To calculate the bond order for the NO2+\\text{NO}_2^+NO2+\u200b ion, we need to consider its molecular structure and resonance. Here’s a step-by-step breakdown:<\/p>\n\n\n\n
Step 1: Lewis Structure<\/h3>\n\n\n\n
For the NO2+\\text{NO}_2^+NO2+\u200b ion:<\/p>\n\n\n\n
- \n
- Nitrogen (N) has 5 valence electrons, and each oxygen (O) has 6 valence electrons.<\/li>\n\n\n\n
- The ion has a positive charge, so it has one fewer electron (total valence electrons = 5 (from N) + 2\u00d76 (from O) \u2212 1 (positive charge) = 16 electrons).<\/li>\n\n\n\n
- In the most common Lewis structure, nitrogen is in the center, double-bonded to both oxygen atoms, with a lone pair on each oxygen.<\/li>\n<\/ol>\n\n\n\n
Step 2: Resonance Structures<\/h3>\n\n\n\n
The NO2+\\text{NO}_2^+NO2+\u200b ion exhibits resonance. There are two major resonance structures:<\/p>\n\n\n\n
- \n
- Structure 1:<\/strong> One oxygen has a double bond with nitrogen, and the other has a single bond. The singly bonded oxygen carries a negative charge, while nitrogen carries a positive charge.<\/li>\n\n\n\n
- Structure 2:<\/strong> The positions of the bonds and charges are flipped.<\/li>\n<\/ol>\n\n\n\n
Both of these structures contribute equally to the actual bonding in the molecule.<\/p>\n\n\n\n
Step 3: Bond Order Calculation<\/h3>\n\n\n\n
Bond order is calculated using the following formula:Bond Order=Number of bonding electrons\u2212Number of antibonding electrons2\\text{Bond Order} = \\frac{\\text{Number of bonding electrons} – \\text{Number of antibonding electrons}}{2}Bond Order=2Number of bonding electrons\u2212Number of antibonding electrons\u200b<\/p>\n\n\n\n
In this case:<\/p>\n\n\n\n
- \n
- The number of bonding electrons in each resonance structure is 8 (from the double bonds).<\/li>\n\n\n\n
- There are no antibonding electrons because we have a stable bonding configuration.<\/li>\n<\/ul>\n\n\n\n
Thus, the bond order can be calculated as:Bond Order=82=4\\text{Bond Order} = \\frac{8}{2} = 4Bond Order=28\u200b=4<\/p>\n\n\n\n
Step 4: Conclusion<\/h3>\n\n\n\n
For NO2+\\text{NO}_2^+NO2+\u200b, considering the resonance structures, the bond order is 2, as the molecule is in between two resonance forms where each nitrogen-oxygen bond has partial double bond character. The bond order is closer to 2 because the electron distribution is more uniform.<\/p>\n\n\n\n
The bond order of the NO2+\\text{NO}_2^+NO2+\u200b ion is 2<\/strong>, indicating that the bonding is stronger than in NO2\\text{NO}_2NO2\u200b, where the bond order is 1.5.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-433.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"calculate the bond order for NO2+ ion, regarding all the resonance structures The Correct Answer and Explanation is: To calculate the bond order for the NO2+\\text{NO}_2^+NO2+\u200b ion, we need to consider its molecular structure and resonance. Here’s a step-by-step breakdown: Step 1: Lewis Structure For the NO2+\\text{NO}_2^+NO2+\u200b ion: Step 2: Resonance Structures The NO2+\\text{NO}_2^+NO2+\u200b ion […]<\/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-244341","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244341","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=244341"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244341\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244341"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244341"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244341"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Structure 2:<\/strong> The positions of the bonds and charges are flipped.<\/li>\n<\/ol>\n\n\n\n
- Structure 1:<\/strong> One oxygen has a double bond with nitrogen, and the other has a single bond. The singly bonded oxygen carries a negative charge, while nitrogen carries a positive charge.<\/li>\n\n\n\n