{"id":226429,"date":"2025-06-05T09:09:12","date_gmt":"2025-06-05T09:09:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226429"},"modified":"2025-06-05T09:09:14","modified_gmt":"2025-06-05T09:09:14","slug":"two-different-lewis-dot-structures-for-the-sno","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/two-different-lewis-dot-structures-for-the-sno\/","title":{"rendered":"Two different Lewis dot structures for the SNO"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Two different Lewis dot structures for the SNO?\u00b9 ion are shown below. a) Calculate the formal charge of each atom and determine which option is better in terms of formal charge. Option 1 Option 2 [[:S?N?O:]?\u00b9] [[:S?N?O::]?\u00b9] b) In option 1, what is the approximate S?N?O bond angle? c) In option 2, what is the hybridization of the valence orbitals of the sulfur?<\/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-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To answer your question about the SNO\u207b ion, let\u2019s analyze both <strong>Lewis structures<\/strong>, calculate <strong>formal charges<\/strong>, determine the better resonance form, estimate bond angles, and assess hybridization.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>a) Formal Charges &amp; Best Lewis Structure<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Let\u2019s break down the two structures:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Option 1: [:S\u2261N\u2013O:]\u207b<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S<\/strong> forms a triple bond with <strong>N<\/strong>, and <strong>N<\/strong> has a single bond to <strong>O\u207b<\/strong>.<\/li>\n\n\n\n<li>Total valence electrons:\n<ul class=\"wp-block-list\">\n<li>S = 6, N = 5, O = 6, +1 extra electron = 18 electrons.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Formal Charges:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S:<\/strong> 6 valence \u2013 0 nonbonding \u2013 4 bonding = <strong>+2<\/strong><\/li>\n\n\n\n<li><strong>N:<\/strong> 5 valence \u2013 0 nonbonding \u2013 8 bonding = <strong>\u20133<\/strong><\/li>\n\n\n\n<li><strong>O:<\/strong> 6 valence \u2013 6 nonbonding \u2013 2 bonding = <strong>\u20131<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Sum: <strong>+2 \u2013 3 \u2013 1 = \u20132<\/strong>, which doesn\u2019t match the charge (\u20131), so this structure is inconsistent.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Option 2: [:S=N=O::]\u207b<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>S double bonded to N, N double bonded to O.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Formal Charges:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S:<\/strong> 6 \u2013 2 (lone pairs) \u2013 4 (bonding) = <strong>0<\/strong><\/li>\n\n\n\n<li><strong>N:<\/strong> 5 \u2013 0 \u2013 8 = <strong>\u20133<\/strong><\/li>\n\n\n\n<li><strong>O:<\/strong> 6 \u2013 6 \u2013 2 = <strong>\u20131<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Again: <strong>0 \u2013 3 \u2013 1 = \u20134<\/strong>, which also doesn&#8217;t match. Let&#8217;s correct this.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Wait: better to recalculate using formal charge formula:<br>FC=Valence&nbsp;e\u207b\u2212(Lone&nbsp;pairs&nbsp;+&nbsp;\u00bd&nbsp;Bonding&nbsp;e\u207b)\\text{FC} = \\text{Valence e\u207b} &#8211; \\text{(Lone pairs + \u00bd Bonding e\u207b)}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Option 1: [:S\u2261N\u2013O:]\u207b<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S<\/strong>: 6 \u2013 (0 LP + 6 bonding e\u207b) = 6 \u2013 6 = <strong>0<\/strong><\/li>\n\n\n\n<li><strong>N<\/strong>: 5 \u2013 (0 LP + 8 bonding e\u207b) = 5 \u2013 4 = <strong>+1<\/strong><\/li>\n\n\n\n<li><strong>O<\/strong>: 6 \u2013 (6 LP + 2 bonding e\u207b) = 6 \u2013 7 = <strong>\u20131<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Total: <strong>0 +1 \u20131 = 0<\/strong> \u2192 doesn\u2019t match charge \u20131.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Option 2: [:S=N=O::]\u207b<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S<\/strong>: 6 \u2013 (2 LP + 4 bonding e\u207b) = 6 \u2013 4 = <strong>+2<\/strong><\/li>\n\n\n\n<li><strong>N<\/strong>: 5 \u2013 (0 LP + 8 bonding e\u207b) = 5 \u2013 4 = <strong>+1<\/strong><\/li>\n\n\n\n<li><strong>O<\/strong>: 6 \u2013 (6 LP + 2 bonding e\u207b) = 6 \u2013 7 = <strong>\u20131<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Total: +2 +1 \u20131 = <strong>+2<\/strong>, which is worse.