What is the molecular geometry of PBr3<\/p>\n\n\n\n
trigonal pyramidal
linear
tetrahedral
trigonal planar<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct molecular geometry of PBr3 (phosphorus tribromide)<\/strong> is trigonal pyramidal<\/strong>.<\/p>\n\n\n\n Thus, the molecular geometry of PBr3 is trigonal pyramidal<\/strong> due to the lone pair on the phosphorus atom and the repulsion between the bonding and non-bonding electron pairs.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the molecular geometry of PBr3 trigonal pyramidallineartetrahedraltrigonal planar The correct answer and explanation is : The correct molecular geometry of PBr3 (phosphorus tribromide) is trigonal pyramidal. Explanation: Thus, the molecular geometry of PBr3 is trigonal pyramidal due to the lone pair on the phosphorus atom and the repulsion between the bonding and non-bonding […]<\/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-203044","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203044","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=203044"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203044\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203044"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203044"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203044"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
Phosphorus (P) has an atomic number of 15, so its electron configuration is [Ne] 3s\u00b2 3p\u00b3. In PBr3, phosphorus forms three single bonds with bromine atoms, and there is one lone pair of electrons on the phosphorus atom. The three bromine atoms are arranged in a way that attempts to minimize electron repulsion, resulting in a shape that is based on a tetrahedral<\/strong> geometry.<\/li>\n\n\n\n
The electron-pair geometry (which considers both bonding and lone pairs of electrons) of PBr3 is tetrahedral<\/strong>, because there are four regions of electron density around the phosphorus atom (three bonds and one lone pair). In a perfect tetrahedron, the bond angles are 109.5\u00b0.<\/li>\n\n\n\n
The molecular geometry<\/strong> of PBr3 is derived from the electron-pair geometry but considers only the positions of the atoms (not the lone pairs). Since there is one lone pair of electrons, it causes the geometry to distort from the ideal tetrahedral shape, resulting in a trigonal pyramidal<\/strong> arrangement. In a trigonal pyramidal geometry, the bond angles are slightly less than the ideal 109.5\u00b0 due to the presence of the lone pair, which exerts greater repulsion than bonding pairs.<\/li>\n\n\n\n
According to Valence Shell Electron Pair Repulsion (VSEPR) theory<\/strong>, electron pairs (bonding and lone pairs) around an atom will arrange themselves to minimize repulsion. For PBr3, there are four regions of electron density (three bonds and one lone pair), so the electron-pair geometry is tetrahedral. However, because the lone pair is not visible in the molecular geometry, the result is a trigonal pyramidal<\/strong> shape for the PBr3 molecule.<\/li>\n<\/ol>\n\n\n\n