{"id":247131,"date":"2025-07-07T14:51:16","date_gmt":"2025-07-07T14:51:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=247131"},"modified":"2025-07-07T14:51:18","modified_gmt":"2025-07-07T14:51:18","slug":"draw-the-lewis-structure-of-no","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/draw-the-lewis-structure-of-no\/","title":{"rendered":"Draw the Lewis structure of NO"},"content":{"rendered":"\n

) Draw the Lewis structure of N
O. Give two arrangement as in N-N-O and N-O-N. (4 marks) b) Using formal charge calculation, determine the most stable structure. Justify your reason. (6 marks)<\/p>\n\n\n\n

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

a) Drawing the Lewis structure of NO (Nitroxyl or azanone):<\/h3>\n\n\n\n

We are dealing with a molecule that can exist in two different arrangements: N-N-O<\/strong> and N-O-N<\/strong>. Here’s the breakdown for both:<\/p>\n\n\n\n

1. N-N-O<\/strong> arrangement:<\/h4>\n\n\n\n
    \n
  • Nitrogen (N) has 5 valence electrons and Oxygen (O) has 6.<\/li>\n\n\n\n
  • The total number of valence electrons is: 5+5+6=165 + 5 + 6 = 165+5+6=16 electrons.<\/li>\n\n\n\n
  • Draw a single bond between the two Nitrogens (N-N) and between Nitrogen and Oxygen (N-O).<\/li>\n\n\n\n
  • Distribute the remaining electrons around the atoms. Steps:<\/strong>\n
      \n
    • Place a single bond between N and N (2 electrons).<\/li>\n\n\n\n
    • Place another bond between N and O (2 electrons).<\/li>\n\n\n\n
    • You now have 12 electrons remaining (16-4 = 12).<\/li>\n\n\n\n
    • Assign lone pairs to satisfy the octet rule for Oxygen first, which uses 6 electrons, leaving 6 electrons to be placed on the Nitrogen atoms.<\/li>\n\n\n\n
    • Place 3 lone pairs on Oxygen and 1 lone pair on the Nitrogen atoms.<\/li>\n\n\n\n
    • The formal charge calculation will help determine if this is stable.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

      2. N-O-N<\/strong> arrangement:<\/h4>\n\n\n\n
        \n
      • Nitrogen (N) bonds to Oxygen (O) and Oxygen bonds to another Nitrogen (N).<\/li>\n\n\n\n
      • The electrons are placed similarly, but the bonding changes the structure. Steps:<\/strong>\n
          \n
        • Place a single bond between N and O, and another bond between O and N.<\/li>\n\n\n\n
        • Place lone pairs on Oxygen and Nitrogens to satisfy the octet rule.<\/li>\n\n\n\n
        • Again, you will perform formal charge calculations.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          b) Using Formal Charge Calculation:<\/h3>\n\n\n\n

          Formal charge (FC) is calculated using the following formula:FC=Valence electrons of atom\u2212(Lone pair electrons+12\u00d7Bonding electrons)FC = \\text{Valence electrons of atom} – (\\text{Lone pair electrons} + \\frac{1}{2} \\times \\text{Bonding electrons})FC=Valence electrons of atom\u2212(Lone pair electrons+21\u200b\u00d7Bonding electrons)<\/p>\n\n\n\n

          Now, let\u2019s calculate the formal charge for both structures:<\/p>\n\n\n\n

          For N-N-O structure:<\/strong><\/h4>\n\n\n\n
            \n
          • Nitrogen (N) on the left: Valence = 5, Lone pairs = 2, Bonding electrons = 2. Formal charge = 5\u2212(2+1)=25 – (2 + 1) = 25\u2212(2+1)=2 (Positive charge).<\/li>\n\n\n\n
          • Nitrogen (N) on the right: Valence = 5, Lone pairs = 2, Bonding electrons = 2. Formal charge = 5\u2212(2+1)=25 – (2 + 1) = 25\u2212(2+1)=2 (Positive charge).<\/li>\n\n\n\n
          • Oxygen (O): Valence = 6, Lone pairs = 6, Bonding electrons = 2. Formal charge = 6\u2212(6+1)=\u221216 – (6 + 1) = -16\u2212(6+1)=\u22121 (Negative charge).<\/li>\n<\/ul>\n\n\n\n

            For N-O-N structure:<\/strong><\/h4>\n\n\n\n
              \n
            • Nitrogen (N) on the left: Valence = 5, Lone pairs = 2, Bonding electrons = 2. Formal charge = 5\u2212(2+1)=25 – (2 + 1) = 25\u2212(2+1)=2 (Positive charge).<\/li>\n\n\n\n
            • Oxygen (O): Valence = 6, Lone pairs = 4, Bonding electrons = 2. Formal charge = 6\u2212(4+1)=16 – (4 + 1) = 16\u2212(4+1)=1 (Neutral charge).<\/li>\n\n\n\n
            • Nitrogen (N) on the right: Valence = 5, Lone pairs = 2, Bonding electrons = 2. Formal charge = 5\u2212(2+1)=25 – (2 + 1) = 25\u2212(2+1)=2 (Positive charge).<\/li>\n<\/ul>\n\n\n\n

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

              The N-O-N structure<\/strong> is the more stable one. Why?<\/p>\n\n\n\n

                \n
              • In the N-N-O structure<\/strong>, the two Nitrogens have positive formal charges, which is less stable because Nitrogens are more electronegative than Oxygen.<\/li>\n\n\n\n
              • In the N-O-N structure<\/strong>, the formal charges are more balanced, with only the Nitrogens carrying a positive formal charge and Oxygen being neutral. This results in a more stable configuration.<\/li>\n<\/ul>\n\n\n\n

                Thus, N-O-N<\/strong> is preferred based on the formal charge distribution.<\/p>\n\n\n\n

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

                ) Draw the Lewis structure of NO. Give two arrangement as in N-N-O and N-O-N. (4 marks) b) Using formal charge calculation, determine the most stable structure. Justify your reason. (6 marks) The Correct Answer and Explanation is: a) Drawing the Lewis structure of NO (Nitroxyl or azanone): We are dealing with a molecule that […]<\/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-247131","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247131","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=247131"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247131\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=247131"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=247131"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=247131"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}