{"id":244345,"date":"2025-07-05T04:34:04","date_gmt":"2025-07-05T04:34:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244345"},"modified":"2025-07-05T04:34:07","modified_gmt":"2025-07-05T04:34:07","slug":"draw-all-possible-resonance-lewis-structures-for-no2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/draw-all-possible-resonance-lewis-structures-for-no2\/","title":{"rendered":"Draw all possible resonance lewis structures for NO2+."},"content":{"rendered":"\n

Draw all possible resonance lewis structures for NO2+. Use formal charges to identify the best lewis structure amongst them.<\/p>\n\n\n\n

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

To determine the best Lewis structure for the NO2+\\text{NO}_2^+NO2+\u200b ion, we need to first draw all the possible resonance structures and use formal charges to choose the most stable one.<\/p>\n\n\n\n

Step 1: Drawing the possible Lewis structures<\/h3>\n\n\n\n

For NO2+\\text{NO}_2^+NO2+\u200b, we know it has 17 valence electrons (Nitrogen has 5, and each Oxygen has 6, but there\u2019s one positive charge, so we subtract 1 electron).<\/p>\n\n\n\n

Structure 1:<\/strong><\/p>\n\n\n\n

    \n
  1. Place the nitrogen in the center and connect it to the two oxygens with single bonds.<\/li>\n\n\n\n
  2. Distribute the remaining electrons as lone pairs on the oxygens.<\/li>\n\n\n\n
  3. There will be a formal positive charge on nitrogen and formal negative charges on the oxygens.<\/li>\n<\/ol>\n\n\n\n

    Structure 2:<\/strong><\/p>\n\n\n\n

      \n
    1. Place the nitrogen in the center, as before, but this time, form a double bond with one oxygen and a single bond with the other.<\/li>\n\n\n\n
    2. Place the remaining electrons as lone pairs.<\/li>\n\n\n\n
    3. The nitrogen will still carry a positive charge, but one oxygen will have a formal negative charge, and the other will be neutral.<\/li>\n<\/ol>\n\n\n\n

      Structure 3:<\/strong><\/p>\n\n\n\n

        \n
      1. Place a double bond between nitrogen and both oxygens.<\/li>\n\n\n\n
      2. Each oxygen will have a formal negative charge, and nitrogen will carry the positive charge.<\/li>\n<\/ol>\n\n\n\n

        Step 2: Calculating formal charges<\/h3>\n\n\n\n

        The formal charge formula is:Formal charge=(Valence electrons)\u2212(Nonbonding electrons)\u2212(Bonding electrons2)\\text{Formal charge} = (\\text{Valence electrons}) – (\\text{Nonbonding electrons}) – \\left( \\frac{\\text{Bonding electrons}}{2} \\right)Formal charge=(Valence electrons)\u2212(Nonbonding electrons)\u2212(2Bonding electrons\u200b)<\/p>\n\n\n\n

          \n
        • Structure 1<\/strong>:
          • Nitrogen: 5\u22120\u221222=+15 – 0 – \\frac{2}{2} = +15\u22120\u221222\u200b=+1<\/li>
          • Oxygen 1: 6\u22126\u221222=\u221216 – 6 – \\frac{2}{2} = -16\u22126\u221222\u200b=\u22121<\/li>
          • Oxygen 2: 6\u22126\u221222=\u221216 – 6 – \\frac{2}{2} = -16\u22126\u221222\u200b=\u22121<\/li><\/ul>Formal charges: Nitrogen +1, both oxygens -1. This is less stable due to two negative charges on oxygens.<\/li>\n\n\n\n
          • Structure 2<\/strong>:
            • Nitrogen: 5\u22120\u221242=+15 – 0 – \\frac{4}{2} = +15\u22120\u221224\u200b=+1<\/li>
            • Oxygen 1 (double bond): 6\u22124\u221242=06 – 4 – \\frac{4}{2} = 06\u22124\u221224\u200b=0<\/li>
            • Oxygen 2 (single bond): 6\u22126\u221222=\u221216 – 6 – \\frac{2}{2} = -16\u22126\u221222\u200b=\u22121<\/li><\/ul>Formal charges: Nitrogen +1, Oxygen 1 neutral, Oxygen 2 -1. This is a more stable arrangement since the charge is distributed more evenly.<\/li>\n\n\n\n
            • Structure 3<\/strong>:
              • Nitrogen: 5\u22120\u221262=+15 – 0 – \\frac{6}{2} = +15\u22120\u221226\u200b=+1<\/li>
              • Oxygen 1 (double bond): 6\u22124\u221242=06 – 4 – \\frac{4}{2} = 06\u22124\u221224\u200b=0<\/li>
              • Oxygen 2 (double bond): 6\u22124\u221242=06 – 4 – \\frac{4}{2} = 06\u22124\u221224\u200b=0<\/li><\/ul>Formal charges: Nitrogen +1, both oxygens neutral. This structure minimizes the formal charges, which is ideal for stability.<\/li>\n<\/ul>\n\n\n\n

                Step 3: Best Resonance Structure<\/h3>\n\n\n\n

                Structure 3, with double bonds between nitrogen and both oxygens, is the most stable. The formal charges are minimized (both oxygens have a formal charge of 0, and nitrogen carries a +1 charge), which makes it the best resonance structure.<\/p>\n\n\n\n

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

                The best Lewis structure for NO2+\\text{NO}_2^+NO2+\u200b is the one where nitrogen forms double bonds with both oxygens, and the formal charges are distributed as follows:<\/p>\n\n\n\n

                  \n
                • Nitrogen: +1<\/li>\n\n\n\n
                • Oxygen 1: 0<\/li>\n\n\n\n
                • Oxygen 2: 0<\/li>\n<\/ul>\n\n\n\n

                  This minimizes formal charges and stabilizes the ion.<\/p>\n\n\n\n

                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Draw all possible resonance lewis structures for NO2+. Use formal charges to identify the best lewis structure amongst them. The Correct Answer and Explanation is: To determine the best Lewis structure for the NO2+\\text{NO}_2^+NO2+\u200b ion, we need to first draw all the possible resonance structures and use formal charges to choose the most stable 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-244345","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244345","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=244345"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244345\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244345"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244345"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244345"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}