what is the molecular geometry around sulfur in SF5- ?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n The molecular geometry around sulfur in the SF\u2085\u207b ion is square pyramidal<\/strong>.<\/p>\n\n\n\n In summary, the molecular geometry around sulfur in SF\u2085\u207b is square pyramidal<\/strong>, resulting from the presence of five bonding pairs of electrons and one lone pair, which creates an electron distribution that minimizes repulsion.<\/p>\n","protected":false},"excerpt":{"rendered":" what is the molecular geometry around sulfur in SF5- ? The correct answer and explanation is : The molecular geometry around sulfur in the SF\u2085\u207b ion is square pyramidal. Explanation: In summary, the molecular geometry around sulfur in SF\u2085\u207b is square pyramidal, resulting from the presence of five bonding pairs of electrons and one lone […]<\/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-201759","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201759","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=201759"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201759\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201759"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201759"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201759"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
Sulfur (S) has the atomic number 16 and thus has the electron configuration [Ne]3s\u00b23p\u2074. When it forms SF\u2085\u207b, sulfur is in the center, bonded to five fluorine atoms. The extra negative charge indicates that sulfur is surrounded by 6 regions of electron density\u2014five single bonds with fluorine atoms and one additional lone pair of electrons.<\/li>\n\n\n\n
In SF\u2085\u207b, the sulfur atom undergoes sp\u00b3d\u00b2 hybridization, which accommodates six electron pairs: five bonding pairs (from the F atoms) and one lone pair. These six regions of electron density are repelled by each other, and according to the VSEPR theory<\/strong> (Valence Shell Electron Pair Repulsion), they will arrange themselves in a way that minimizes repulsion.<\/li>\n\n\n\n
The ideal arrangement for six electron pairs around a central atom is octahedral<\/strong>. However, in SF\u2085\u207b, there is one lone pair. Lone pairs occupy more space than bonding pairs, and they tend to distort the molecular geometry. In this case, the lone pair will occupy one of the axial positions, while the five fluorine atoms will occupy the remaining five positions. This leads to a square pyramidal<\/strong> shape, where four fluorine atoms are placed at the equatorial positions, and one fluorine is at the axial position. The lone pair occupies the other axial position, resulting in this geometry.<\/li>\n\n\n\n
In the square pyramidal geometry, the bond angles between the equatorial fluorine atoms are 90\u00b0, and between the axial and equatorial fluorine atoms, the angle is 81.9\u00b0 (slightly less than 90\u00b0 due to the presence of the lone pair).<\/li>\n<\/ol>\n\n\n\n