{"id":245252,"date":"2025-07-05T19:56:12","date_gmt":"2025-07-05T19:56:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245252"},"modified":"2025-07-05T19:56:14","modified_gmt":"2025-07-05T19:56:14","slug":"identify-the-polarity-and-intermolecular-forces-imf-for-each-pure-liquid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/identify-the-polarity-and-intermolecular-forces-imf-for-each-pure-liquid\/","title":{"rendered":"Identify the polarity and intermolecular forces (IMF) for each pure liquid."},"content":{"rendered":"\n<p>Question 38 Identify the polarity and intermolecular forces (IMF) for each pure liquid. Strongly recommend drawing Lewis structure first, then based on VSEPR theory determine the molecular geometry (shape) and polarity: Molecular Polarity: Hydrogen bonding: [Select] CH4 [Select] HBr [Select] polar nonpolar [Select] NF3 [Select] CH2O [Select] [Select] CH3OH [Select] [Select] CS2 [Select] [Select] SCl2<\/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\/image-209.png\" alt=\"\" class=\"wp-image-245253\"\/><\/figure>\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<p>Here are the correct selections for each molecule based on its polarity and ability to form hydrogen bonds.<\/p>\n\n\n\n<p><strong>CH\u2084 (Methane)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>nonpolar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>HBr (Hydrogen Bromide)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>polar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>NF\u2083 (Nitrogen Trifluoride)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>polar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>CH\u2082O (Formaldehyde)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>polar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>CH\u2083OH (Methanol)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>polar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>yes<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>CS\u2082 (Carbon Disulfide)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>nonpolar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>SCl\u2082 (Sulfur Dichloride)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular Polarity:\u00a0<strong>polar<\/strong><\/li>\n\n\n\n<li>Hydrogen bonding:\u00a0<strong>no<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The polarity of a molecule is determined by both the polarity of its individual bonds and its overall molecular geometry. A molecule with polar bonds can be nonpolar if its shape is symmetrical, causing the bond dipoles to cancel out. Hydrogen bonding, a strong type of intermolecular force, occurs in pure substances where molecules contain a hydrogen atom covalently bonded to a highly electronegative atom\u2014specifically nitrogen (N), oxygen (O), or fluorine (F).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>CH\u2084 (Methane):<\/strong>\u00a0Methane has a symmetrical tetrahedral geometry. Although the C-H bonds have a very small electronegativity difference, their symmetrical arrangement causes the bond dipoles to cancel, making the molecule\u00a0<strong>nonpolar<\/strong>. It lacks H bonded to N, O, or F, so it does\u00a0<strong>not<\/strong>\u00a0exhibit hydrogen bonding.<\/li>\n\n\n\n<li><strong>HBr (Hydrogen Bromide):<\/strong>\u00a0As a diatomic molecule with a significant electronegativity difference between hydrogen and bromine, the H-Br bond is polar. This makes the entire molecule\u00a0<strong>polar<\/strong>. Hydrogen is bonded to bromine, not N, O, or F, so there is\u00a0<strong>no<\/strong>\u00a0hydrogen bonding.<\/li>\n\n\n\n<li><strong>NF\u2083 (Nitrogen Trifluoride):<\/strong>\u00a0The N-F bonds are polar. Due to a lone pair on the central nitrogen atom, the molecule has a trigonal pyramidal shape. This asymmetrical geometry prevents the bond dipoles from canceling, making NF\u2083\u00a0<strong>polar<\/strong>. It does not contain hydrogen, so it\u00a0<strong>cannot<\/strong>\u00a0hydrogen bond.<\/li>\n\n\n\n<li><strong>CH\u2082O (Formaldehyde):<\/strong>\u00a0The C=O double bond is very polar. The molecule&#8217;s trigonal planar geometry is asymmetrical because the C=O and C-H bonds are different. Thus, there is a net dipole moment, and the molecule is\u00a0<strong>polar<\/strong>. Hydrogen is bonded to carbon, not oxygen, so pure formaldehyde does\u00a0<strong>not<\/strong>\u00a0form hydrogen bonds.<\/li>\n\n\n\n<li><strong>CH\u2083OH (Methanol):<\/strong>\u00a0The molecule contains polar C-O and O-H bonds and has an asymmetrical structure, making it\u00a0<strong>polar<\/strong>. Crucially, it has an O-H group, which allows methanol molecules to act as both hydrogen bond donors and acceptors. Therefore, it\u00a0<strong>does<\/strong>\u00a0exhibit hydrogen bonding.<\/li>\n\n\n\n<li><strong>CS\u2082 (Carbon Disulfide):<\/strong>\u00a0The molecule has a linear, symmetrical structure (S=C=S). Any small polarity in the C-S bonds is canceled out, making the molecule\u00a0<strong>nonpolar<\/strong>. It contains no hydrogen atoms, so it\u00a0<strong>cannot<\/strong>\u00a0hydrogen bond.<\/li>\n\n\n\n<li><strong>SCl\u2082 (Sulfur Dichloride):<\/strong>\u00a0The S-Cl bonds are polar. The central sulfur atom has two lone pairs, giving the molecule a bent geometry. This asymmetrical shape results in a net dipole moment, making SCl\u2082\u00a0<strong>polar<\/strong>. It has no hydrogen atoms and thus\u00a0<strong>cannot<\/strong>\u00a0hydrogen bond.<\/li>\n<\/ul>\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-banner5-612.jpeg\" alt=\"\" class=\"wp-image-245254\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Question 38 Identify the polarity and intermolecular forces (IMF) for each pure liquid. Strongly recommend drawing Lewis structure first, then based on VSEPR theory determine the molecular geometry (shape) and polarity: Molecular Polarity: Hydrogen bonding: [Select] CH4 [Select] HBr [Select] polar nonpolar [Select] NF3 [Select] CH2O [Select] [Select] CH3OH [Select] [Select] CS2 [Select] [Select] SCl2 [&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-245252","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245252","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=245252"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245252\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245252"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245252"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245252"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}