{"id":26418,"date":"2025-06-19T14:25:09","date_gmt":"2025-06-19T14:25:09","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26418"},"modified":"2025-06-19T14:25:16","modified_gmt":"2025-06-19T14:25:16","slug":"what-is-the-strongest-intermolecular-force-that-would-act-between-ch4-molecules","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-strongest-intermolecular-force-that-would-act-between-ch4-molecules\/","title":{"rendered":"What is the strongest intermolecular force that would act between CH4 molecules"},"content":{"rendered":"\n

What is the strongest intermolecular force that would act between CH4 molecules? Dipole-dipole interactions, hydrogen bonding, London dispersion forces Draw the Lewis structure of CHF2; Select Draw More Erase 3 Rings<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

The strongest intermolecular force acting between CH\u2084 (methane) molecules is London dispersion forces<\/strong>.<\/p>\n\n\n\n

CH\u2084 is a nonpolar molecule. Carbon and hydrogen have a small electronegativity difference, and the tetrahedral symmetry of CH\u2084 ensures that any dipole moments cancel out, resulting in no permanent dipole. Because CH\u2084 is nonpolar and has no hydrogen atoms bonded to highly electronegative atoms like nitrogen, oxygen, or fluorine, neither dipole-dipole interactions nor hydrogen bonding applies. That leaves only London dispersion forces, also known as instantaneous dipole-induced dipole interactions. These arise from momentary shifts in electron distribution that induce weak attractions between adjacent molecules. Although relatively weak, London forces are universally present in all molecules, including nonpolar ones like CH\u2084.<\/p>\n\n\n\n

Lewis Structure of CHF\u2082 (difluoromethane)<\/strong><\/p>\n\n\n\n

    \n
  1. Count valence electrons<\/strong>:\n
      \n
    • Carbon: 4<\/li>\n\n\n\n
    • Hydrogen: 1<\/li>\n\n\n\n
    • Fluorine (\u00d72): 7 \u00d7 2 = 14<\/li>\n\n\n\n
    • Total = 4 + 1 + 14 = 19 electrons (but that\u2019s odd\u2014so actually the correct molecular formula is likely CH\u2082F\u2082<\/strong>, which matches a stable even count of 20 valence electrons)<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

      So assuming the intent was CH\u2082F\u2082:<\/p>\n\n\n\n

        \n
      1. Determine central atom<\/strong>: Carbon will be at the center.<\/li>\n\n\n\n
      2. Attach outer atoms<\/strong>:\n
          \n
        • Carbon forms single bonds with two hydrogen atoms and two fluorine atoms.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Distribute remaining electrons<\/strong>:\n
            \n
          • Each fluorine gets three lone pairs to complete its octet.<\/li>\n\n\n\n
          • Hydrogen atoms need no lone pairs.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Structure<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                H\n    |\nF \u2013 C \u2013 H\n    |\n    F\n<\/code><\/pre>\n\n\n\n
            This tetrahedral geometry minimizes electron repulsion, obeys the octet rule, and reflects correct valence electron distribution. The polar C\u2013F bonds introduce a molecular dipole, making CH\u2082F\u2082 a polar molecule, but CH\u2084 remains nonpolar. Understanding these distinctions helps predict boiling points, solubility, and behavior in polar versus nonpolar solvents. Want to explore how this molecular polarity affects real-world applications like refrigerants or solvents? I\u2019d be happy to dive in.<\/p>\n\n\n\n
            \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
            What is the strongest intermolecular force that would act between CH4 molecules? Dipole-dipole interactions, hydrogen bonding, London dispersion forces Draw the Lewis structure of CHF2; Select Draw More Erase 3 Rings The Correct Answer and Explanation is: The strongest intermolecular force acting between CH\u2084 (methane) molecules is London dispersion forces. CH\u2084 is a nonpolar molecule. […]<\/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-26418","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26418","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=26418"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26418\/revisions"}],"predecessor-version":[{"id":26422,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26418\/revisions\/26422"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26418"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26418"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26418"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}