{"id":261721,"date":"2025-07-20T08:08:42","date_gmt":"2025-07-20T08:08:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261721"},"modified":"2025-07-20T08:08:46","modified_gmt":"2025-07-20T08:08:46","slug":"draw-the-lewis-structure-for-phosphite-po3-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/draw-the-lewis-structure-for-phosphite-po3-2\/","title":{"rendered":"Draw the Lewis structure for phosphite, PO3-"},"content":{"rendered":"\n

Draw the Lewis structure for phosphite, PO3-. What is the molecular geometry of this ion? a) Trigonal planar b) Trigonal pyramidal c) Trigonal bipyramidal d) None of the above 22. Draw the Lewis structure for iodine tribromide, IBr3. What is the molecular geometry of this molecule? a) Trigonal planar b) Trigonal pyramidal c) Trigonal bipyramidal d) None of the above<\/p>\n\n\n\n

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

Phosphite Ion (PO\u2083\u00b3\u207b) Lewis Structure and Molecular Geometry<\/h3>\n\n\n\n
    \n
  1. Lewis Structure for PO\u2083\u00b3\u207b:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Phosphite (PO\u2083\u00b3\u207b) consists of a phosphorus (P) atom centrally bonded to three oxygen (O) atoms.<\/li>\n\n\n\n
    • The phosphorus atom has 5 valence electrons, and each oxygen atom has 6 valence electrons. Since the ion has a charge of -3, there are 3 additional electrons.<\/li>\n\n\n\n
    • Total number of valence electrons = (5 from P) + (3 \u00d7 6 from O) + 3 (due to the charge) = 24 electrons.<\/li>\n\n\n\n
    • The phosphorus atom forms single bonds with each oxygen atom, and one of the oxygen atoms carries a negative charge due to the extra electrons. Two of the oxygen atoms form double bonds with phosphorus.<\/li>\n<\/ul>\n\n\n\n

      The structure of PO\u2083\u00b3\u207b will have:<\/p>\n\n\n\n

        \n
      • One P atom in the center.<\/li>\n\n\n\n
      • Three O atoms around the phosphorus, one double-bonded and two single-bonded. The single-bonded oxygens carry the negative charge.<\/li>\n\n\n\n
      • The molecule adopts a trigonal planar geometry<\/strong> since there are three electron pairs around the central phosphorus atom, and there is no lone pair on phosphorus.<\/li>\n<\/ul>\n\n\n\n
          \n
        1. Molecular Geometry for PO\u2083\u00b3\u207b:<\/strong><\/li>\n<\/ol>\n\n\n\n

          The molecular geometry of PO\u2083\u00b3\u207b is Trigonal Planar<\/strong> because there are no lone pairs on phosphorus and the electron pairs around phosphorus form an equal, symmetrical arrangement. This results in a flat, triangular structure.<\/p>\n\n\n\n

          Correct answer: a) Trigonal planar.<\/strong><\/p>\n\n\n\n

          \n\n\n\n

          Iodine Tribromide (IBr\u2083) Lewis Structure and Molecular Geometry<\/h3>\n\n\n\n
            \n
          1. Lewis Structure for IBr\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Iodine (I) has 7 valence electrons, and each bromine (Br) atom also has 7 valence electrons. Total valence electrons = (7 from I) + (3 \u00d7 7 from Br) = 28 electrons.<\/li>\n\n\n\n
            • Iodine forms single bonds with three bromine atoms, and there are lone pairs of electrons on each bromine.<\/li>\n\n\n\n
            • The iodine atom has three single bonds to bromine atoms, and there are three lone pairs of electrons on iodine.<\/li>\n<\/ul>\n\n\n\n

              The structure of IBr\u2083 will have:<\/p>\n\n\n\n

                \n
              • Iodine (I) at the center with three single bonds to the bromine atoms.<\/li>\n\n\n\n
              • Three lone pairs of electrons on iodine.<\/li>\n\n\n\n
              • This creates a T-shaped<\/strong> molecular geometry.<\/li>\n<\/ul>\n\n\n\n
                  \n
                1. Molecular Geometry for IBr\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n

                  The molecular geometry of IBr\u2083 is T-shaped<\/strong> due to the three bonding pairs and two lone pairs on the iodine atom. The lone pairs of electrons repel the bonding pairs, forcing the molecule into this T-shaped arrangement. Since iodine is in Period 5 of the periodic table, it can accommodate more than an octet of electrons.<\/p>\n\n\n\n

                  Correct answer: d) None of the above.<\/strong><\/p>\n\n\n\n

                  (Note: Although iodine tribromide is T-shaped, none of the provided options in the question exactly match the geometry.)<\/p>\n\n\n\n

                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Draw the Lewis structure for phosphite, PO3-. What is the molecular geometry of this ion? a) Trigonal planar b) Trigonal pyramidal c) Trigonal bipyramidal d) None of the above 22. Draw the Lewis structure for iodine tribromide, IBr3. What is the molecular geometry of this molecule? a) Trigonal planar b) Trigonal pyramidal c) Trigonal bipyramidal […]<\/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-261721","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261721","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=261721"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261721\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261721"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261721"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261721"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}