{"id":224176,"date":"2025-06-02T18:45:08","date_gmt":"2025-06-02T18:45:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224176"},"modified":"2025-06-02T18:45:11","modified_gmt":"2025-06-02T18:45:11","slug":"pair-geometry","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/pair-geometry\/","title":{"rendered":"pair geometry"},"content":{"rendered":"\n<p>pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow. Draw the Lewis structure for hydrogen cyanide, HCN. What is the electron pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow.<br>Draw the Lewis structure for hydrogen sulfide, H2S. What is the electron pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow. Draw the Lewis structure for hydrogen cyanide, HCN. What is the electron pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-40.png\" alt=\"\" class=\"wp-image-224177\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Hydrogen sulfide (H\u2082S)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: H\u2013S\u2013H with two lone-pair dots on S (total 8 valence e\u207b).<\/li>\n\n\n\n<li><strong>Electron-pair geometry (VSEPR)<\/strong>: <strong>tetrahedral<\/strong> (4 regions around S: 2 lone pairs + 2 bond pairs).<\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: <strong>bent (V-shaped)<\/strong> because the two lone pairs occupy two vertices of the tetrahedron, leaving the two S\u2013H bonds at the other vertices.<\/li>\n\n\n\n<li><strong>Estimated H\u2013S\u2013H bond angle<\/strong>: compressed well below the ideal 109.5 \u00b0; experimentally \u2248 92 \u00b0, usually quoted as <strong>~95 \u00b0<\/strong>.<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: <strong>polar<\/strong>. Each S\u2013H bond dipole points from H toward the more-electronegative S; the two dipoles reinforce, giving a net vector from the midpoint of the H\u2013H line toward S (draw the arrow from between the H atoms toward S, bar at the H end).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Hydrogen cyanide (HCN)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: H\u2013C\u2261N with a lone pair on N (10 valence e\u207b).<\/li>\n\n\n\n<li><strong>Electron-pair geometry about C<\/strong>: <strong>linear<\/strong> (2 regions: one \u03c3 bond to H and one \u03c3 bond of the C\u2261N triple).<\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: <strong>linear<\/strong>.<\/li>\n\n\n\n<li><strong>Estimated H\u2013C\u2013N bond angle<\/strong>: <strong>180 \u00b0<\/strong>.<\/li>\n\n\n\n<li><strong>Polarity<\/strong>: <strong>polar<\/strong>. The C\u2261N bond dipole (points toward N) is larger than the opposite-directed C\u2013H dipole, so the overall molecular dipole points from the H end through C toward N (arrow head at N, plus sign near H).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Why these answers<\/h3>\n\n\n\n<p>VSEPR theory relates molecular shape to the number of electron-density regions (bonding + lone pairs) around a central atom. Each region repels the others, adopting a geometry that maximizes separation.<\/p>\n\n\n\n<p><em>In H\u2082S<\/em> the sulfur atom supplies six valence electrons; two are used for the S\u2013H \u03c3 bonds, leaving two lone pairs. Four regions around S give a <strong>tetrahedral electron framework<\/strong>. Lone pairs repel more strongly than bonding pairs, pushing the S\u2013H bonds closer together, so the observed angle (\u2248 92 \u00b0) is much smaller than the ideal 109.5 \u00b0. Unequal electronegativities (\u03c7 S = 2.58 &gt; \u03c7 H = 2.20) make each S\u2013H bond polar. Because the vector sum of the two bond dipoles is non-zero, H\u2082S has a net dipole (\u2248 0.97 D), explaining its moderate water solubility and higher boiling point compared with H\u2082.<\/p>\n\n\n\n<p><em>In HCN<\/em> carbon contributes four electrons, forming one \u03c3 bond to H and one \u03c3 + two \u03c0 bonds to N (a triple bond counts as a single region in VSEPR). Two regions demand a <strong>linear arrangement<\/strong>, so the H\u2013C\u2013N axis is 180 \u00b0. Although the C\u2013H bond dipole points slightly toward C (\u03c7 C = 2.55), the much stronger C\u2261N dipole (\u03c7 N = 3.04) dominates, giving a net moment (\u2248 3.0 D) toward nitrogen. The polarity, plus the linear shape, accounts for HCN\u2019s high toxicity: its small, polar molecules cross biological membranes readily and bind to cytochrome oxidase.<\/p>\n\n\n\n<p>Thus, VSEPR, electronegativity differences, and dipole-vector addition together explain both molecular shapes and polarities.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-91.jpeg\" alt=\"\" class=\"wp-image-224178\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow. Draw the Lewis structure for hydrogen cyanide, HCN. What is the electron pair geometry? What is the molecular geometry? Estimate the bond angle. Is this molecule polar? If so, draw the dipole arrow.Draw the [&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-224176","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224176","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=224176"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224176\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224176"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224176"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224176"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}