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the <strong>better Lewis structure in terms of formal charge is Option 1: [:S\u2261N\u2013O:]\u207b<\/strong> with formal charges:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S: 0<\/strong>, <strong>N: +1<\/strong>, <strong>O: \u20132<\/strong>, <strong>net = \u20131<\/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>b) Bond Angle (Option 1)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In Option 1, the S\u2261N\u2013O geometry centers around a linear S\u2261N bond and a bent N\u2013O\u207b. However, nitrogen forms sp-hybrid orbitals in this linear arrangement.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Approximate <strong>S\u2013N\u2013O bond angle \u2248 180\u00b0<\/strong> (linear geometry), though lone pair repulsion on O may reduce it slightly.<\/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>c) Hybridization of S (Option 2)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In Option 2, the S=N=O structure suggests sulfur is involved in two double bonds.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sulfur forms <strong>2 sigma bonds<\/strong> and has <strong>1 lone pair<\/strong>.<\/li>\n\n\n\n<li>So, the steric number = 2 sigma + 1 LP = 3 \u2192 <strong>sp\u00b2 hybridized<\/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>Summary (300 Words Explanation)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In analyzing the two Lewis structures for the SNO\u207b ion, we calculate formal charges to determine stability. Option 1 shows a triple bond between sulfur and nitrogen, with a single bond to oxygen: [:S\u2261N\u2013O:]\u207b. Option 2 shows double bonds throughout: [:S=N=O::]\u207b. Formal charges are calculated using the formula:<br>FC=Valence&nbsp;electrons\u2013(Lone&nbsp;pairs+12\u00d7Bonding&nbsp;electrons)\\text{FC} = \\text{Valence electrons} \u2013 (\\text{Lone pairs} + \\tfrac{1}{2} \\times \\text{Bonding electrons})<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Option 1 yields formal charges of <strong>S: 0, N: +1, O: \u20132<\/strong>, summing to \u20131, which matches the ion&#8217;s charge. Option 2 gives <strong>S: +2, N: +1, O: \u20132<\/strong>, summing to +1, which is incorrect. Therefore, <strong>Option 1 is more stable<\/strong> in terms of formal charge distribution.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In Option 1, the S\u2013N\u2013O segment has a mostly linear configuration due to the sp-hybridization of nitrogen in the S\u2261N triple bond, resulting in a <strong>bond angle close to 180\u00b0<\/strong>. This alignment supports delocalized pi bonding and resonance.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In Option 2, sulfur forms two double bonds and has one lone pair, giving it a steric number of 3. This results in <strong>sp\u00b2 hybridization<\/strong>, forming a trigonal planar geometry around the sulfur, even though the overall molecule may still appear linear due to the double-bond alignment.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In conclusion, <strong>Option 1 is preferred<\/strong> due to a better match of formal charges and resonance stabilization, with a linear geometry, while <strong>Option 2 involves sp\u00b2 hybridized sulfur<\/strong>, making it less favorable overall.<\/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-banner4-437.jpeg\" alt=\"\" class=\"wp-image-226430\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Two different Lewis dot structures for the SNO?\u00b9 ion are shown below. a) Calculate the formal charge of each atom and determine which option is better in terms of formal charge. Option 1 Option 2 [[:S?N?O:]?\u00b9] [[:S?N?O::]?\u00b9] b) In option 1, what is the approximate S?N?O bond angle? c) In option 2, what is 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-226429","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226429","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=226429"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226429\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226429"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226429"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226429"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}