{"id":220088,"date":"2025-05-27T13:23:02","date_gmt":"2025-05-27T13:23:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220088"},"modified":"2025-05-27T13:23:04","modified_gmt":"2025-05-27T13:23:04","slug":"consider-an-ionic-compound","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/consider-an-ionic-compound\/","title":{"rendered":"Consider an ionic compound"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Consider an ionic compound, MX_2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX_2 is delta H_t degree = -957 kj\/mol. The enthalpy of sublimation of M is delta H_sub = 151 kJ\/mol. The first and second ionization energies of M are IE_1 = 675 kj\/mol and IE_2 = 1361 kJ\/mol. The electron affinity of X is delta H_ea = -347 kj\/mol. The bond energy of X_2 is BE = 187 kj\/mol. Determine the lattice energy of MX_2.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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<p class=\"wp-block-paragraph\">To determine the <strong>lattice energy (U)<\/strong> of the ionic compound <strong>MX\u2082<\/strong>, we apply the <strong>Born\u2013Haber cycle<\/strong>, which relates the standard enthalpy of formation of an ionic compound to the energies involved in its formation from elements in their standard states.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given Data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394H_f\u00b0(MX\u2082) = -957 kJ\/mol (formation of solid MX\u2082 from M(s) and X\u2082(g))<\/li>\n\n\n\n<li>\u0394H_sub(M) = +151 kJ\/mol (sublimation of M from solid to gas)<\/li>\n\n\n\n<li>IE\u2081(M) = +675 kJ\/mol (first ionization energy)<\/li>\n\n\n\n<li>IE\u2082(M) = +1361 kJ\/mol (second ionization energy)<\/li>\n\n\n\n<li>BE(X\u2082) = +187 kJ\/mol (bond dissociation energy of X\u2082 molecule to 2X atoms)<\/li>\n\n\n\n<li>EA(X) = -347 kJ\/mol (electron affinity for one X atom, for two X atoms: 2 \u00d7 EA = -694 kJ\/mol)<\/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\">Step-by-step Born\u2013Haber Cycle:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We are forming MX\u2082(s) from its elements in their standard states:<br>M(s) + X\u2082(g) \u2192 MX\u2082(s)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We break this into steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Sublimation of M<\/strong>:<br>M(s) \u2192 M(g)\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = +151 kJ\/mol<\/li>\n\n\n\n<li><strong>Ionization of M to M\u00b2\u207a<\/strong>:<br>M(g) \u2192 M\u207a(g)\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = +675 kJ\/mol<br>M\u207a(g) \u2192 M\u00b2\u207a(g)\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = +1361 kJ\/mol<br>Total = +2036 kJ\/mol<\/li>\n\n\n\n<li><strong>Dissociation of X\u2082<\/strong>:<br>X\u2082(g) \u2192 2X(g)\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = +187 kJ\/mol<\/li>\n\n\n\n<li><strong>Electron affinity of X atoms<\/strong>:<br>2X(g) + 2e\u207b \u2192 2X\u207b(g)\u2003\u2003\u2003\u2003\u0394H = 2 \u00d7 (-347) = -694 kJ\/mol<\/li>\n\n\n\n<li><strong>Formation of MX\u2082 from gaseous ions (lattice energy)<\/strong>:<br>M\u00b2\u207a(g) + 2X\u207b(g) \u2192 MX\u2082(s)\u2003\u2003\u2003\u0394H = <strong>U<\/strong><\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Now, apply Hess\u2019s Law: \u0394Hf\u2218=\u0394Hsub+IE1+IE2+BE1+2\u00d7EA+U\\Delta H_f^\\circ = \\Delta H_{sub} + IE_1 + IE_2 + \\frac{BE}{1} + 2 \\times EA + U \u2212957=151+675+1361+187\u2212694+U-957 = 151 + 675 + 1361 + 187 &#8211; 694 + U \u2212957=1680+U-957 = 1680 + U U=\u2212957\u22121680=\u22122637&nbsp;kJ\/molU = -957 &#8211; 1680 = -2637\\ \\text{kJ\/mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Lattice Energy (U) = -2637 kJ\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The <strong>lattice energy<\/strong> of an ionic compound is the energy released when gaseous ions combine to form the solid crystal lattice. It reflects the strength of electrostatic attraction between ions. For MX\u2082, a compound consisting of a metal ion (M\u00b2\u207a) and two halide ions (X\u207b), we calculate the lattice energy using the <strong>Born\u2013Haber cycle<\/strong>\u2014a thermochemical pathway that breaks down the formation of the solid into several steps.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We begin by sublimating the metal M from its solid form to a gas. Next, we remove two electrons from M to form M\u00b2\u207a; this requires the sum of the first and second ionization energies. The halogen X exists naturally as a diatomic gas (X\u2082), so we split one X\u2082 molecule into two atoms using its bond dissociation energy. Each X atom then gains an electron, releasing energy equal to its electron affinity. Finally, the M\u00b2\u207a and 2X\u207b gaseous ions come together to form the ionic crystal MX\u2082, releasing lattice energy.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By adding all the energy changes\u2014sublimation, ionizations, bond dissociation, and electron affinities\u2014we calculate the total energy input before lattice formation. Since the total enthalpy change of formation of MX\u2082 is known, we can rearrange the equation to solve for the lattice energy.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This method shows that lattice energy is crucial in stabilizing ionic compounds. A large negative lattice energy (in this case, -2637 kJ\/mol) indicates a strong electrostatic interaction, which typically leads to high melting points, hardness, and stability of the compound.<\/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-banner5-55.jpeg\" alt=\"\" class=\"wp-image-220089\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Consider an ionic compound, MX_2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX_2 is delta H_t degree = -957 kj\/mol. The enthalpy of sublimation of M is delta H_sub = 151 kJ\/mol. The first and second ionization energies of M are IE_1 = 675 kj\/mol and IE_2 [&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-220088","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220088","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=220088"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220088\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220088"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220088"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}