CBr4 has a zero dipole moment and a boiling point of 189.5\u00b0C; CH3Br has a dipole moment of 0.05 D and a boiling point of 3.56\u00b0C.Briefly explain why this polar molecule has a lower boiling point than the non-polar molecule.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The difference in boiling points of CBr\u2084<\/strong> and CH\u2083Br<\/strong> despite their differing dipole moments can be attributed to the nature and strength of the intermolecular forces acting on these molecules, as well as their molecular sizes.<\/p>\n\n\n\n CBr\u2084 (Carbon Tetrafluoride) is a non-polar molecule, despite the fact that the individual C-Br bonds are polar. The molecular geometry of CBr\u2084 is tetrahedral, and the dipoles of the individual bonds cancel out due to the symmetry of the molecule, resulting in a net dipole moment of zero. The primary intermolecular force in CBr\u2084 is London dispersion forces<\/strong>, which arise from the temporary fluctuations in electron distribution that induce dipoles in neighboring molecules. These dispersion forces are generally weaker than the dipole-dipole interactions present in polar molecules. However, the size and mass of the CBr\u2084 molecule (due to the presence of large bromine atoms) contribute to relatively stronger dispersion forces compared to smaller, less polar molecules. This allows CBr\u2084 to have a relatively high boiling point of 189.5\u00b0C.<\/p>\n\n\n\n CH\u2083Br (Methyl Bromide) is a polar molecule because of the significant difference in electronegativity between carbon and bromine, which results in a permanent dipole moment of 0.05 D. The molecule is asymmetrical, meaning the dipoles do not cancel out. As a result, dipole-dipole interactions<\/strong> between CH\u2083Br molecules exist, which are generally stronger than London dispersion forces. However, the boiling point of CH\u2083Br is much lower (3.56\u00b0C) compared to CBr\u2084, despite the presence of dipole-dipole forces. This can be explained by the fact that CH\u2083Br is much smaller than CBr\u2084 and its molecular mass is lower. The overall weaker London dispersion forces in CH\u2083Br (due to its smaller size) are unable to compensate for the effects of weaker dipole-dipole interactions, resulting in a lower boiling point.<\/p>\n\n\n\n While CH\u2083Br has a permanent dipole and experiences dipole-dipole interactions, the significantly larger size of CBr\u2084 leads to stronger London dispersion forces, ultimately giving CBr\u2084 the higher boiling point. Thus, the boiling point of a molecule is not determined by dipole moment alone but by a combination of all intermolecular forces and the size and mass of the molecules involved.<\/p>\n","protected":false},"excerpt":{"rendered":" CBr4 has a zero dipole moment and a boiling point of 189.5\u00b0C; CH3Br has a dipole moment of 0.05 D and a boiling point of 3.56\u00b0C.Briefly explain why this polar molecule has a lower boiling point than the non-polar molecule. The Correct Answer and Explanation is : The difference in boiling points of CBr\u2084 and […]<\/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-189655","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189655","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=189655"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189655\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189655"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189655"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189655"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}CBr\u2084: Non-Polar Molecule<\/strong><\/h3>\n\n\n\n
CH\u2083Br: Polar Molecule<\/strong><\/h3>\n\n\n\n
Conclusion<\/strong><\/h3>\n\n\n\n