{"id":187182,"date":"2025-02-03T05:34:35","date_gmt":"2025-02-03T05:34:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187182"},"modified":"2025-02-03T05:34:37","modified_gmt":"2025-02-03T05:34:37","slug":"draw-the-lewis-structure-of-given-compound","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/03\/draw-the-lewis-structure-of-given-compound\/","title":{"rendered":"Draw the Lewis structure of given compound"},"content":{"rendered":"\n<p>Q1. Draw the Lewis structure of given compound; a) CH3CH2NH2 (etil amine) b) CH3COOH (acetic acid) c) HCSNH2 d) NH CONH2 e) CICH2CH2NH2 Q2. Draw Lewis structure of the following species, indicate formal charges and resonance where applicable. a) H2SO4 b) H2NCN c) FCO d) HCO3 e) FSO3 Q3. Use the VSEPR theory to predict the shape of a) The molecule OSF2 b) The ion CIO3 c) The ion of S2032- d) Th ion of BrF4* . . . Q4. For given compound, drawn an appropriate Lewis structure, determine the electron and molecular geometry using VSEPR, indicate the VSEPR notation, determine whether molecule is polar or not, identify the hybridization of all interior atom make a sketch of the molecules based on valance bond theory a) COCI2 b) S2F2 C) SF4 d) CH3SH . . Q5. What is the molecular shape of propyne (CH3CCH) and propene (CH3CHCH2)? Q6. Write the hybridization scheme of central atom of each molecule and sketch the orbital overlapping structure. a) HONO2 b) N2H4 c) CCl4 d) COCl2 e) C2N2 Q7. Draw the molecular orbital diagram of given compound and predict the bond order and comment on whether molecule can exit or not. Please indicate whether they are paramagnetic or diamagnetic. a) F2 b) Li22- c) Bez d) C22+<\/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<p>That&#8217;s a lot of chemistry to cover! I&#8217;ll break it down step by step:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Q1: Lewis Structures<\/strong><\/h3>\n\n\n\n<p>The Lewis structures of the following molecules involve drawing all valence electrons and ensuring that each atom obeys the octet rule where applicable.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>a) CH\u2083CH\u2082NH\u2082 (Ethylamine)<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon forms 4 bonds (tetrahedral structure).<\/li>\n\n\n\n<li>Nitrogen has a lone pair.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>b) CH\u2083COOH (Acetic Acid)<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The carboxyl (-COOH) group has a double-bonded oxygen and an OH group.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>c) HCSNH\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon forms a double bond with sulfur and a single bond with nitrogen.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>d) NHCONH\u2082 (Urea)<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Central carbon forms single bonds with two NH\u2082 groups and a double bond with oxygen.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>e) ClCH\u2082CH\u2082NH\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ethylamine with one hydrogen replaced by a chlorine atom.<\/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>Q2: Lewis Structures with Formal Charges and Resonance<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a) H\u2082SO\u2084<\/strong>: Sulfur in the center, resonance in sulfate anion.<\/li>\n\n\n\n<li><strong>b) H\u2082NCN<\/strong>: Carbon triple bonded to nitrogen.<\/li>\n\n\n\n<li><strong>c) FCO<\/strong>: Resonance structures due to carbonyl.<\/li>\n\n\n\n<li><strong>d) HCO\u2083\u207b<\/strong>: Bicarbonate ion, delocalized electrons.<\/li>\n\n\n\n<li><strong>e) FSO\u2083\u207b<\/strong>: Fluorosulfate ion, resonance.<\/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>Q3: Molecular Geometry Using VSEPR<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a) OSF\u2082<\/strong> \u2192 <strong>Bent<\/strong> (AX\u2082E)<\/li>\n\n\n\n<li><strong>b) ClO\u2083\u207b<\/strong> \u2192 <strong>Trigonal pyramidal<\/strong> (AX\u2083E)<\/li>\n\n\n\n<li><strong>c) S\u2082O\u2083\u00b2\u207b<\/strong> \u2192 <strong>Bent &amp; Trigonal pyramidal<\/strong><\/li>\n\n\n\n<li><strong>d) BrF\u2084\u207b<\/strong> \u2192 <strong>Square planar<\/strong> (AX\u2084E\u2082)<\/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>Q4: Geometry, Hybridization &amp; Polarity<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a) COCl\u2082<\/strong> \u2192 Trigonal planar, <strong>sp\u00b2<\/strong>, polar.