanate ion (OCN\u207b) helps us understand the bonding and charge distribution in the molecule. OCN\u207b consists of an oxygen atom, a carbon atom, and a nitrogen atom. Since it\u2019s an ion with a charge of -1, we must account for one extra electron when distributing valence electrons.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We are told to evaluate the structure where the carbon forms <strong>double bonds with both oxygen and nitrogen<\/strong> (O=C=N\u207b). This satisfies the octet rule for all atoms and uses up all 16 valence electrons efficiently.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Formal charge calculations help identify the most stable structure and verify if the charges align with the overall charge of the ion. For this double-bonded structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Oxygen<\/strong> has 6 valence electrons. With two lone pairs and two bonds, its formal charge is 0.<\/li>\n\n\n\n<li><strong>Carbon<\/strong> shares 4 bonds and has no lone pairs, making its formal charge also 0.<\/li>\n\n\n\n<li><strong>Nitrogen<\/strong> has two lone pairs and two bonds, giving it a formal charge of -1.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">The sum of formal charges (0 + 0 + -1) equals the total ionic charge of -1, confirming the stability and correctness of this Lewis structure.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the formal charges on the oxygen, carbon, and nitrogen atoms are <strong>0, 0, and -1<\/strong>, respectively. This matches choice <strong>a<\/strong>, making it the correct answer. The use of formal charges is essential in choosing the most representative resonance structure for polyatomic ions like OCN\u207b.<\/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-banner5-115.jpeg\" alt=\"\" class=\"wp-image-227001\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the OCN- Lewis structure: One way of writing the Lewis structure of the cyanate ion OCN- places one double bond between the carbon atom and the oxygen atom and another double bond between the carbon atom and the nitrogen atom; Don&#8217;t forget the lone pair of electrons. What are the formal charges on the [&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-226995","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226995","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=226995"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226995\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226995"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226995"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226995"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}