{"id":188716,"date":"2025-02-07T08:35:50","date_gmt":"2025-02-07T08:35:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188716"},"modified":"2025-02-07T08:36:08","modified_gmt":"2025-02-07T08:36:08","slug":"the-lewis-structure-for-the-chlorate-ion-cio3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/the-lewis-structure-for-the-chlorate-ion-cio3\/","title":{"rendered":"The Lewis Structure For The Chlorate Ion, CIO3"},"content":{"rendered":"\n

The Lewis Structure For The Chlorate Ion, CIO3, That Expands The Octet It Is Shown In The Figure Below. Answer The Following Questions: (16 Pts) <\/p>\n\n\n\n

a. Include the electron counts: valence, bonding, nonbonding. (3 pts)
b. Calculate the formal charge for each atom in the molecule. (6 pts)
c. Determine the hybridization on the central atom. (3 pts)
d. Determine the number of sigma bonds and pi bonds. (4 pts)<\/p>\n\n\n\n

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

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

It looks like I can’t access the image you’ve linked. However, I can walk you through the Lewis structure and answer the questions about the chlorate ion, ( \\text{ClO}_3^- ), based on what you described. Here’s a breakdown:<\/p>\n\n\n\n

A. Electron Counts:<\/strong><\/h3>\n\n\n\n
    \n
  1. Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Chlorine (Cl) is in group 7, so it contributes 7 electrons.<\/li>\n\n\n\n
    • Oxygen (O) is in group 6, so each oxygen atom contributes 6 electrons. There are 3 oxygen atoms, so this gives ( 3 \\times 6 = 18 ) electrons.<\/li>\n\n\n\n
    • The ion has a negative charge, which adds 1 more electron.<\/li>\n\n\n\n
    • Total valence electrons<\/strong> = ( 7 (\\text{from Cl}) + 18 (\\text{from O}) + 1 (\\text{from the negative charge}) = 26 ) valence electrons.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Bonding and Nonbonding Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • Chlorine forms a central atom bonded to 3 oxygen atoms. Assuming it uses single bonds with each oxygen, the bonding electrons will account for ( 3 \\times 2 = 6 ) electrons.<\/li>\n\n\n\n
        • The remaining 20 electrons will be placed as lone pairs on the oxygen atoms, considering that each oxygen will typically have 3 lone pairs (6 electrons), and the chlorine may have lone pairs as well.<\/li>\n<\/ul>\n\n\n\n

          B. Formal Charge Calculation:<\/strong><\/h3>\n\n\n\n

          The formula to calculate the formal charge for each atom is:
          [
          \\text{Formal charge} = \\text{Valence electrons} – (\\text{Bonding electrons} \/ 2) – \\text{Nonbonding electrons}
          ]
          For the chlorate ion, you’ll follow this procedure for each atom:<\/p>\n\n\n\n

            \n
          • Chlorine (Cl):<\/strong><\/li>\n\n\n\n
          • Valence = 7<\/li>\n\n\n\n
          • Bonding = 6 electrons (from 3 single bonds)<\/li>\n\n\n\n
          • Nonbonding = 2 electrons (1 lone pair)<\/li>\n\n\n\n
          • Formal charge = ( 7 – (6 \/ 2) – 2 = 7 – 3 – 2 = 2 ) (This should be 0, indicating resonance helps reduce charge delocalization.)<\/li>\n\n\n\n
          • Oxygen (O):<\/strong><\/li>\n\n\n\n
          • For each oxygen:\n
              \n
            • Valence = 6<\/li>\n\n\n\n
            • Bonding = 2 electrons (1 single bond)<\/li>\n\n\n\n
            • Nonbonding = 6 electrons (3 lone pairs)<\/li>\n\n\n\n
            • Formal charge = ( 6 – (2 \/ 2) – 6 = 6 – 1 – 6 = -1 )<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

              C. Hybridization on the Central Atom (Cl):<\/strong><\/h3>\n\n\n\n
                \n
              • Chlorine in ( \\text{ClO}_3^- ) has 3 bonding regions (with 3 oxygen atoms) and 1 lone pair. Thus, chlorine is sp^3 hybridized<\/strong>. The lone pair accounts for one of the regions, and the three bonds to oxygen contribute to the bonding regions.<\/li>\n<\/ul>\n\n\n\n

                D. Expanding the Octet:<\/strong><\/h3>\n\n\n\n
                  \n
                • Chlorine can expand its octet since it is in period 3, where elements can accommodate more than 8 electrons. In ( \\text{ClO}_3^- ), chlorine uses 12 electrons, which is consistent with the expanded octet.<\/li>\n<\/ul>\n\n\n\n

                  This structure and formal charges result in a resonance structure where the negative charge is delocalized over the oxygens, helping to stabilize the ion.<\/p>\n\n\n\n

                  Let me know if you’d like any further details!<\/p>\n","protected":false},"excerpt":{"rendered":"

                  The Lewis Structure For The Chlorate Ion, CIO3, That Expands The Octet It Is Shown In The Figure Below. Answer The Following Questions: (16 Pts) a. Include the electron counts: valence, bonding, nonbonding. (3 pts)b. Calculate the formal charge for each atom in the molecule. (6 pts)c. Determine the hybridization on the central atom. (3 […]<\/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-188716","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188716","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=188716"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188716\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188716"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188716"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188716"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}