{"id":258820,"date":"2025-07-18T13:51:59","date_gmt":"2025-07-18T13:51:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258820"},"modified":"2025-07-18T13:52:02","modified_gmt":"2025-07-18T13:52:02","slug":"for-the-molecule-seh2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/for-the-molecule-seh2\/","title":{"rendered":"For the molecule SeH2"},"content":{"rendered":"\n

For the molecule SeH2: Email the correct Lewis structure (including resonance forms, if any to ec What is the geometry of the molecule? Is it polar or non-polar? What is the hybridization of the central atom? QUESTION 8 For the molecule (CO3)2-: Email the correct Lewis structure (including resonance forms, if any to edo What is the geometry of the molecule? polar or non-polar? e and Submit to save and submit. Click Save All Answers to save all<\/p>\n\n\n\n

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

SeH\u2082 (Selenium Hydride)<\/h3>\n\n\n\n

Lewis Structure:<\/h4>\n\n\n\n

The Lewis structure of SeH\u2082 involves placing two hydrogen atoms bonded to a central selenium (Se) atom. Selenium, like oxygen, belongs to Group 16 and typically forms two bonds (as it has two valence electrons available for bonding). The lone pairs on selenium are crucial for determining the structure. Here\u2019s a step-by-step guide:<\/p>\n\n\n\n

    \n
  1. Valence Electrons<\/strong>:\n
      \n
    • Selenium (Se) has 6 valence electrons.<\/li>\n\n\n\n
    • Each hydrogen (H) atom has 1 valence electron, so the total valence electrons in SeH\u2082 are:
      6+2(1)=8\u00a0electrons6 + 2(1) = 8 \\text{ electrons}6+2(1)=8\u00a0electrons<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Bonding<\/strong>:\n
        \n
      • The two hydrogen atoms will each form a single bond with selenium, using two of the 8 electrons.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Lone Pairs<\/strong>:\n
          \n
        • After bonding with hydrogen, selenium has 4 remaining electrons, which will be placed as two lone pairs on selenium.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

          The structure will look like this:<\/p>\n\n\n\n

              H
               |
          H - Se
               |
               :
          <\/code><\/pre>\n\n\n\n
          (Selenium has two lone pairs of electrons, denoted by the colon).<\/p>\n\n\n\n
          Geometry:<\/h4>\n\n\n\n
          The geometry of SeH\u2082 is bent (V-shaped)<\/strong>. This is due to the lone pairs of electrons on selenium that create repulsion, causing the bonding pairs of electrons to be pushed closer together, forming an angle less than 109.5\u00b0 (approximately 91.9\u00b0).<\/p>\n\n\n\n
          Polarity:<\/h4>\n\n\n\n
          SeH\u2082 is polar<\/strong>. Selenium has a higher electronegativity than hydrogen, creating a dipole moment with the electron density concentrated towards selenium, making the molecule asymmetrical.<\/p>\n\n\n\n
          Hybridization:<\/h4>\n\n\n\n
          The hybridization of selenium in SeH\u2082 is sp\u00b3<\/strong>. Selenium has two single bonds with hydrogen and two lone pairs, resulting in a tetrahedral electron domain geometry, which corresponds to sp\u00b3 hybridization.<\/p>\n\n\n\n
          \n\n\n\n
          (CO\u2083)\u00b2\u207b (Carbonate Ion)<\/h3>\n\n\n\n
          Lewis Structure:<\/h4>\n\n\n\n
          The carbonate ion has 24 valence electrons:<\/p>\n\n\n\n
            \n
          • Carbon (C) has 4 valence electrons.<\/li>\n\n\n\n
          • Each oxygen (O) has 6 valence electrons, and there are three oxygens in the ion.<\/li>\n\n\n\n
          • The 2 negative charges contribute 2 additional electrons.<\/li>\n<\/ul>\n\n\n\n
            Step-by-step process:<\/p>\n\n\n\n
              \n
            1. Valence Electrons<\/strong>:
              4+3(6)+2=24\u00a0electrons4 + 3(6) + 2 = 24 \\text{ electrons}4+3(6)+2=24\u00a0electrons<\/li>\n\n\n\n
            2. Bonding<\/strong>:\n
                \n
              • Carbon is the central atom, and it will form a double bond with one oxygen and single bonds with the other two oxygens.<\/li>\n\n\n\n
              • To fulfill the octet rule, the singly bonded oxygens will have lone pairs of electrons, and the doubly bonded oxygen will have no lone pairs.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
                The structure is:<\/p>\n\n\n\n
                    O
                    ||
                C - O
                    |
                    O
                <\/code><\/pre>\n\n\n\n
                This structure has resonance, where the double bond can be alternated between any of the three oxygen atoms, meaning all three C-O bonds are equivalent, with partial double bond character.<\/p>\n\n\n\n
                Geometry:<\/h4>\n\n\n\n
                The geometry of the carbonate ion is trigonal planar<\/strong>. The bond angles are approximately 120\u00b0, and the electron groups around carbon are symmetrically arranged in a plane.<\/p>\n\n\n\n
                Polarity:<\/h4>\n\n\n\n
                (CO\u2083)\u00b2\u207b is non-polar<\/strong>. Although individual C-O bonds are polar due to the difference in electronegativity, the molecule is symmetrical, and the dipoles cancel each other out.<\/p>\n\n\n\n
                Hybridization:<\/h4>\n\n\n\n
                The hybridization of carbon in (CO\u2083)\u00b2\u207b is sp\u00b2<\/strong>. Carbon forms three equivalent bonds with oxygen atoms, and the electron domains around carbon are arranged in a trigonal planar geometry, which corresponds to sp\u00b2 hybridization.<\/p>\n\n\n\n
                \n\n\n\n
                These explanations should provide a thorough understanding of both molecules! Let me know if you need any further clarifications.<\/p>\n\n\n\n
                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
                For the molecule SeH2: Email the correct Lewis structure (including resonance forms, if any to ec What is the geometry of the molecule? Is it polar or non-polar? What is the hybridization of the central atom? QUESTION 8 For the molecule (CO3)2-: Email the correct Lewis structure (including resonance forms, if any to edo What […]<\/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-258820","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258820","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=258820"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258820\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258820"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258820"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258820"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}