{"id":222730,"date":"2025-05-31T17:04:45","date_gmt":"2025-05-31T17:04:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222730"},"modified":"2025-05-31T17:04:47","modified_gmt":"2025-05-31T17:04:47","slug":"the-lewis-diagram-for-nobr-is-br-the-electron-pair-geometry-around-the-n-atom-in-nobr-is-there-is-are-lone-pairs-around-the-central-atom","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/the-lewis-diagram-for-nobr-is-br-the-electron-pair-geometry-around-the-n-atom-in-nobr-is-there-is-are-lone-pairs-around-the-central-atom\/","title":{"rendered":"The Lewis diagram for NOBr is: Br The electron-pair geometry around the N atom in NOBr is: There is\/are lone pair(s) around the central atom."},"content":{"rendered":"\n

The Lewis diagram for NOBr is: Br The electron-pair geometry around the N atom in NOBr is: There is\/are lone pair(s) around the central atom. So the geometry of NOBr is: The Lewis diagram for NOF is: 6-N=& The electron-pair geometry around the atom in NOF is: There is\/are lone pair(s) around the central atom, so the geometry of NOF is:
The Lewis diagram for NOBr is: Br The electron-pair geometry around the N atom in NOBr is: There is\/are lone pair(s) around the central atom. So the geometry of NOBr is: The Lewis diagram for NOF is: 6-N=& The electron-pair geometry around the atom in NOF is: There is\/are lone pair(s) around the central atom, so the geometry of NOF is:<\/p>\n\n\n\n

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

Correct Answers<\/strong><\/h3>\n\n\n\n

1. NOBr:<\/strong><\/p>\n\n\n\n

    \n
  • Lewis diagram:<\/strong>
    :Br\u2013N=O
    (with a lone pair on N)<\/li>\n\n\n\n
  • Electron-pair geometry around the N atom:<\/strong> Trigonal planar<\/strong><\/li>\n\n\n\n
  • There is\/are lone pair(s) around the central atom:<\/strong> 1<\/strong><\/li>\n\n\n\n
  • Molecular geometry of NOBr:<\/strong> Bent<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    2. NOF:<\/strong><\/p>\n\n\n\n

      \n
    • Lewis diagram:<\/strong>
      F\u2013N=O
      (with a lone pair on N)<\/li>\n\n\n\n
    • Electron-pair geometry around the N atom:<\/strong> Trigonal planar<\/strong><\/li>\n\n\n\n
    • There is\/are lone pair(s) around the central atom:<\/strong> 1<\/strong><\/li>\n\n\n\n
    • Molecular geometry of NOF:<\/strong> Bent<\/strong><\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Explanation <\/strong><\/h3>\n\n\n\n

      To determine molecular geometry, we start with the Lewis structures<\/strong> of the molecules. For both NOBr<\/strong> and NOF<\/strong>, nitrogen (N) is the central atom, bonded to oxygen (O) and either bromine (Br) or fluorine (F).<\/p>\n\n\n\n

      1. NOBr:<\/strong>
      Nitrogen has 5 valence electrons. It forms a double bond<\/strong> with oxygen (which prefers to form two bonds to complete its octet) and a single bond<\/strong> with bromine (a halogen, which usually forms one bond). After bonding, nitrogen has one lone pair remaining. This gives a total of three regions of electron density<\/strong> (a double bond, a single bond, and a lone pair), which corresponds to a trigonal planar electron geometry<\/strong> according to VSEPR theory. However, because one region is a lone pair, the actual molecular geometry is bent<\/strong>. The lone pair pushes the bonding atoms slightly closer together, resulting in a bond angle slightly less than 120\u00b0.<\/p>\n\n\n\n

      2. NOF:<\/strong>
      Similar to NOBr, NOF has nitrogen at the center bonded to fluorine and oxygen. Nitrogen again forms a double bond with oxygen and a single bond with fluorine. It also retains one lone pair. This leads to three regions of electron density<\/strong>, just like in NOBr. Therefore, the electron-pair geometry is trigonal planar<\/strong>, but the molecular geometry is bent<\/strong> due to the presence of the lone pair.<\/p>\n\n\n\n

      In both molecules, the lone pair on nitrogen distorts the shape away from perfect trigonal planar geometry. The bent molecular shape affects the polarity of the molecules, making both polar<\/strong>, with uneven electron distribution across the molecule.<\/p>\n\n\n\n

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

      The Lewis diagram for NOBr is: Br The electron-pair geometry around the N atom in NOBr is: There is\/are lone pair(s) around the central atom. So the geometry of NOBr is: The Lewis diagram for NOF is: 6-N=& The electron-pair geometry around the atom in NOF is: There is\/are lone pair(s) around the central atom, […]<\/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-222730","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222730","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=222730"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222730\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222730"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222730"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222730"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}