How many lone pairs are on the Br atom in BrCl?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n In the molecule BrCl<\/strong>, the bromine (Br) atom has 3 lone pairs<\/strong> of electrons.<\/p>\n\n\n\n To determine the number of lone pairs on the Br atom in BrCl<\/strong>, we need to consider the following:<\/p>\n\n\n\n In BrCl<\/strong>, the Br atom has 3 lone pairs of electrons. This is determined by the number of valence electrons (7 for Br) and the bonding with chlorine, leaving 6 electrons, which are arranged as 3 lone pairs.<\/p>\n","protected":false},"excerpt":{"rendered":" How many lone pairs are on the Br atom in BrCl? The correct answer and explanation is : In the molecule BrCl, the bromine (Br) atom has 3 lone pairs of electrons. Explanation: To determine the number of lone pairs on the Br atom in BrCl, we need to consider the following: Conclusion: In BrCl, […]<\/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-209782","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209782","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=209782"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209782\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209782"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209782"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
Bromine (Br) is in Group 17 of the periodic table (also known as the halogens). It has 7 valence electrons, as all elements in Group 17 have 7 electrons in their outermost shell.<\/li>\n\n\n\n
In BrCl<\/strong>, bromine forms a single covalent bond with chlorine (Cl). Chlorine, like bromine, is also in Group 17 and has 7 valence electrons. When bromine and chlorine form a bond, they share one electron each, which constitutes a single covalent bond.<\/li>\n\n\n\n
After forming the single bond, each atom still has electrons left to complete their octet. The bonding between Br and Cl accounts for two electrons, one from each atom.<\/li>\n\n\n\n
Since bromine originally has 7 valence electrons and uses 1 of them to bond with chlorine, it has 6 remaining electrons. These 6 electrons will be arranged as 3 lone pairs<\/strong> (each lone pair consists of 2 electrons).<\/li>\n\n\n\n
Therefore, after the bonding, the Br atom will have 3 lone pairs of electrons, meaning that the 6 remaining electrons are not involved in bonding and are localized on the Br atom as lone pairs.<\/li>\n<\/ol>\n\n\n\nConclusion:<\/h3>\n\n\n\n