{"id":220176,"date":"2025-05-27T19:25:37","date_gmt":"2025-05-27T19:25:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220176"},"modified":"2025-05-27T19:25:39","modified_gmt":"2025-05-27T19:25:39","slug":"geometry-indicate-the-molecular-shape-determine-hybridization-and-determine-if-polar-nonpolar","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/geometry-indicate-the-molecular-shape-determine-hybridization-and-determine-if-polar-nonpolar\/","title":{"rendered":"Geometry, indicate the molecular shape, determine hybridization, and determine if polar\/nonpolar,"},"content":{"rendered":"\n<p>geometry, indicate the molecular shape, determine hybridization, and determine if polar\/nonpolar,<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A. NH<sub>4<\/sub><sup>+<\/sup><\/li>\n\n\n\n<li>B. H<sub>2<\/sub>NF<sub>2<\/sub><sup>+<\/sup><\/li>\n\n\n\n<li>C. HCO<sub>3<\/sub><sup>&#8211;<\/sup><\/li>\n\n\n\n<li>D. CO<sub>3<\/sub><sup>2-<\/sup><\/li>\n\n\n\n<li>E. NO<sub>3<\/sub><sup>&#8211;<\/sup><\/li>\n\n\n\n<li>F. NO<sub>2<\/sub><sup>&#8211;<\/sup><\/li>\n\n\n\n<li>G. SO<sub>4<\/sub><sup>2-<\/sup><\/li>\n\n\n\n<li>H. SO<sub>3<\/sub><sup>2-<\/sup><\/li>\n\n\n\n<li>I. SO<sub>3<\/sub><\/li>\n\n\n\n<li>J. SO<sub>2<\/sub><\/li>\n\n\n\n<li>K. PO<sub>4<\/sub><sup>3-<\/sup><\/li>\n\n\n\n<li>L. PO<sub>3<\/sub><sup>3-<\/sup><\/li>\n<\/ul>\n\n\n\n<p>For the following polyatomic ions and compounds with varying numbers of oxygens, draw the correct Lewis structure, indicate electron pair geometry, indicate the molecular shape, determine hybridization, and determine if polar\/nonpolar,<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">A. <strong>NH\u2084\u207a (Ammonium)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: N single bonded to four H, one positive charge.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">B. <strong>H\u2082NF\u2082\u207a<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: N central, bonded to 2 H and 2 F, one positive charge.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Bent (due to lone pair on N)<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Polar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">C. <strong>HCO\u2083\u207b (Bicarbonate)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: Central C bonded to OH and two O atoms (one with a double bond), one negative charge.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Polar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">D. <strong>CO\u2083\u00b2\u207b (Carbonate)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: C central, bonded to three O atoms, resonance structure, two negative charges.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">E. <strong>NO\u2083\u207b (Nitrate)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: N central, bonded to 3 O atoms, one with double bond, resonance, one negative charge.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">F. <strong>NO\u2082\u207b (Nitrite)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: N bonded to 2 O atoms, one lone pair, one negative charge.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Bent<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Polar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">G. <strong>SO\u2084\u00b2\u207b (Sulfate)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: S central, four O atoms (two with double bonds), two negative charges.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">H. <strong>SO\u2083\u00b2\u207b (Sulfite)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: S central, three O atoms, one lone pair on S, two negative charges.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal pyramidal<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Polar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">I. <strong>SO\u2083 (Sulfur Trioxide)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: S central, three double-bonded O atoms (resonance).<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">J. <strong>SO\u2082 (Sulfur Dioxide)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: S central, two double bonds to O, one lone pair on S.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Trigonal planar<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Bent<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b2<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Polar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">K. <strong>PO\u2084\u00b3\u207b (Phosphate)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: P central, four O atoms, three with single bonds, one with double, three negative charges.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: Nonpolar<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">L. <strong>PO\u2083\u00b3\u207b (Phosphite)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong>: P central, three O atoms, one lone pair, three negative charges.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: Tetrahedral<\/li>\n\n\n\n<li><strong>Molecular Shape<\/strong>: Trigonal pyramidal<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: sp\u00b3<\/li>\n\n\n\n<li><strong>Polarity<\/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> Explanation<\/strong><\/h3>\n\n\n\n<p>Understanding molecular structure involves analyzing electron configurations and geometry. Lewis structures show how atoms and electrons are arranged. Molecular geometry is determined by the number of bonding and lone pairs using VSEPR theory. Hybridization explains orbital overlap forming specific molecular shapes.<\/p>\n\n\n\n<p>For example, <strong>NH\u2084\u207a<\/strong> is tetrahedral and nonpolar because its symmetric geometry cancels out dipoles. In contrast, <strong>H\u2082NF\u2082\u207a<\/strong> has a bent shape due to lone pairs on nitrogen, making it polar. <strong>Carbonate (CO\u2083\u00b2\u207b)<\/strong> and <strong>nitrate (NO\u2083\u207b)<\/strong> are resonance-stabilized and trigonal planar, resulting in nonpolarity due to symmetry.<\/p>\n\n\n\n<p>Ions like <strong>SO\u2084\u00b2\u207b<\/strong> and <strong>PO\u2084\u00b3\u207b<\/strong> exhibit tetrahedral geometry due to four regions of electron density. However, when lone pairs exist, as in <strong>SO\u2083\u00b2\u207b<\/strong> or <strong>PO\u2083\u00b3\u207b<\/strong>, the shape becomes trigonal pyramidal and results in a polar molecule. The lone pair contributes to molecular asymmetry.<\/p>\n\n\n\n<p><strong>Sulfur compounds<\/strong> like <strong>SO\u2082<\/strong> and <strong>SO\u2083<\/strong> differ in shape and polarity due to the number of oxygen atoms and lone pairs on sulfur. SO\u2082 is bent and polar, while SO\u2083 is trigonal planar and nonpolar.<\/p>\n\n\n\n<p>Hybridization reflects the number of bonding domains: sp\u00b3 for tetrahedral shapes (e.g., NH\u2084\u207a, SO\u2084\u00b2\u207b), and sp\u00b2 for trigonal planar (e.g., NO\u2083\u207b). Polar molecules have unequal charge distributions, typically from lone pairs or asymmetric shapes, affecting their physical and chemical properties, including solubility and reactivity.<\/p>\n\n\n\n<p>This understanding is essential in predicting molecular behavior in reactions, bonding strength, and interactions in biological and industrial processes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner8-45.jpeg\" alt=\"\" class=\"wp-image-220177\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>geometry, indicate the molecular shape, determine hybridization, and determine if polar\/nonpolar, For the following polyatomic ions and compounds with varying numbers of oxygens, draw the correct Lewis structure, indicate electron pair geometry, indicate the molecular shape, determine hybridization, and determine if polar\/nonpolar, The Correct Answer and Explanation is: A. NH\u2084\u207a (Ammonium) B. H\u2082NF\u2082\u207a C. HCO\u2083\u207b [&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-220176","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220176","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=220176"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220176\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220176"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220176"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220176"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}