{"id":185082,"date":"2025-01-22T05:51:15","date_gmt":"2025-01-22T05:51:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185082"},"modified":"2025-01-22T05:51:21","modified_gmt":"2025-01-22T05:51:21","slug":"which-of-the-following-best-describes-the-bonding-scheme-consistent-with-the-molecular-geometry-predicted-by-vsepr","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/which-of-the-following-best-describes-the-bonding-scheme-consistent-with-the-molecular-geometry-predicted-by-vsepr\/","title":{"rendered":"Which of the following best describes the bonding scheme consistent with the molecular geometry predicted by VSEPR"},"content":{"rendered":"\n

Consider the molecule CIF3. Which of the following best describes the bonding scheme consistent with the molecular geometry predicted by VSEPR? The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining unhybridized d orbitals. The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals. The central atom is dsp3 hybridized; it forms 1 sigma bond and 2 pi bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals. The central atom is sp2 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and there are no lone pairs remaining on the central atom. The central atom is sp2 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining unhybridized p orbitals.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

The correct answer is:<\/p>\n\n\n\n

The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining hybrid orbitals.<\/strong><\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n
    \n
  1. Molecular Geometry and Hybridization<\/strong>:\n
      \n
    • The molecular formula ClF3\\text{ClF}_3 has chlorine (Cl) as the central atom bonded to three fluorine (F) atoms. Chlorine has 7 valence electrons, and each fluorine contributes one electron for bonding, resulting in 7+3=107 + 3 = 10 total valence electrons around the central atom.<\/li>\n\n\n\n
    • These 10 electrons form 5 electron domains: 3 bonding pairs (sigma bonds with F atoms) and 2 lone pairs.<\/li>\n\n\n\n
    • According to VSEPR (Valence Shell Electron Pair Repulsion) theory, 5 electron domains correspond to a trigonal bipyramidal electron geometry<\/strong>. The molecular geometry, considering the lone pairs, becomes T-shaped<\/strong> because the lone pairs occupy equatorial positions to minimize repulsion.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Hybridization<\/strong>:\n
        \n
      • To accommodate 5 electron domains, the chlorine atom must use dsp3^3<\/strong> hybrid orbitals, which are formed by combining one ss, three pp, and one dd orbital.<\/li>\n\n\n\n
      • In this hybridization, the 3 bonding pairs occupy three hybrid orbitals, forming sigma bonds with the 2p2p orbitals of fluorine atoms. The remaining two hybrid orbitals contain lone pairs of electrons.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Lone Pairs<\/strong>:\n
          \n
        • The two lone pairs reside in the equatorial positions of the trigonal bipyramidal geometry, which minimizes electron-electron repulsion because equatorial lone pairs are further apart than axial lone pairs.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Incorrect Options<\/strong>:\n
            \n
          • sp2sp^2 hybridization is incorrect because it accommodates only 3 electron domains, not 5.<\/li>\n\n\n\n
          • Lone pairs do not occupy unhybridized dd-orbitals; all orbitals used are part of the hybridization. dsp3dsp^3 hybrid orbitals include both bonding and lone pair electrons.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

            This explanation aligns with the bonding and geometry predicted by VSEPR theory and hybridization concepts.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Consider the molecule CIF3. Which of the following best describes the bonding scheme consistent with the molecular geometry predicted by VSEPR? The central atom is dsp3 hybridized; it forms 3 sigma bonds by overlapping with 2p orbitals on F atoms, and the two sets of lone pairs reside in the remaining unhybridized d orbitals. The […]<\/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-185082","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185082","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=185082"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185082\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185082"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185082"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185082"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}