{"id":257497,"date":"2025-07-17T13:48:30","date_gmt":"2025-07-17T13:48:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257497"},"modified":"2025-07-17T13:48:33","modified_gmt":"2025-07-17T13:48:33","slug":"draw-the-lewis-structure-for-phosphite-po3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/draw-the-lewis-structure-for-phosphite-po3\/","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

Let’s break down both molecules and their molecular geometries.<\/p>\n\n\n\n

1. Phosphite Ion, PO\u2083\u00b3\u207b:<\/strong><\/h3>\n\n\n\n

To begin, the phosphite ion<\/strong> (PO\u2083\u00b3\u207b) consists of one phosphorus (P) atom and three oxygen (O) atoms. Phosphorus in the center, as it is less electronegative than oxygen, forms single bonds with the oxygen atoms. The ion carries a -3 charge, so we need to account for the extra electrons in the structure.<\/p>\n\n\n\n

Lewis Structure for PO\u2083\u00b3\u207b:<\/strong><\/p>\n\n\n\n

    \n
  • Step 1:<\/strong> Phosphorus (P) is the central atom since it is less electronegative.<\/li>\n\n\n\n
  • Step 2:<\/strong> Each oxygen atom forms a single bond with phosphorus. Phosphorus will need three bonds to satisfy the valency of three bonds.<\/li>\n\n\n\n
  • Step 3:<\/strong> After bonding, oxygen atoms each have lone pairs to complete their octets. Due to the -3 charge, one of the oxygen atoms will likely have an extra lone pair to carry the negative charge.<\/li>\n<\/ul>\n\n\n\n

    After placing all the electrons, you will notice that the phosphite ion has three bonding regions<\/strong> around the central phosphorus atom. Each oxygen atom is involved in a single bond with phosphorus, and there are lone pairs on oxygen atoms. The phosphite ion has a trigonal planar electron domain geometry<\/strong> around phosphorus. Since there are no lone pairs on phosphorus, the molecular geometry is also trigonal planar<\/strong>.<\/p>\n\n\n\n

    Thus, the correct answer for phosphite ion (PO\u2083\u00b3\u207b)<\/strong> is:<\/p>\n\n\n\n

      \n
    • a) Trigonal planar.<\/strong><\/li>\n<\/ul>\n\n\n\n

      2. Iodine Tribromide, IBr\u2083:<\/strong><\/h3>\n\n\n\n

      Iodine tribromide (IBr\u2083) is a molecule where iodine (I) is the central atom, and three bromine (Br) atoms are attached to it. Iodine is in period 5, so it can accommodate more than 8 electrons in its valence shell.<\/p>\n\n\n\n

      Lewis Structure for IBr\u2083:<\/strong><\/p>\n\n\n\n

        \n
      • Step 1:<\/strong> Iodine is the central atom, as it is less electronegative than bromine.<\/li>\n\n\n\n
      • Step 2:<\/strong> Iodine forms single bonds with three bromine atoms, using up three valence electrons for bonding.<\/li>\n\n\n\n
      • Step 3:<\/strong> Each bromine atom gets three lone pairs, and iodine completes its octet with the bonds. However, iodine can expand its valence shell to accommodate more than 8 electrons, forming an expanded octet<\/strong>.<\/li>\n\n\n\n
      • Step 4:<\/strong> The three bromine atoms are placed around iodine, and there are no lone pairs on iodine.<\/li>\n<\/ul>\n\n\n\n

        With three bonding regions and no lone pairs on the iodine, the electron domain geometry<\/strong> around iodine is trigonal bipyramidal<\/strong>. The molecular geometry<\/strong> is also trigonal bipyramidal<\/strong>, as there are no lone pairs on iodine to distort the shape.<\/p>\n\n\n\n

        Thus, the correct answer for iodine tribromide (IBr\u2083)<\/strong> is:<\/p>\n\n\n\n

          \n
        • d) None of the above<\/strong> (since it is trigonal bipyramidal, but the option does not specify this).<\/li>\n<\/ul>\n\n\n\n

          Summary of Answers:<\/h3>\n\n\n\n
            \n
          1. For PO\u2083\u00b3\u207b<\/strong>: a) Trigonal planar<\/strong>.<\/li>\n\n\n\n
          2. For IBr\u2083<\/strong>: d) None of the above<\/strong> (because the geometry is trigonal bipyramidal).<\/li>\n<\/ol>\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-257497","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257497","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=257497"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257497\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}