{"id":183211,"date":"2025-01-16T05:29:08","date_gmt":"2025-01-16T05:29:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183211"},"modified":"2025-01-16T05:29:14","modified_gmt":"2025-01-16T05:29:14","slug":"how-many-total-valence-electrons-are-on-the-silicate-sio3-2-ion","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/how-many-total-valence-electrons-are-on-the-silicate-sio3-2-ion\/","title":{"rendered":"How many total valence electrons are on the silicate (SiO3 2-) ion"},"content":{"rendered":"\n<p>Please answer and explain!<\/p>\n\n\n\n<p>1) How many total valence electrons are on the silicate (SiO3 2-) ion?<\/p>\n\n\n\n<p>2) How many lone pairs are on the entire structure for the correct Lewis structure of carbon dioxide?<\/p>\n\n\n\n<p>3) What is the electron geometry of a molecule with two bonding domains and three pairs of electrons on the central atom?<\/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<h3 class=\"wp-block-heading\">1) <strong>How many total valence electrons are on the silicate (SiO\u2083\u00b2\u207b) ion?<\/strong><\/h3>\n\n\n\n<p>The silicate ion, ( \\text{SiO}_3^{2-} ), consists of a silicon (Si) atom and three oxygen (O) atoms with a 2- charge. Here&#8217;s how to calculate the total number of valence electrons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Silicon (Si)<\/strong> is in group 14 of the periodic table, so it has 4 valence electrons.<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong> is in group 16, so each oxygen atom has 6 valence electrons. Since there are 3 oxygen atoms, this gives ( 6 \\times 3 = 18 ) valence electrons.<\/li>\n\n\n\n<li>The 2- charge on the ion means the ion has gained 2 extra electrons.<\/li>\n<\/ul>\n\n\n\n<p><strong>Total valence electrons<\/strong>:<br>[<br>\\text{Valence electrons} = 4 (\\text{from Si}) + 18 (\\text{from O}) + 2 (\\text{charge}) = 24 \\text{ electrons}<br>]<\/p>\n\n\n\n<p>Thus, the silicate ion ( \\text{SiO}_3^{2-} ) has <strong>24 valence electrons<\/strong> in total.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2) <strong>How many lone pairs are on the entire structure for the correct Lewis structure of carbon dioxide (CO\u2082)?<\/strong><\/h3>\n\n\n\n<p>Carbon dioxide (CO\u2082) has a central carbon atom double-bonded to two oxygen atoms. Here&#8217;s the process to determine the lone pairs:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Valence electrons in CO\u2082<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) is in group 14, so it has 4 valence electrons.<\/li>\n\n\n\n<li>Each oxygen (O) is in group 16, so each has 6 valence electrons. There are two oxygens, so ( 6 \\times 2 = 12 ) valence electrons from oxygen.<\/li>\n<\/ul>\n\n\n\n<p>Total valence electrons = 4 (from C) + 12 (from O) = 16 valence electrons.<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Drawing the Lewis structure<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is the central atom, and each oxygen is double-bonded to carbon. Each double bond uses 4 electrons, 2 for each bond.<\/li>\n\n\n\n<li>The remaining electrons are placed as lone pairs on the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n<p>After forming the double bonds, there are 4 electrons left for the oxygen atoms (2 electrons for each lone pair). Thus, each oxygen atom will have <strong>2 lone pairs<\/strong>.<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Counting lone pairs<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each oxygen has 2 lone pairs.<\/li>\n\n\n\n<li>Since there are two oxygens, this gives ( 2 \\times 2 = 4 ) lone pairs in total.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the <strong>correct number of lone pairs<\/strong> in the entire structure of CO\u2082 is <strong>4 lone pairs<\/strong> (2 on each oxygen atom).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3) <strong>What is the electron geometry of a molecule with two bonding domains and three pairs of electrons on the central atom?<\/strong><\/h3>\n\n\n\n<p>In a molecule with <strong>two bonding domains<\/strong> and <strong>three pairs of electrons<\/strong> on the central atom, the total number of electron pairs around the central atom is <strong>five<\/strong> (2 bonding + 3 lone pairs). This corresponds to a <strong>trigonal bipyramidal electron geometry<\/strong>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Explanation:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Electron Geometry<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The electron geometry is determined by the total number of electron pairs around the central atom, both bonding and lone pairs.<\/li>\n\n\n\n<li>For 5 electron pairs, the electron geometry is trigonal bipyramidal. In this geometry, the central atom is surrounded by 5 positions: 3 in a plane (equatorial positions) and 2 above and below the plane (axial positions).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Molecular Shape<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>bonding domains<\/strong> are the regions where atoms are bonded. In this case, there are 2 bonding domains (from the 2 bonds).<\/li>\n\n\n\n<li>The <strong>lone pairs<\/strong> occupy positions that minimize electron repulsion. Lone pairs preferentially occupy the equatorial positions (the plane) because they experience less repulsion compared to the axial positions.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Electron Pair Arrangement<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>With 3 lone pairs and 2 bonding domains, the lone pairs will occupy 3 of the 5 positions (preferably the equatorial positions), and the bonding pairs will occupy the remaining 2 positions.<\/li>\n\n\n\n<li>Although the <strong>molecular geometry<\/strong> (the arrangement of atoms) may differ based on the number of lone pairs, the <strong>electron geometry<\/strong> remains trigonal bipyramidal due to the total number of electron pairs being 5.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the <strong>electron geometry<\/strong> of the molecule is <strong>trigonal bipyramidal<\/strong>.<\/p>\n\n\n\n<p>This geometry results in a less symmetrical molecular shape, such as <strong>linear<\/strong> if only 2 atoms are bonded (depending on how the bonding pairs are oriented). The lone pairs cause deviations from the ideal bonding angles.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Please answer and explain! 1) How many total valence electrons are on the silicate (SiO3 2-) ion? 2) How many lone pairs are on the entire structure for the correct Lewis structure of carbon dioxide? 3) What is the electron geometry of a molecule with two bonding domains and three pairs of electrons 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-183211","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183211","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=183211"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183211\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183211"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183211"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183211"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}