{"id":201072,"date":"2025-03-14T19:01:45","date_gmt":"2025-03-14T19:01:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201072"},"modified":"2025-03-14T19:01:47","modified_gmt":"2025-03-14T19:01:47","slug":"identify-the-lewis-acid-and-lewis-base-from-among-the-reactants-in-the-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/14\/identify-the-lewis-acid-and-lewis-base-from-among-the-reactants-in-the-equation\/","title":{"rendered":"Identify the Lewis acid and Lewis base from among the reactants in the equation"},"content":{"rendered":"\n<p>Identify the Lewis acid and Lewis base from among the reactants in the equation.<\/p>\n\n\n\n<p>Fe3+ (aq) + 6H2O (l) = Fe(H2O)63+ (aq)<\/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\">Identifying the Lewis Acid and Lewis Base<\/h3>\n\n\n\n<p>In the given chemical reaction:<\/p>\n\n\n\n<p>[<br>\\text{Fe}^{3+} (aq) + 6H_2O (l) \\rightarrow \\text{Fe(H}_2O)_6^{3+} (aq)<br>]<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Acid:<\/strong> <strong>Fe\u00b3\u207a (aq)<\/strong><\/li>\n\n\n\n<li><strong>Lewis Base:<\/strong> <strong>H\u2082O (l)<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Understanding Lewis Acid and Lewis Base<\/strong><\/h4>\n\n\n\n<p>The <strong>Lewis acid-base theory<\/strong>, proposed by Gilbert N. Lewis, defines:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A <strong>Lewis acid<\/strong> as a species that can accept an electron pair.<\/li>\n\n\n\n<li>A <strong>Lewis base<\/strong> as a species that donates an electron pair.<\/li>\n<\/ul>\n\n\n\n<p>This theory focuses on electron pair transfer rather than proton exchange (as in the Br\u00f8nsted-Lowry definition).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Why Fe\u00b3\u207a is a Lewis Acid?<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>iron(III) ion (Fe\u00b3\u207a)<\/strong> is highly <strong>electron-deficient<\/strong> due to its high positive charge.<\/li>\n\n\n\n<li>It has empty <strong>d orbitals<\/strong> that can accept electron pairs.<\/li>\n\n\n\n<li>It readily interacts with species that can donate electron pairs to form <strong>coordinate bonds<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Thus, Fe\u00b3\u207a acts as a <strong>Lewis acid<\/strong> because it <strong>accepts<\/strong> electron pairs from water molecules.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Why H\u2082O is a Lewis Base?<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each water molecule (H\u2082O) has two <strong>lone pairs<\/strong> on the oxygen atom.<\/li>\n\n\n\n<li>Water can <strong>donate<\/strong> these electron pairs to Fe\u00b3\u207a to form coordinate covalent bonds.<\/li>\n\n\n\n<li>Since it donates electron pairs, water acts as a <strong>Lewis base<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Formation of the Complex Ion [Fe(H\u2082O)\u2086]\u00b3\u207a<\/strong><\/h4>\n\n\n\n<p>When Fe\u00b3\u207a is added to water, it gets surrounded by <strong>six<\/strong> water molecules, forming the <strong>hexaaquairon(III) complex<\/strong> ([Fe(H_2O)_6]^{3+}). Each water molecule <strong>donates an electron pair<\/strong> to Fe\u00b3\u207a, creating <strong>Fe\u2013O coordinate covalent bonds<\/strong>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Conclusion<\/strong><\/h4>\n\n\n\n<p>In this reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Fe\u00b3\u207a acts as the Lewis acid<\/strong> because it accepts electron pairs.<\/li>\n\n\n\n<li><strong>H\u2082O acts as the Lewis base<\/strong> because it donates electron pairs.<br>This illustrates a classic example of Lewis acid-base interaction leading to the formation of a complex ion.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Identify the Lewis acid and Lewis base from among the reactants in the equation. Fe3+ (aq) + 6H2O (l) = Fe(H2O)63+ (aq) The correct answer and explanation is : Identifying the Lewis Acid and Lewis Base In the given chemical reaction: [\\text{Fe}^{3+} (aq) + 6H_2O (l) \\rightarrow \\text{Fe(H}_2O)_6^{3+} (aq)] Explanation Understanding Lewis Acid and Lewis [&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-201072","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201072","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=201072"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201072\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201072"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201072"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201072"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}