9.50 (a)<\/strong> BF\u2083 is nonpolar<\/strong>. 9.51 Hybridization of the central atom:<\/strong> Molecular Polarity and Geometry<\/strong><\/p>\n\n\n\n The polarity of a molecule depends on both the polarity of its individual bonds and its molecular geometry. BF\u2083 (boron trifluoride)<\/strong> has three identical B\u2013F bonds, each polar due to the electronegativity difference between B and F. However, the molecule adopts a trigonal planar<\/strong> shape with 120\u00b0 bond angles, and the three bond dipoles cancel each other out. Therefore, BF\u2083 is nonpolar<\/strong> overall despite having polar bonds.<\/p>\n\n\n\n When BF\u2083 reacts with a fluoride ion to form BF\u2084\u207b<\/strong>, the molecule becomes tetrahedral<\/strong>, with four equivalent B\u2013F bonds at angles of 109.5\u00b0. This shape is not planar<\/strong>, and the added electron pair is delocalized over the structure. The ion is symmetrical and nonpolar.<\/p>\n\n\n\n In BF\u2082Cl<\/strong>, however, the situation is different. It has a trigonal planar geometry like BF\u2083, but one of the fluorine atoms is replaced by a chlorine atom. Since Cl and F have different electronegativities, the dipoles do not<\/strong> cancel out. As a result, the molecule has a net dipole moment<\/strong> and is polar<\/strong>.<\/p>\n\n\n\n Hybridization of Central Atoms<\/strong><\/p>\n\n\n\n In H\u2082S<\/strong>, sulfur forms two sigma bonds and has two lone pairs, giving it a bent shape. This requires sp\u00b3 hybridization<\/strong>. These hybridizations explain the geometries and bonding observed in each compound.<\/p>\n\n\n\n Is the molecule BF3 polar or nonpolar? (b) If you react BF3 to make the ion BF4-, is this ion planar? (c) Does the molecule BF2Cl have a dipole moment? 9.51 Indicate the hybridization of the central atom in (a) H2S, (b) SeF4, (c) POH3, (d) AlCl3. The Correct Answer and Explanation is: 9.50 (a) […]<\/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-232942","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232942","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=232942"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232942\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
(b)<\/strong> The ion BF\u2084\u207b is not planar<\/strong>.
(c)<\/strong> The molecule BF\u2082Cl has a dipole moment<\/strong>.<\/p>\n\n\n\n
(a)<\/strong> H\u2082S \u2192 sp\u00b3<\/strong>
(b)<\/strong> SeF\u2084 \u2192 sp\u00b3d<\/strong>
(c)<\/strong> POH\u2083 \u2192 sp\u00b3<\/strong>
(d)<\/strong> AlCl\u2083 \u2192 sp\u00b2<\/strong><\/p>\n\n\n\n
\n\n\n\nConceptual Explanation <\/h3>\n\n\n\n
In SeF\u2084<\/strong>, selenium forms four bonds and has one lone pair, leading to a see-saw<\/strong> geometry, requiring sp\u00b3d hybridization<\/strong>.
In POH\u2083<\/strong>, phosphorus forms three single bonds with hydrogen and one with oxygen (which has a lone pair on O), requiring sp\u00b3 hybridization<\/strong>.
In AlCl\u2083<\/strong>, aluminum forms three sigma bonds with chlorine atoms and has no lone pairs, adopting a trigonal planar<\/strong> geometry, so it is sp\u00b2 hybridized<\/strong>.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-380.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"