{"id":26258,"date":"2025-06-19T11:40:13","date_gmt":"2025-06-19T11:40:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26258"},"modified":"2025-06-19T11:40:15","modified_gmt":"2025-06-19T11:40:15","slug":"molecular-polarity-assignment-this-assignment-involves-using-lewis-theory-valence-bond-theory-and-vsepr-theory-to-describe-the-bonding-and-shape-of-two-molecules","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/molecular-polarity-assignment-this-assignment-involves-using-lewis-theory-valence-bond-theory-and-vsepr-theory-to-describe-the-bonding-and-shape-of-two-molecules\/","title":{"rendered":"Molecular Polarity Assignment This assignment involves using Lewis theory, valence bond theory, and VSEPR theory to describe the bonding and shape of two molecules"},"content":{"rendered":"\n

Molecular Polarity Assignment This assignment involves using Lewis theory, valence bond theory, and VSEPR theory to describe the bonding and shape of two molecules. The concepts of bond and molecular polarity will be used to predict if the assigned molecules possess a molecular dipole. For each of the assigned molecular formulas: e) Identify the hybrid state of all the central atom(s) in each molecule based on the orbital representation and electron geometry. [A2] f) Determine if bond polarity exists using electronegativity values. If so, draw dipole vectors on the shape to show the bond dipoles. [A2] g) Determine if molecular polarity exists. If so, draw a dashed vector to indicate the direction of the molecular dipole. [A2] Assigned compounds: I) carbonyl fluoride, COF2 II) diboron tetrafluoride, B2F4<\/p>\n\n\n\n

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

Here is the analysis of the two assigned compounds\u2014carbonyl fluoride (COF\u2082)<\/strong> and diboron tetrafluoride (B\u2082F\u2084)<\/strong>\u2014using Lewis theory, valence bond theory, VSEPR theory, and concepts of polarity:<\/p>\n\n\n\n

\n\n\n\n

I) Carbonyl Fluoride, COF\u2082<\/strong><\/h3>\n\n\n\n

(e) Hybridization of the central atom (Carbon):<\/strong><\/p>\n\n\n\n

    \n
  • The central atom is carbon<\/strong>, bonded to one oxygen<\/strong> (double bond) and two fluorine<\/strong> atoms (single bonds).<\/li>\n\n\n\n
  • There are three regions of electron density<\/strong> around carbon.<\/li>\n\n\n\n
  • According to VSEPR theory, this gives a trigonal planar<\/strong> geometry.<\/li>\n\n\n\n
  • The hybridization of carbon is sp\u00b2<\/strong>.<\/li>\n<\/ul>\n\n\n\n

    (f) Bond polarity:<\/strong><\/p>\n\n\n\n

      \n
    • Electronegativity values:\n
        \n
      • C = 2.55<\/li>\n\n\n\n
      • O = 3.44<\/li>\n\n\n\n
      • F = 3.98<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • The C=O bond<\/strong> and each C\u2013F bond<\/strong> are polar since the bonded atoms have differing electronegativities.<\/li>\n\n\n\n
      • Draw dipole vectors from carbon toward oxygen and fluorine<\/strong>, as both are more electronegative than carbon.<\/li>\n<\/ul>\n\n\n\n

        (g) Molecular polarity:<\/strong><\/p>\n\n\n\n

          \n
        • The bond dipoles do not<\/strong> cancel out because of the differing strengths and directions (O is less electronegative than F, but the double bond adds electron density).<\/li>\n\n\n\n
        • Thus, COF\u2082 is a polar molecule<\/strong>.<\/li>\n\n\n\n
        • A dashed arrow representing the molecular dipole should point from carbon toward the more negative side<\/strong>, likely between the direction of the fluorines and oxygen.<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          II) Diboron Tetrafluoride, B\u2082F\u2084<\/strong><\/h3>\n\n\n\n

          (e) Hybridization of the central atoms (each Boron):<\/strong><\/p>\n\n\n\n

            \n
          • Each boron is bonded to two fluorine atoms<\/strong> and to the other boron atom<\/strong>.<\/li>\n\n\n\n
          • That gives three regions of electron density<\/strong> around each boron.<\/li>\n\n\n\n
          • Geometry is trigonal planar<\/strong> around each boron.<\/li>\n\n\n\n
          • So, hybridization of each boron is sp\u00b2<\/strong>.<\/li>\n<\/ul>\n\n\n\n

            (f) Bond polarity:<\/strong><\/p>\n\n\n\n

              \n
            • Electronegativity:\n
                \n
              • B = 2.04<\/li>\n\n\n\n
              • F = 3.98<\/li>\n<\/ul>\n<\/li>\n\n\n\n
              • The B\u2013F bonds<\/strong> are highly polar<\/strong> with dipole vectors pointing toward fluorine<\/strong>.<\/li>\n\n\n\n
              • The B\u2013B bond<\/strong> is nonpolar (same atoms).<\/li>\n<\/ul>\n\n\n\n

                (g) Molecular polarity:<\/strong><\/p>\n\n\n\n

                  \n
                • The B\u2013F bond dipoles<\/strong> are arranged symmetrically<\/strong> in a planar molecule.<\/li>\n\n\n\n
                • Since dipoles point in opposite directions and are equal in magnitude, they cancel out<\/strong>.<\/li>\n\n\n\n
                • Therefore, B\u2082F\u2084 is nonpolar overall<\/strong>.<\/li>\n\n\n\n
                • No molecular dipole vector<\/strong> is drawn.<\/li>\n<\/ul>\n\n\n\n
                  \n\n\n\n

                  Summary:<\/strong><\/h3>\n\n\n\n
                  Molecule<\/th>Hybridization<\/th>Electron Geometry<\/th>Molecular Geometry<\/th>Bond Polarity<\/th>Molecular Polarity<\/th><\/tr><\/thead>
                  COF\u2082<\/td>sp\u00b2<\/td>Trigonal Planar<\/td>Trigonal Planar<\/td>Yes<\/td>Polar<\/strong><\/td><\/tr>
                  B\u2082F\u2084<\/td>sp\u00b2 (both B)<\/td>Trigonal Planar<\/td>Planar<\/td>Yes<\/td>Nonpolar<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Molecular Polarity Assignment This assignment involves using Lewis theory, valence bond theory, and VSEPR theory to describe the bonding and shape of two molecules. The concepts of bond and molecular polarity will be used to predict if the assigned molecules possess a molecular dipole. For each of the assigned molecular formulas: e) Identify the hybrid […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26258","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26258","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26258"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26258\/revisions"}],"predecessor-version":[{"id":26260,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26258\/revisions\/26260"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26258"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26258"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26258"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}