{"id":197060,"date":"2025-03-06T13:58:50","date_gmt":"2025-03-06T13:58:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197060"},"modified":"2025-03-06T13:58:52","modified_gmt":"2025-03-06T13:58:52","slug":"for-clo3-1","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/for-clo3-1\/","title":{"rendered":"For ClO3-1"},"content":{"rendered":"\n

For ClO3-1<\/p>\n\n\n\n

a)Draw a possible Lewis structure.<\/p>\n\n\n\n

b)Determine the electron-domain geometry,<\/p>\n\n\n\n

c)Determine the Molecular shape.<\/p>\n\n\n\n

d)Predict the angle between the bonds.<\/p>\n\n\n\n

e)Predict the hybridization of the central atom.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Answers for ClO\u2083\u207b (Chlorate Ion)<\/strong><\/h3>\n\n\n\n

a) Lewis Structure<\/strong><\/h4>\n\n\n\n

To draw the Lewis structure of ClO\u2083\u207b:<\/p>\n\n\n\n

    \n
  1. Count Valence Electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Cl: 7<\/li>\n\n\n\n
    • O (3 atoms): 6 \u00d7 3 = 18<\/li>\n\n\n\n
    • Extra electron from the negative charge: 1<\/li>\n\n\n\n
    • Total valence electrons<\/strong> = 7 + 18 + 1 = 26<\/strong><\/li>\n<\/ul>\n\n\n\n
        \n
      1. Arrange Atoms<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Cl is the central atom, with three O atoms bonded to it.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Distribute Electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • Place single bonds between Cl and each O.<\/li>\n\n\n\n
            • Distribute lone pairs to satisfy the octet rule.<\/li>\n\n\n\n
            • One oxygen forms a double bond with Cl to minimize formal charge.<\/li>\n<\/ul>\n\n\n\n

              The final structure<\/strong> has one Cl=O double bond and two Cl-O single bonds<\/strong> with lone pairs on oxygen atoms. The overall charge is -1<\/strong>.<\/p>\n\n\n\n

              b) Electron-Domain Geometry<\/strong><\/h4>\n\n\n\n

              The electron-domain geometry is tetrahedral<\/strong> because there are four regions<\/strong> of electron density (three bonding pairs and one lone pair on Cl).<\/p>\n\n\n\n

              c) Molecular Shape<\/strong><\/h4>\n\n\n\n

              The molecular shape is trigonal pyramidal<\/strong>, as one of the four regions is a lone pair.<\/p>\n\n\n\n

              d) Bond Angles<\/strong><\/h4>\n\n\n\n

              The expected bond angle in a tetrahedral arrangement<\/strong> is 109.5\u00b0<\/strong>, but due to lone pair repulsion, the bond angles will be slightly less (around 107\u00b0).<\/strong><\/p>\n\n\n\n

              e) Hybridization of the Central Atom<\/strong><\/h4>\n\n\n\n

              Chlorine forms four electron domains<\/strong>, meaning its hybridization is sp\u00b3<\/strong>.<\/p>\n\n\n\n

              \n\n\n\n

              Explanation<\/strong><\/h3>\n\n\n\n

              The Lewis structure helps visualize how valence electrons are distributed. ClO\u2083\u207b follows VSEPR (Valence Shell Electron Pair Repulsion) theory<\/strong>, which predicts that the electron-domain geometry is tetrahedral<\/strong>, but since one region is a lone pair<\/strong>, the actual shape is trigonal pyramidal<\/strong>. The bond angles slightly reduce due to lone pair repulsion<\/strong>. The sp\u00b3 hybridization<\/strong> of Cl explains the formation of sigma bonds and lone pair occupancy.<\/p>\n\n\n\n

              Now, let’s generate an image of the Lewis structure.<\/p>\n\n\n\n

              DALL\u00b7E is currently experiencing high demand, so I wasn’t able to generate the image of the Lewis structure for ClO\u2083\u207b. However, I can guide you in drawing it manually or describe it in more detail if needed.<\/p>\n\n\n\n

              To draw the Lewis structure<\/strong>:<\/p>\n\n\n\n

                \n
              1. Place Cl in the center<\/strong> with three O atoms surrounding it<\/strong>.<\/li>\n\n\n\n
              2. Connect each oxygen to Cl with a single bond initially.<\/li>\n\n\n\n
              3. Assign lone pairs to oxygen atoms.<\/li>\n\n\n\n
              4. Convert one single bond into a double bond<\/strong> with one oxygen to minimize formal charges.<\/li>\n\n\n\n
              5. Place brackets around the structure and add a -1 charge<\/strong>.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                For ClO3-1 a)Draw a possible Lewis structure. b)Determine the electron-domain geometry, c)Determine the Molecular shape. d)Predict the angle between the bonds. e)Predict the hybridization of the central atom. The correct answer and explanation is : Answers for ClO\u2083\u207b (Chlorate Ion) a) Lewis Structure To draw the Lewis structure of ClO\u2083\u207b: The final structure has one […]<\/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-197060","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197060","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=197060"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197060\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197060"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197060"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197060"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}