{"id":239265,"date":"2025-07-02T19:24:50","date_gmt":"2025-07-02T19:24:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239265"},"modified":"2025-07-02T19:24:53","modified_gmt":"2025-07-02T19:24:53","slug":"write-the-lewis-structure-of-the-nitrite-ion-no2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/write-the-lewis-structure-of-the-nitrite-ion-no2\/","title":{"rendered":"Write the Lewis structure of the nitrite ion, NO2^-"},"content":{"rendered":"\n

Write the Lewis structure of the nitrite ion, NO2^-<\/p>\n\n\n\n

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

To write the Lewis structure of the nitrite ion (NO\u2082\u207b), we need to follow a few steps:<\/p>\n\n\n\n

Step 1: Count the Total Valence Electrons<\/h3>\n\n\n\n

Nitrogen (N) is in group 15 of the periodic table, so it has 5 valence electrons. Oxygen (O) is in group 16, so each oxygen atom has 6 valence electrons. The nitrite ion has a negative charge, meaning it has gained one extra electron. Therefore, the total number of valence electrons is:5\u2009(from N)+2\u00d76\u2009(from two O)+1\u2009(from the negative charge)=18\u2009electrons.5 \\, (\\text{from N}) + 2 \\times 6 \\, (\\text{from two O}) + 1 \\, (\\text{from the negative charge}) = 18 \\, \\text{electrons}.5(from N)+2\u00d76(from two O)+1(from the negative charge)=18electrons.<\/p>\n\n\n\n

Step 2: Determine the Central Atom<\/h3>\n\n\n\n

In this case, nitrogen (N) will be the central atom because it is less electronegative than oxygen and can form more bonds.<\/p>\n\n\n\n

Step 3: Connect the Atoms with Single Bonds<\/h3>\n\n\n\n

We place single bonds between nitrogen and each oxygen atom. A single bond consists of two electrons, so for two bonds, we have already used 4 electrons.N\u2212OandN\u2212ON – O \\quad \\text{and} \\quad N – ON\u2212OandN\u2212O<\/p>\n\n\n\n

Step 4: Distribute the Remaining Electrons<\/h3>\n\n\n\n

After forming the single bonds, we have 18 total electrons, and we’ve used 4 for the bonds. This leaves us with 14 electrons to distribute. Place lone pairs on the oxygen atoms to satisfy their octet rule. Each oxygen atom needs 6 more electrons to complete its octet, so we place 3 lone pairs (6 electrons) on each oxygen atom.<\/p>\n\n\n\n

Step 5: Check Nitrogen\u2019s Octet<\/h3>\n\n\n\n

At this point, nitrogen only has 4 electrons in its valence shell (one from each bond). Since nitrogen can hold more than 8 electrons, we will form a double bond with one of the oxygen atoms. This will move two electrons from the lone pairs of that oxygen into the bonding region, creating a double bond.<\/p>\n\n\n\n

Step 6: Final Structure<\/h3>\n\n\n\n

Now we have:<\/p>\n\n\n\n

    \n
  • A double bond between nitrogen and one oxygen.<\/li>\n\n\n\n
  • A single bond between nitrogen and the other oxygen.<\/li>\n\n\n\n
  • 2 lone pairs on the oxygen with the single bond.<\/li>\n\n\n\n
  • 3 lone pairs on the oxygen with the double bond.<\/li>\n\n\n\n
  • The negative charge is placed on the oxygen with the single bond.<\/li>\n<\/ul>\n\n\n\n

    The final structure of NO\u2082\u207b is:O=N\u2212O\u2212O = N – O^-O=N\u2212O\u2212<\/p>\n\n\n\n

    Step 7: Verify<\/h3>\n\n\n\n
      \n
    • Each oxygen atom has a full octet of electrons.<\/li>\n\n\n\n
    • Nitrogen has an expanded octet (it can have up to 10 electrons in its valence shell).<\/li>\n\n\n\n
    • The total number of electrons used (18) matches the count from step 1.<\/li>\n\n\n\n
    • The negative charge is correctly placed on one of the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n

      Thus, the Lewis structure for the nitrite ion is correct.<\/p>\n\n\n\n

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

      Write the Lewis structure of the nitrite ion, NO2^- The Correct Answer and Explanation is: To write the Lewis structure of the nitrite ion (NO\u2082\u207b), we need to follow a few steps: Step 1: Count the Total Valence Electrons Nitrogen (N) is in group 15 of the periodic table, so it has 5 valence electrons. […]<\/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-239265","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239265","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=239265"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239265\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239265"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239265"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239265"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}