<\/li>\n\n\n\n<li><strong>b) S\u2082F\u2082<\/strong> \u2192 Bent, <strong>sp\u00b3<\/strong>, polar.<\/li>\n\n\n\n<li><strong>c) SF\u2084<\/strong> \u2192 See-saw, <strong>sp\u00b3d<\/strong>, polar.<\/li>\n\n\n\n<li><strong>d) CH\u2083SH<\/strong> \u2192 Bent at S, <strong>sp\u00b3<\/strong>, polar.<\/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>Q5: Molecular Shape<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Propyne (CH\u2083C\u2261CH)<\/strong> \u2192 Linear at C\u2261C.<\/li>\n\n\n\n<li><strong>Propene (CH\u2083CH=CH\u2082)<\/strong> \u2192 Trigonal planar at C=C.<\/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>Q6: Hybridization &amp; Orbital Overlap<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HONO\u2082<\/strong> \u2192 Central N is <strong>sp\u00b2<\/strong>.<\/li>\n\n\n\n<li><strong>N\u2082H\u2084<\/strong> \u2192 N is <strong>sp\u00b3<\/strong>.<\/li>\n\n\n\n<li><strong>CCl\u2084<\/strong> \u2192 C is <strong>sp\u00b3<\/strong>.<\/li>\n\n\n\n<li><strong>COCl\u2082<\/strong> \u2192 C is <strong>sp\u00b2<\/strong>.<\/li>\n\n\n\n<li><strong>C\u2082N\u2082<\/strong> \u2192 C is <strong>sp<\/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>Q7: Molecular Orbital Theory &amp; Bond Order<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a) F\u2082<\/strong> \u2192 Bond order = 1, diamagnetic.<\/li>\n\n\n\n<li><strong>b) Li\u2082\u00b2\u207b<\/strong> \u2192 Bond order = 0 (doesn\u2019t exist).<\/li>\n\n\n\n<li><strong>c) Be\u2082<\/strong> \u2192 Bond order = 0 (doesn\u2019t exist).<\/li>\n\n\n\n<li><strong>d) C\u2082\u00b2\u207a<\/strong> \u2192 Bond order = 2, paramagnetic.<\/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>Explanation (300 Words)<\/strong><\/h3>\n\n\n\n<p>Lewis structures represent the bonding in molecules, ensuring all atoms obey the octet rule where applicable. Formal charges help determine stability, and resonance structures show electron delocalization.<\/p>\n\n\n\n<p>VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular shapes based on electron domain repulsion. For example, COCl\u2082 has a trigonal planar shape due to its <strong>sp\u00b2<\/strong> hybridization, while SF\u2084 has a see-saw shape due to lone pair repulsions.<\/p>\n\n\n\n<p>Hybridization describes how atomic orbitals mix to form bonds. In <strong>propene<\/strong>, carbon at the double bond is <strong>sp\u00b2<\/strong>, leading to a trigonal planar shape, while in <strong>propyne<\/strong>, the triple-bonded carbons are <strong>sp<\/strong> hybridized, resulting in a linear geometry.<\/p>\n\n\n\n<p>Molecular Orbital (MO) theory explains bonding at a quantum level. <strong>F\u2082<\/strong>, with a bond order of 1, is stable and diamagnetic, while <strong>Be\u2082<\/strong> has a bond order of 0 and does not exist. <strong>C\u2082\u00b2\u207a<\/strong> has a bond order of 2 and is paramagnetic.<\/p>\n\n\n\n<p>Through this analysis, we predict molecular behavior, polarity, and reactivity.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Q1. Draw the Lewis structure of given compound; a) CH3CH2NH2 (etil amine) b) CH3COOH (acetic acid) c) HCSNH2 d) NH CONH2 e) CICH2CH2NH2 Q2. Draw Lewis structure of the following species, indicate formal charges and resonance where applicable. a) H2SO4 b) H2NCN c) FCO d) HCO3 e) FSO3 Q3. Use the VSEPR theory to predict [&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-187182","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187182","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=187182"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187182\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187182"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187182"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187182"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}