{"id":187789,"date":"2025-02-06T06:40:47","date_gmt":"2025-02-06T06:40:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187789"},"modified":"2025-02-06T06:40:50","modified_gmt":"2025-02-06T06:40:50","slug":"the-substance-chlorine-monoxide-clog","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/the-substance-chlorine-monoxide-clog\/","title":{"rendered":"The substance chlorine monoxide, ClO(g)"},"content":{"rendered":"\n

8.86 The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has a dipole moment of 1.24 D and the Cl-O bond length is 1.60 \u00c5. (a) Determine the magnitude of the charges on the Cl and O atoms in units of the electronic charge, e. (b) Based on the electronegativities of the elements, which atom would you expect to have a negative charge in the ClO molecule? (c) By using formal charges as a guide, propose the dominant Lewis structure for the molecule. Are the formal charges consistent with your answers to parts (a) and (b)? Can you reconcile any differences you find?<\/p>\n\n\n\n

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

Solution: Chlorine Monoxide (ClO) Analysis<\/strong><\/h3>\n\n\n\n

(a) Determining the Magnitude of the Charges on Cl and O in Units of Electronic Charge (e)<\/strong><\/h4>\n\n\n\n

The dipole moment ((\\mu)) of ClO is given as 1.24 D<\/strong> (Debye). The bond length ((r)) is 1.60 \u00c5<\/strong> (1.60 \u00d7 10\u207b\u00b9\u2070 m). The relationship between dipole moment and charge separation is:<\/p>\n\n\n\n

[
\\mu = q \\cdot r
]<\/p>\n\n\n\n

where:<\/p>\n\n\n\n

    \n
  • ( \\mu = 1.24 ) D ( = 1.24 \\times 3.336 \\times 10^{-30} ) C\u00b7m<\/li>\n\n\n\n
  • ( r = 1.60 \\times 10^{-10} ) m<\/li>\n\n\n\n
  • ( q ) is the charge in Coulombs.<\/li>\n<\/ul>\n\n\n\n

    Solving for ( q ):<\/p>\n\n\n\n

    [
    q = \\frac{\\mu}{r} = \\frac{(1.24 \\times 3.336 \\times 10^{-30})}{(1.60 \\times 10^{-10})}
    ]<\/p>\n\n\n\n

    [
    q = 2.59 \\times 10^{-20} \\text{ C}
    ]<\/p>\n\n\n\n

    Since the elementary charge ( e ) is 1.602 \u00d7 10\u207b\u00b9\u2079 C<\/strong>, the charge in units of ( e ) is:<\/p>\n\n\n\n

    [
    q = \\frac{2.59 \\times 10^{-20}}{1.602 \\times 10^{-19}} = 0.162 e
    ]<\/p>\n\n\n\n

    Thus, the magnitude of the charge on Cl and O is 0.162 e<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    (b) Which Atom Carries the Negative Charge?<\/strong><\/h4>\n\n\n\n

    Electronegativity values:<\/p>\n\n\n\n

      \n
    • Chlorine (Cl):<\/strong> 3.16<\/li>\n\n\n\n
    • Oxygen (O):<\/strong> 3.44<\/li>\n<\/ul>\n\n\n\n

      Since oxygen has a higher electronegativity<\/strong> than chlorine, it will attract more electron density, meaning Oxygen (O) carries the negative charge<\/strong> while Chlorine (Cl) carries the positive charge<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      (c) Dominant Lewis Structure and Formal Charges<\/strong><\/h4>\n\n\n\n

      The dominant Lewis structure<\/strong> for ClO consists of a single bond<\/strong> between Cl and O, with one lone pair on Cl<\/strong> and three lone pairs on O<\/strong>. A resonance structure<\/strong> with a Cl=O double bond is also possible.<\/p>\n\n\n\n

      To calculate formal charges<\/strong>:<\/p>\n\n\n\n

      [
      \\text{Formal Charge} = \\text{Valence Electrons} – (\\text{Lone Pair Electrons} + \\frac{\\text{Bonding Electrons}}{2})
      ]<\/p>\n\n\n\n

      For Cl<\/strong> (7 valence electrons):<\/p>\n\n\n\n

        \n
      • Cl in Cl\u2013O (single bond): Formal charge = 7 – (4 + 1) = +2<\/strong><\/li>\n\n\n\n
      • Cl in Cl=O (double bond): Formal charge = 7 – (2 + 2) = +1<\/strong><\/li>\n<\/ul>\n\n\n\n

        For O<\/strong> (6 valence electrons):<\/p>\n\n\n\n

          \n
        • O in Cl\u2013O: Formal charge = 6 – (6 + 1) = -1<\/strong><\/li>\n\n\n\n
        • O in Cl=O: Formal charge = 6 – (4 + 2) = 0<\/strong><\/li>\n<\/ul>\n\n\n\n

          The single-bond structure (Cl\u2013O)<\/strong> with a negative formal charge on O<\/strong> and positive charge on Cl<\/strong> aligns best with part (b), but a Cl=O resonance structure also contributes.<\/p>\n\n\n\n

          Reconciling Formal Charges with (a) and (b):<\/strong><\/p>\n\n\n\n

            \n
          • The dipole calculation suggests partial charge separation (\u00b10.162 e)<\/strong>, which is much lower than full ionic charges.<\/li>\n\n\n\n
          • The formal charge suggests Cl is positive and O is negative<\/strong>, which agrees<\/strong> with the electronegativity argument.<\/li>\n\n\n\n
          • The reality is that Cl\u2013O bonding is covalent, but with some ionic character<\/strong>, explaining why the formal charge is larger than the calculated partial charges.<\/li>\n<\/ul>\n\n\n\n

            Thus, the dominant Lewis structure (Cl\u2013O) with a negative charge on O is consistent with electronegativity trends and dipole moment calculations<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

            8.86 The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has a dipole moment of 1.24 D and the Cl-O bond length is 1.60 \u00c5. (a) Determine the magnitude of the charges on the Cl and O atoms in units of the electronic […]<\/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-187789","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187789","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=187789"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187789\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}