{"id":170883,"date":"2024-11-20T01:06:50","date_gmt":"2024-11-20T01:06:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=170883"},"modified":"2024-11-20T01:06:53","modified_gmt":"2024-11-20T01:06:53","slug":"draw-the-lewis-structure-for-bef2-in-the-box-at-the-right-including-lone-pairs","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/20\/draw-the-lewis-structure-for-bef2-in-the-box-at-the-right-including-lone-pairs\/","title":{"rendered":"Draw the lewis structure for bef2 in the box at the right, including lone pairs."},"content":{"rendered":"\n<p>Draw the lewis structure for bef2 in the box at the right, including lone pairs.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To draw the Lewis structure of <strong>BeF\u2082<\/strong> (beryllium fluoride), we need to follow a few steps that adhere to the rules of electron bonding and valence electron sharing. Here\u2019s how you approach it:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Steps for Drawing the Lewis Structure of BeF\u2082:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Beryllium (Be)<\/strong> is in Group 2 of the periodic table, so it has <strong>2 valence electrons<\/strong>.<\/li>\n\n\n\n<li><strong>Fluorine (F)<\/strong> is in Group 17 of the periodic table, so each fluorine atom has <strong>7 valence electrons<\/strong>.<\/li>\n\n\n\n<li>Since there are two fluorine atoms in BeF\u2082, the total number of valence electrons is:<br>[<br>(2 \\text{ from Be}) + (2 \\times 7 \\text{ from F}) = 16 \\text{ valence electrons}.<br>]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the Central Atom:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Beryllium (Be) is the least electronegative element (except hydrogen) and will therefore be the central atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Connect the Atoms:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place Be in the center and connect each fluorine atom to it with a single bond (which uses up 2 electrons per bond).<\/li>\n\n\n\n<li>This leaves <strong>16 &#8211; 4 = 12 electrons<\/strong> to be distributed as lone pairs.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Distribute the Remaining Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place the remaining 12 electrons as lone pairs on the fluorine atoms. Since each fluorine needs 8 electrons to complete its octet, place 6 electrons (3 lone pairs) on each fluorine atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Check the Octets:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Beryllium is an exception to the octet rule. It will only have <strong>4 electrons<\/strong> in its valence shell (from the two single bonds with fluorine), which is acceptable for beryllium.<\/li>\n\n\n\n<li>Each fluorine has 8 electrons in its valence shell (2 from the bond with Be and 6 from lone pairs).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Final Lewis Structure:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The structure is as follows:\n<ul class=\"wp-block-list\">\n<li>Beryllium (Be) is in the center with two single bonds to two fluorine (F) atoms.<\/li>\n\n\n\n<li>Each fluorine atom has three lone pairs.<\/li>\n\n\n\n<li>The structure is linear, with the fluorine atoms positioned at a 180\u00b0 angle relative to each other.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>In BeF\u2082, the beryllium atom has only 4 electrons in its valence shell due to its bonding with two fluorine atoms, which is acceptable since beryllium can accommodate fewer than 8 electrons. Each fluorine atom has 8 electrons in its valence shell, fulfilling the octet rule. The Lewis structure reflects this electron distribution, ensuring that all electrons are accounted for and that the bonding conforms to the elements&#8217; typical bonding behavior.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the lewis structure for bef2 in the box at the right, including lone pairs. The Correct Answer and Explanation is: To draw the Lewis structure of BeF\u2082 (beryllium fluoride), we need to follow a few steps that adhere to the rules of electron bonding and valence electron sharing. Here\u2019s how you approach it: Steps [&hellip;]<\/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-170883","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/170883","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=170883"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/170883\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=170883"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=170883"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=170883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}