Answer: XeF\u2082 (xenon difluoride) is a nonpolar molecule.<\/strong><\/p>\n\n\n\n To determine whether XeF\u2082<\/strong> is polar or nonpolar, we need to examine both its molecular geometry<\/strong> and the electronegativity differences<\/strong> between the atoms.<\/p>\n\n\n\n 1. Electronegativity Consideration:<\/strong><\/p>\n\n\n\n This suggests that the Xe\u2013F bonds are polar<\/strong>, as fluorine is significantly more electronegative and will pull electron density toward itself. However, the polarity of individual bonds does not automatically<\/strong> mean the molecule is polar.<\/p>\n\n\n\n 2. Molecular Geometry:<\/strong><\/p>\n\n\n\n XeF\u2082 has five electron domains<\/strong> around the xenon atom (two bonding pairs for the fluorines and three lone pairs). According to VSEPR theory<\/strong>, this results in a trigonal bipyramidal<\/strong> electron geometry. The three lone pairs occupy the equatorial positions<\/strong> to minimize repulsion, and the two fluorine atoms occupy the axial positions<\/strong>, 180\u00b0 apart.<\/p>\n\n\n\n This gives XeF\u2082 a linear molecular shape<\/strong>.<\/p>\n\n\n\n 3. Net Dipole Moment:<\/strong><\/p>\n\n\n\n Even though each Xe\u2013F bond is polar, the linear geometry<\/strong> causes the two bond dipoles to point in exactly opposite directions. As a result, these dipoles cancel each other out<\/strong>, and there is no net dipole moment<\/strong> in the molecule.<\/p>\n\n\n\n 4. Conclusion:<\/strong><\/p>\n\n\n\n Despite having polar bonds, the symmetrical linear shape<\/strong> of XeF\u2082 ensures that the molecule is nonpolar overall<\/strong>. This is a classic example where geometry plays a crucial role in molecular polarity. The cancellation of equal and opposite dipole moments makes XeF\u2082 a nonpolar molecule<\/strong>, which affects its physical properties such as solubility and boiling point<\/strong>.<\/p>\n\n\n\n Is the molecule XeF2 polar or nonpolar? The electronegativity of Xe is 2.6 The correct Answer and Explanation is: Answer: XeF\u2082 (xenon difluoride) is a nonpolar molecule. Explanation To determine whether XeF\u2082 is polar or nonpolar, we need to examine both its molecular geometry and the electronegativity differences between the atoms. 1. Electronegativity Consideration: This […]<\/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-225891","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225891","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=225891"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225891\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225891"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225891"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation<\/h3>\n\n\n\n
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