{"id":223996,"date":"2025-06-02T15:39:43","date_gmt":"2025-06-02T15:39:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223996"},"modified":"2025-06-02T15:39:45","modified_gmt":"2025-06-02T15:39:45","slug":"draw-the-lewis-structure-of-krf4-then-answer-the-following-questions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/draw-the-lewis-structure-of-krf4-then-answer-the-following-questions\/","title":{"rendered":"Draw the Lewis structure of KrF4, then answer the following questions."},"content":{"rendered":"\n

Draw the Lewis structure of KrF4, then answer the following questions. The electron group geometry of KrF4 is – A. trigonal bipyramid B. seesaw or sawhorse C. T-shape D. octahedral E.square pyramid F. square planar. The molecular shape of KrF4 is — A. trigonal bipyramid B. seesaw or sawhorse C. T-shape D. octahedral E. square pyramid F. square planar. The Kr-F bond is — A. polar B. nonpolar. The F-Kr-F bond angle is approximately — A. 90? B. 120? C. 109? D. 90? and 120? E. 180?. A KrF4 molecule is — A. nonpolar B. polar.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/strong>1<\/a><\/sup><\/p>\n\n\n\n\n\n

Answers<\/strong><\/p>\n\n\n\n

Property<\/th>Correct choice<\/th><\/tr><\/thead>
Electron-group geometry of KrF\u2084<\/td>D. octahedral<\/strong><\/td><\/tr>
Molecular shape (actual geometry)<\/td>F. square planar<\/strong><\/td><\/tr>
Individual Kr\u2013F bond<\/td>A. polar<\/strong><\/td><\/tr>
F\u2013Kr\u2013F bond angle (between adjacent F atoms)<\/td>A. \u2248 90 \u00b0<\/strong><\/td><\/tr>
Overall polarity of the KrF\u2084 molecule<\/td>A. nonpolar<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Why those answers are correct <\/h3>\n\n\n\n
    \n
  1. Lewis structure and electron counting<\/strong>
    Krypton brings 8 valence electrons and each of the four fluorine atoms 7, for a total of
    8+4\u00d77=368 + 4\\times7 = 36 e\u207b.\n
      \n
    • Place Kr in the center, bond it to four F atoms (8 e\u207b).<\/li>\n\n\n\n
    • Give each F three lone pairs (24 e\u207b).<\/li>\n\n\n\n
    • The remaining 4 e\u207b stay on Kr as two lone pairs.
      Six regions of electron density (4 \u03c3-bonds + 2 lone pairs) surround Kr.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Electron-group geometry: octahedral<\/strong>
      VSEPR treats every region\u2014bonding or lone pair\u2014the same when defining the electron-group<\/em> (or steric) geometry. Six regions arrange themselves at the corners of an octahedron to minimize repulsion, so the correct electron-group geometry is octahedral<\/strong>.<\/li>\n\n\n\n
    • Molecular shape: square planar<\/strong>
      Two of the six positions in an octahedron can be occupied by the lone pairs. They adopt opposite (axial) sites to keep as far apart as possible, leaving four fluorines in the equatorial plane. The visible shape formed by the atoms is therefore square planar<\/strong>.<\/li>\n\n\n\n
    • Bond polarity vs. molecular polarity<\/strong>
      A Kr\u2013F bond is polar because fluorine (\u03c7 \u2248 3.98) is appreciably more electronegative than krypton (\u03c7 \u2248 3.0). However, in a perfect square planar arrangement the four bond dipoles point toward the corners of a square and cancel vectorially. Consequently, the whole molecule is non-polar<\/strong> despite having four polar bonds.<\/li>\n\n\n\n
    • Bond angles<\/strong>
      In a square plane, the angle between adjacent fluorines is 90 \u00b0, while diametrically opposite fluorines are 180 \u00b0 apart. Textbook multiple-choice questions usually focus on the characteristic 90 \u00b0 value, so \u2248 90 \u00b0<\/strong> is the best single answer among the options provided.<\/li>\n<\/ol>\n\n\n\n

      Thus, octahedral electron grouping, square-planar molecular shape, polar bonds, 90 \u00b0 F\u2013Kr\u2013F angles, and net non-polarity together describe KrF\u2084 accurately<\/p>\n\n\n\n

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

      Draw the Lewis structure of KrF4, then answer the following questions. The electron group geometry of KrF4 is – A. trigonal bipyramid B. seesaw or sawhorse C. T-shape D. octahedral E.square pyramid F. square planar. The molecular shape of KrF4 is — A. trigonal bipyramid B. seesaw or sawhorse C. T-shape D. octahedral E. square […]<\/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-223996","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223996","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=223996"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223996\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223996"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223996"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}