{"id":246133,"date":"2025-07-06T19:29:32","date_gmt":"2025-07-06T19:29:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246133"},"modified":"2025-07-06T19:29:34","modified_gmt":"2025-07-06T19:29:34","slug":"there-are-hybrid-orbitals-represented-by-this-picture","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/there-are-hybrid-orbitals-represented-by-this-picture\/","title":{"rendered":"There are hybrid orbitals represented by this picture."},"content":{"rendered":"\n

There are hybrid orbitals represented by this picture. They are composed of s p d (number) atomic orbitals, corresponding to hybridization. They have an electron pair geometry of . With bond angles of (If more than one bond angle is possible, separate each with a space). \u00b0 . B-For each hybridization type, choose a picture that corresponds to the correct orbital picture and specify the electon pair geometry. \u2026 If no picture matches, enter N. Hybridization \u2026 Picture \u2026 Electron Pair Geometry sp3 sp3d2 sp A \u2026 B \u2026 C C- A molecule has sp3 hybridization with 2 lone pairs. \u2026 The electron pair geometry of this molecule is: \u2026 The geometry of this molecule is: . \u2026 This molecule will have an approximate bond angle of (If more than one bond angle is possible, separate each with a space.): \u00b0. D- 1. What is the hybridization of the central atom in BeI2? Hybridization = What are the approximate bond angles in this substance? Bond angles = \u00b0 2. What is the hybridization of the central atom in POCl3? Hybridization = What are the approximate bond angles in this substance? Bond angles = \u00b0<\/p>\n\n\n\n

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

Let’s break this down step-by-step.<\/p>\n\n\n\n

Part A: Hybridization and Electron Pair Geometry<\/strong><\/h3>\n\n\n\n

1. sp3 Hybridization<\/strong><\/h4>\n\n\n\n
    \n
  • Hybridized Orbitals:<\/strong> One s orbital and three p orbitals combine to form four sp3 hybrid orbitals.<\/li>\n\n\n\n
  • Electron Pair Geometry:<\/strong> Tetrahedral (because there are four regions of electron density around the central atom).<\/li>\n\n\n\n
  • Bond Angles:<\/strong> 109.5\u00b0 (ideal bond angle in a tetrahedral structure).<\/li>\n<\/ul>\n\n\n\n

    2. sp3d2 Hybridization<\/strong><\/h4>\n\n\n\n
      \n
    • Hybridized Orbitals:<\/strong> One s orbital, three p orbitals, and two d orbitals combine to form six sp3d2 hybrid orbitals.<\/li>\n\n\n\n
    • Electron Pair Geometry:<\/strong> Octahedral (because there are six regions of electron density).<\/li>\n\n\n\n
    • Bond Angles:<\/strong> 90\u00b0 and 180\u00b0 (in an octahedral structure, all angles between bonds are either 90\u00b0 or 180\u00b0).<\/li>\n<\/ul>\n\n\n\n

      3. sp Hybridization<\/strong><\/h4>\n\n\n\n
        \n
      • Hybridized Orbitals:<\/strong> One s orbital and one p orbital combine to form two sp hybrid orbitals.<\/li>\n\n\n\n
      • Electron Pair Geometry:<\/strong> Linear (because there are two regions of electron density around the central atom).<\/li>\n\n\n\n
      • Bond Angles:<\/strong> 180\u00b0 (ideal bond angle in a linear structure).<\/li>\n<\/ul>\n\n\n\n

        Part B: Choose the Correct Picture<\/strong><\/h3>\n\n\n\n

        For each hybridization type, match the appropriate picture to the hybrid orbital configuration, if the pictures are provided. If none fit, the answer would be N<\/strong>.<\/p>\n\n\n\n

        Part C: Molecule with sp3 Hybridization and 2 Lone Pairs<\/strong><\/h3>\n\n\n\n
          \n
        • Electron Pair Geometry:<\/strong> Tetrahedral (the electron pairs still arrange in a tetrahedral geometry).<\/li>\n\n\n\n
        • Molecular Geometry:<\/strong> Bent<\/strong> (due to the lone pairs that repel the bonding pairs, the shape is bent rather than linear or trigonal).<\/li>\n\n\n\n
        • Bond Angles:<\/strong> Approximately 104.5\u00b0 (slightly smaller than 109.5\u00b0 due to lone pair repulsion).<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Part D: Questions on Specific Molecules<\/strong><\/h3>\n\n\n\n

          1. Hybridization of the Central Atom in BeI2<\/strong><\/h4>\n\n\n\n
            \n
          • Hybridization:<\/strong> sp (since Be in BeI2 has two bonding regions and no lone pairs).<\/li>\n\n\n\n
          • Bond Angles:<\/strong> 180\u00b0 (BeI2 has a linear structure).<\/li>\n<\/ul>\n\n\n\n

            2. Hybridization of the Central Atom in POCl3<\/strong><\/h4>\n\n\n\n
              \n
            • Hybridization:<\/strong> sp3d (since P has five regions of electron density: four bonding pairs with chlorine atoms and one lone pair).<\/li>\n\n\n\n
            • Bond Angles:<\/strong> Approximately 120\u00b0 and 90\u00b0 (POCl3 adopts a trigonal bipyramidal structure, so the angles vary between 90\u00b0 and 120\u00b0 depending on the position of the bonds).<\/li>\n<\/ul>\n\n\n\n
              \n\n\n\n

              Explanation:<\/h3>\n\n\n\n
                \n
              • Hybridization<\/strong> refers to the mixing of atomic orbitals to form a set of equivalent hybrid orbitals that are oriented in a way that minimizes electron repulsion. Each hybridization type has a corresponding geometry and bond angle, as shown above.<\/li>\n\n\n\n
              • The bond angles in a molecule are influenced by the electron pair geometry (which includes both bonding pairs and lone pairs), and how these pairs repel each other to minimize energy.<\/li>\n<\/ul>\n\n\n\n
                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                There are hybrid orbitals represented by this picture. They are composed of s p d (number) atomic orbitals, corresponding to hybridization. They have an electron pair geometry of . With bond angles of (If more than one bond angle is possible, separate each with a space). \u00b0 . B-For each hybridization type, choose a picture […]<\/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-246133","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246133","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=246133"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246133\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246133"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246133"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246133"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}