{"id":181490,"date":"2025-01-10T20:04:26","date_gmt":"2025-01-10T20:04:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181490"},"modified":"2025-01-10T20:04:29","modified_gmt":"2025-01-10T20:04:29","slug":"kno3-draw-the-lewis-dot-structure-for-kno3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/10\/kno3-draw-the-lewis-dot-structure-for-kno3\/","title":{"rendered":"KNO3 Draw the Lewis dot structure for KNO3"},"content":{"rendered":"\n

KNO3<\/p>\n\n\n\n

Draw the Lewis dot structure for KNO3. Include all nonbonding electrons. Show the formal charges of all atoms. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom.<\/p>\n\n\n\n

(I already tried N+, N2+, N and it was wrong…and this question want us to put the charge on each element instead of using bracket)<\/p>\n\n\n\n

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

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

The molecular compound potassium nitrate (KNO\u2083) consists of the potassium cation (K\u207a), the nitrate anion (NO\u2083\u207b), and the associated bonds between nitrogen and oxygen atoms. To draw the Lewis structure of KNO\u2083, we must focus on the NO\u2083\u207b anion and then place the potassium ion.<\/p>\n\n\n\n

Steps for drawing the Lewis structure of NO\u2083\u207b:<\/h3>\n\n\n\n
    \n
  1. Count valence electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Nitrogen (N) is in group 15, so it has 5 valence electrons.<\/li>\n\n\n\n
    • Oxygen (O) is in group 16, so each oxygen atom has 6 valence electrons. There are three oxygen atoms, so 6 * 3 = 18 electrons from oxygen.<\/li>\n\n\n\n
    • The extra negative charge on the nitrate ion (NO\u2083\u207b) means we add 1 more electron.<\/li>\n\n\n\n
    • Total valence electrons: 5 (N) + 18 (O) + 1 (charge) = 24 electrons.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Construct the skeletal structure<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Nitrogen is the central atom, with three oxygen atoms around it.<\/li>\n\n\n\n
        • Draw single bonds between nitrogen and each oxygen atom. This accounts for 6 electrons (2 electrons per bond).<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Distribute the remaining electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • After placing the bonds, we have 24 – 6 = 18 electrons remaining.<\/li>\n\n\n\n
            • Place these 18 electrons as lone pairs around the oxygen atoms, ensuring each oxygen gets 8 electrons (two electrons in a single bond and 6 electrons in lone pairs).<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Adjust for formal charges<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • Formal charge = (Valence electrons) – (Nonbonding electrons) – (1\/2 Bonding electrons)<\/li>\n\n\n\n
                • For nitrogen<\/strong>:\n
                    \n
                  • Nitrogen has 5 valence electrons, 0 nonbonding electrons, and shares 6 bonding electrons (2 per bond to oxygen).<\/li>\n\n\n\n
                  • Formal charge on nitrogen = 5 – 0 – (6\/2) = 5 – 3 = +2.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • For oxygen<\/strong> (in a single bond):\n
                      \n
                    • Each oxygen atom has 6 valence electrons, 6 nonbonding electrons, and shares 2 bonding electrons.<\/li>\n\n\n\n
                    • Formal charge on each oxygen = 6 – 6 – (2\/2) = 6 – 6 – 1 = -1.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                    • For oxygen<\/strong> (in a double bond):\n
                        \n
                      • This oxygen shares 4 electrons in the double bond and has 4 nonbonding electrons.<\/li>\n\n\n\n
                      • Formal charge = 6 – 4 – (4\/2) = 6 – 4 – 2 = 0.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
                          \n
                        1. Final structure<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                            \n
                          • The nitrogen atom has a formal charge of +1, the doubly bonded oxygen has no formal charge, and the two singly bonded oxygens each have a formal charge of -1.<\/li>\n\n\n\n
                          • This gives a net charge of -1 for the nitrate ion (NO\u2083\u207b).<\/li>\n<\/ul>\n\n\n\n

                            Conclusion:<\/h3>\n\n\n\n

                            The final structure of the nitrate ion (NO\u2083\u207b) is:<\/p>\n\n\n\n

                              \n
                            • Nitrogen (N) in the center with a formal charge of +1.<\/li>\n\n\n\n
                            • One oxygen (O) double-bonded to nitrogen with a formal charge of 0.<\/li>\n\n\n\n
                            • Two oxygens (O) single-bonded to nitrogen, each with a formal charge of -1.<\/li>\n<\/ul>\n\n\n\n

                              K\u207a ion<\/strong>: The potassium ion (K\u207a) is simply a potassium atom that has lost one electron, giving it a +1 charge. It is not part of the Lewis structure for NO\u2083\u207b but associates with the nitrate ion to form KNO\u2083.<\/p>\n\n\n\n

                              The final Lewis structure of KNO\u2083 is:<\/p>\n\n\n\n

                              K\u207a [O=N-O]\u207b (O)<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
                              KNO3 Draw the Lewis dot structure for KNO3. Include all nonbonding electrons. Show the formal charges of all atoms. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom. (I already tried N+, N2+, N and it was wrong…and this question want us to put the […]<\/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-181490","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181490","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=181490"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181490\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}