{"id":221711,"date":"2025-05-30T12:52:47","date_gmt":"2025-05-30T12:52:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221711"},"modified":"2025-05-30T12:52:50","modified_gmt":"2025-05-30T12:52:50","slug":"draw-on-paper-a-lewis-structure-for-bcl3-and-answer-the-following-questions-based-on-your-drawing","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/draw-on-paper-a-lewis-structure-for-bcl3-and-answer-the-following-questions-based-on-your-drawing\/","title":{"rendered":"Draw (on paper) a Lewis structure for BCl3 and answer the following questions based on your drawing."},"content":{"rendered":"\n

Draw (on paper) a Lewis structure for BCl3 and answer the following questions based on your drawing. 1. For the central boron atom: The number of lone pairs The number of single bonds = The number of double bonds = 2. The central boron atom A. obeys the octet rule. B. has an incomplete octet. C. has an expanded octet. Draw (on paper) a Lewis structure for PO43- and answer the following questions based on your drawing. Do not draw double bonds unless they are needed for the central atom to obey the octet rule. 1. For the central phosphorus atom: the number of lone pairs the number of single bonds the number of double bonds = 2. The central phosphorus atom A. obeys the octet rule. B. has an incomplete octet. C. has an expanded octet.<\/p>\n\n\n\n

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

\n\n\n\n

1. BCl\u2083 (Boron Trichloride)<\/strong><\/h3>\n\n\n\n

Lewis Structure Summary<\/strong>:<\/p>\n\n\n\n

    \n
  • Boron (B) is the central atom.<\/li>\n\n\n\n
  • Each chlorine (Cl) atom forms a single bond<\/strong> with boron.<\/li>\n\n\n\n
  • Boron has 3 single bonds<\/strong> and no lone pairs<\/strong>.<\/li>\n\n\n\n
  • Each Cl atom has 3 lone pairs<\/strong> to complete its octet.<\/li>\n<\/ul>\n\n\n\n

    Answers<\/strong>:<\/p>\n\n\n\n

      \n
    1. For the central boron atom:\n
        \n
      • Number of lone pairs = 0<\/strong><\/li>\n\n\n\n
      • Number of single bonds = 3<\/strong><\/li>\n\n\n\n
      • Number of double bonds = 0<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • The central boron atom:\n
          \n
        • B. has an incomplete octet.<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

          Explanation<\/strong>:
          Boron is an exception to the octet rule. In BCl\u2083, boron forms only 3 single covalent bonds, sharing a total of 6 electrons<\/strong>, not 8. While this might seem unstable, BCl\u2083 is actually stable due to its small size and ability to accept electron pairs (making it a Lewis acid). Since there are no double bonds or lone pairs on boron, it clearly has an incomplete octet. No double bonds are needed because the molecule is already in its most stable form.<\/p>\n\n\n\n

          \n\n\n\n

          2. PO\u2084\u00b3\u207b (Phosphate Ion)<\/strong><\/h3>\n\n\n\n

          Lewis Structure Summary<\/strong>:<\/p>\n\n\n\n

            \n
          • Phosphorus (P) is the central atom.<\/li>\n\n\n\n
          • There are 4 oxygen atoms<\/strong>, each ideally bonded via single bonds.<\/li>\n\n\n\n
          • The total valence electrons = 5 (P) + 4\u00d76 (O) + 3 (charge) = 32 electrons<\/strong>.<\/li>\n\n\n\n
          • To obey the octet rule, phosphorus must form one double bond<\/strong> with an oxygen and three single bonds with the others. This allows formal charges<\/strong> to be minimized.<\/li>\n<\/ul>\n\n\n\n

            Answers<\/strong>:<\/p>\n\n\n\n

              \n
            1. For the central phosphorus atom:\n
                \n
              • Number of lone pairs = 0<\/strong><\/li>\n\n\n\n
              • Number of single bonds = 3<\/strong><\/li>\n\n\n\n
              • Number of double bonds = 1<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
              • The central phosphorus atom:\n
                  \n
                • C. has an expanded octet.<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

                  Explanation<\/strong>:
                  Phosphorus is in period 3<\/strong> of the periodic table and can expand its octet beyond 8 electrons. In the optimal Lewis structure, phosphorus forms one P=O double bond<\/strong> and three P\u2013O single bonds<\/strong>, for a total of 10 electrons<\/strong> around P (2 from the double bond, 6 from the three single bonds). This expanded octet is stable for phosphorus. No lone pairs are present on the phosphorus atom itself\u2014its valence shell is fully occupied by bonding electrons. Using one double bond also helps reduce formal charges, giving a more stable resonance structure for PO\u2084\u00b3\u207b.<\/p>\n\n\n\n

                  \n\n\n\n

                  Conclusion<\/strong>:<\/h3>\n\n\n\n
                    \n
                  • BCl\u2083<\/strong>: Boron has an incomplete octet<\/strong> (6 electrons, 3 single bonds, 0 lone pairs).<\/li>\n\n\n\n
                  • PO\u2084\u00b3\u207b<\/strong>: Phosphorus has an expanded octet<\/strong> (10 electrons, 3 single bonds, 1 double bond, 0 lone pairs).<\/li>\n<\/ul>\n\n\n\n
                    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                    Draw (on paper) a Lewis structure for BCl3 and answer the following questions based on your drawing. 1. For the central boron atom: The number of lone pairs The number of single bonds = The number of double bonds = 2. The central boron atom A. obeys the octet rule. B. has an incomplete octet. […]<\/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-221711","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221711","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=221711"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221711\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221711"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221711"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}