{"id":246279,"date":"2025-07-07T04:26:58","date_gmt":"2025-07-07T04:26:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246279"},"modified":"2025-07-07T04:27:00","modified_gmt":"2025-07-07T04:27:00","slug":"determine-ph2-electron-pair-geometry","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/determine-ph2-electron-pair-geometry\/","title":{"rendered":"Determine PH2- electron pair geometry"},"content":{"rendered":"\n<p>Determine PH2- electron pair geometry: What is electron pair geometry for TeCl42-? What is electron pair geometry of CH3+?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0);color:#0693e3\" class=\"has-inline-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the electron pair geometry for both <strong>TeCl4\u00b2\u207b<\/strong> and <strong>CH3\u207a<\/strong>, we need to analyze their molecular structures using the <strong>VSEPR theory (Valence Shell Electron Pair Repulsion)<\/strong>, which helps predict the geometry of molecules based on the repulsion between electron pairs.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>TeCl4\u00b2\u207b (Tellurium Tetrachloride Ion)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Central Atom<\/strong>: Tellurium (Te)<\/li>\n\n\n\n<li><strong>Charge<\/strong>: The charge is 2- (TeCl4\u00b2\u207b), meaning the molecule has 2 extra electrons compared to a neutral TeCl4.<\/li>\n\n\n\n<li><strong>Electron Configuration of Te<\/strong>: Tellurium has an atomic number of 52 and an electron configuration of [Kr]5s\u00b24d\u00b9\u20705p\u2074.<\/li>\n\n\n\n<li><strong>Electron Pairs<\/strong>: In TeCl4\u00b2\u207b, Tellurium forms four bonds with chlorine atoms, and the extra 2 electrons contribute to lone pairs on Te. This gives <strong>4 bonding pairs<\/strong> and <strong>2 lone pairs<\/strong> on the Te atom.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: According to VSEPR theory, 6 electron pairs around the central atom (4 bonding pairs and 2 lone pairs) will arrange themselves in an <strong>octahedral<\/strong> geometry. The lone pairs will occupy opposite positions to minimize repulsion, but the electron pair geometry remains octahedral.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>CH3\u207a (Methylium Ion)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Central Atom<\/strong>: Carbon (C)<\/li>\n\n\n\n<li><strong>Charge<\/strong>: The positive charge on the molecule (CH3\u207a) indicates that one electron has been removed from the carbon atom.<\/li>\n\n\n\n<li><strong>Electron Configuration of C<\/strong>: Carbon has an atomic number of 6, and its neutral electron configuration is 1s\u00b22s\u00b22p\u00b2. In CH3\u207a, after losing an electron, carbon has 3 bonding pairs (one for each hydrogen).<\/li>\n\n\n\n<li><strong>Electron Pairs<\/strong>: In CH3\u207a, there are <strong>3 bonding pairs<\/strong> and <strong>0 lone pairs<\/strong> on carbon.<\/li>\n\n\n\n<li><strong>Electron Pair Geometry<\/strong>: The <strong>3 bonding pairs<\/strong> will arrange themselves in a <strong>trigonal planar<\/strong> geometry to minimize electron repulsion, according to VSEPR theory.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Electron Pair Geometries:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>TeCl4\u00b2\u207b<\/strong> has an <strong>octahedral<\/strong> electron pair geometry.<\/li>\n\n\n\n<li><strong>CH3\u207a<\/strong> has a <strong>trigonal planar<\/strong> electron pair geometry.<\/li>\n<\/ul>\n\n\n\n<p>These geometries are based on minimizing the repulsion between bonding and lone electron pairs around the central atom.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-832.jpeg\" alt=\"\" class=\"wp-image-246280\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine PH2- electron pair geometry: What is electron pair geometry for TeCl42-? What is electron pair geometry of CH3+? The Correct Answer and Explanation is: To determine the electron pair geometry for both TeCl4\u00b2\u207b and CH3\u207a, we need to analyze their molecular structures using the VSEPR theory (Valence Shell Electron Pair Repulsion), which helps 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-246279","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246279","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=246279"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246279\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246279"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246279"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}