{"id":197727,"date":"2025-03-07T14:58:13","date_gmt":"2025-03-07T14:58:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197727"},"modified":"2025-03-07T14:58:16","modified_gmt":"2025-03-07T14:58:16","slug":"for-each-molecule-draw-its-shape-using-lewis-structures-and-vsepr-theory","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/for-each-molecule-draw-its-shape-using-lewis-structures-and-vsepr-theory\/","title":{"rendered":"For each molecule, draw its shape using Lewis structures and VSEPR theory"},"content":{"rendered":"\n<p>For each molecule, draw its shape using Lewis structures and VSEPR theory. Include your steps in creating each sketch. Each diagram should include labelled electronegativities and partial charges for each atom within the molecule, and vector arrows representing the magnitude and direction of each bond dipole and net dipole (for polar molecules). Finally, determine whether the molecule is polar or non-polar and explain your rationale based on bond polarity and molecular shape.<\/p>\n\n\n\n<p>CH2Cl2<br>NI3<br>CH2O<br>BeF2<br>SF6<br>SO2<br>HBr<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s go through the steps for each molecule one by one and explain the results:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>CH\u2082Cl\u2082 (Dichloromethane)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) in the center, bonded to two hydrogen (H) atoms and two chlorine (Cl) atoms.<\/li>\n\n\n\n<li>Each bond is a single bond, and carbon follows the octet rule.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The central atom (C) has 4 regions of electron density (2 single bonds with H and 2 with Cl), resulting in a <strong>tetrahedral geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C:<\/strong> 2.55, <strong>H:<\/strong> 2.20, <strong>Cl:<\/strong> 3.16<\/li>\n\n\n\n<li>Chlorine is more electronegative than both carbon and hydrogen, pulling electron density away from C-H bonds. This creates partial charges on each Cl atom (\u03b4-) and partial positive charges on H (\u03b4+).<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Because of the asymmetric distribution of electronegativities, CH\u2082Cl\u2082 has a <strong>net dipole<\/strong> that points from the H atoms towards the Cl atoms.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar molecule<\/strong> because of the net dipole from the difference in electronegativities and the tetrahedral shape.<\/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\">2. <strong>NI\u2083 (Nitrogen triiodide)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen (N) in the center, bonded to three iodine (I) atoms. Each bond is a single bond.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen has 3 bonding pairs and 1 lone pair of electrons, resulting in a <strong>trigonal pyramidal geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>N:<\/strong> 3.04, <strong>I:<\/strong> 2.66<\/li>\n\n\n\n<li>Iodine atoms are less electronegative than nitrogen, so there is a partial negative charge on I and a partial positive charge on N.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule is <strong>polar<\/strong> due to the asymmetry of the electronegativity difference, and the trigonal pyramidal shape creates a net dipole pointing towards nitrogen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar molecule<\/strong> because of the net dipole due to the trigonal pyramidal geometry and electronegativity difference.<\/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\">3. <strong>CH\u2082O (Formaldehyde)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) bonded to two hydrogen (H) atoms and one oxygen (O) atom.<\/li>\n\n\n\n<li>Oxygen has two lone pairs, and the bonds are single and double bonds.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon has 3 regions of electron density, resulting in a <strong>trigonal planar geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C:<\/strong> 2.55, <strong>H:<\/strong> 2.20, <strong>O:<\/strong> 3.44<\/li>\n\n\n\n<li>Oxygen is more electronegative than carbon and hydrogen, resulting in a partial negative charge on oxygen and partial positive charges on the hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule has a <strong>net dipole<\/strong> pointing towards oxygen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar molecule<\/strong> due to the bond dipoles and the trigonal planar shape.<\/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\">4. <strong>BeF\u2082 (Beryllium fluoride)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Beryllium (Be) is the central atom, bonded to two fluorine (F) atoms with single bonds.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule has 2 bonding regions, leading to a <strong>linear geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Be:<\/strong> 1.57, <strong>F:<\/strong> 3.98<\/li>\n\n\n\n<li>Fluorine is significantly more electronegative than beryllium, so there is a large partial negative charge on the fluorine atoms and a partial positive charge on the beryllium atom.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule has a <strong>net dipole<\/strong>, but the linear geometry causes the dipoles to cancel each other out, making the molecule <strong>non-polar<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Non-polar molecule<\/strong> due to symmetry, even though the bonds are polar.<\/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\">5. <strong>SF\u2086 (Sulfur hexafluoride)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sulfur (S) in the center, bonded to six fluorine (F) atoms.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Six bonding pairs of electrons result in an <strong>octahedral geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S:<\/strong> 2.58, <strong>F:<\/strong> 3.98<\/li>\n\n\n\n<li>Fluorine is more electronegative, so the fluorine atoms have a partial negative charge, and sulfur has a partial positive charge.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Despite the polar bonds, the <strong>octahedral shape<\/strong> causes the bond dipoles to cancel out.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Non-polar molecule<\/strong> because the dipoles cancel out due to symmetry.<\/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\">6. <strong>SO\u2082 (Sulfur dioxide)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sulfur (S) bonded to two oxygen (O) atoms with double bonds.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule has 2 bonding pairs and 1 lone pair on sulfur, leading to a <strong>bent geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>S:<\/strong> 2.58, <strong>O:<\/strong> 3.44<\/li>\n\n\n\n<li>Oxygen is more electronegative than sulfur, resulting in partial negative charges on oxygen and a partial positive charge on sulfur.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The bent geometry leads to a <strong>net dipole<\/strong> pointing from sulfur to oxygen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar molecule<\/strong> due to the bent shape and the electronegativity difference.<\/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\">7. <strong>HBr (Hydrogen bromide)<\/strong><\/h3>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen (H) bonded to bromine (Br) with a single bond.<\/li>\n<\/ul>\n\n\n\n<p><strong>VSEPR Theory:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule has a <strong>linear geometry<\/strong> due to only one bond.<\/li>\n<\/ul>\n\n\n\n<p><strong>Electronegativity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>H:<\/strong> 2.20, <strong>Br:<\/strong> 2.96<\/li>\n\n\n\n<li>Bromine is more electronegative than hydrogen, leading to a partial negative charge on bromine and a partial positive charge on hydrogen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole Moment:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>There is a <strong>net dipole<\/strong> pointing from hydrogen to bromine.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar molecule<\/strong> due to the electronegativity difference.<\/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\">Conclusion<\/h3>\n\n\n\n<p>In summary:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar Molecules:<\/strong> CH\u2082Cl\u2082, NI\u2083, CH\u2082O, SO\u2082, HBr<\/li>\n\n\n\n<li><strong>Non-polar Molecules:<\/strong> BeF\u2082, SF\u2086<\/li>\n<\/ul>\n\n\n\n<p>Polarity arises from the combination of electronegativity differences and molecular shape. Symmetric molecules like BeF\u2082 and SF\u2086, despite having polar bonds, cancel out their dipoles and remain non-polar. In contrast, molecules with asymmetric shapes, such as CH\u2082Cl\u2082 or HBr, have net dipoles and are polar.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-552.png\" alt=\"\" class=\"wp-image-197728\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>For each molecule, draw its shape using Lewis structures and VSEPR theory. Include your steps in creating each sketch. Each diagram should include labelled electronegativities and partial charges for each atom within the molecule, and vector arrows representing the magnitude and direction of each bond dipole and net dipole (for polar molecules). Finally, determine whether [&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-197727","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197727","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=197727"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197727\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197727"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197727"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197727"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}