{"id":241749,"date":"2025-07-04T03:02:10","date_gmt":"2025-07-04T03:02:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=241749"},"modified":"2025-07-04T03:02:12","modified_gmt":"2025-07-04T03:02:12","slug":"using-lewis-dot-structures-indicate-the-movement-of-electrons-during-the-formation-of-the-ionic-compound-calcium-oxide-cao-from-calcium-ca-and-oxygen-o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/using-lewis-dot-structures-indicate-the-movement-of-electrons-during-the-formation-of-the-ionic-compound-calcium-oxide-cao-from-calcium-ca-and-oxygen-o\/","title":{"rendered":"Using Lewis dot structures, indicate the movement of electrons during the formation of the ionic compound calcium oxide (CaO) from calcium (Ca) and oxygen (O)"},"content":{"rendered":"\n<p>CF\u2084 (6) (b) Using Lewis dot structures, indicate the movement of electrons during the formation of the ionic compound calcium oxide (CaO) from calcium (Ca) and oxygen (O). (c) Sketch the geometry of the NF\u2083 molecule. Indicate lone pairs of electrons on nitrogen. (d) Give the molecular geometry of the atoms for i), ii), and iii) in the structure below: H H H H i) ii) iii) (6) (Total \/ Jumlah: 20<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-87.png\" alt=\"\" class=\"wp-image-241750\"\/><\/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>Of course. Here is the solution to the chemistry questions.<\/p>\n\n\n\n<p><strong>Correct Answer<\/strong><\/p>\n\n\n\n<p><strong>(a) Lewis Structures and Molecular Geometries:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>i. NH\u2083 (Ammonia):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure:<\/strong>\u00a0Nitrogen is the central atom, bonded to three hydrogen atoms, with one lone pair of electrons on the nitrogen.Generated code<code>H | :N - H | H<\/code>content_copydownloadUse code\u00a0<a href=\"https:\/\/support.google.com\/legal\/answer\/13505487\" target=\"_blank\" rel=\"noreferrer noopener\">with caution<\/a>.<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong>\u00a0Trigonal pyramidal<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>ii. H\u2082O (Water):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure:<\/strong>\u00a0Oxygen is the central atom, bonded to two hydrogen atoms, with two lone pairs of electrons on the oxygen.Generated code<code>H - O - H : :<\/code>content_copydownloadUse code\u00a0<a href=\"https:\/\/support.google.com\/legal\/answer\/13505487\" target=\"_blank\" rel=\"noreferrer noopener\">with caution<\/a>.(Note: the lone pairs are on the oxygen atom)<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong>\u00a0Bent (or V-shaped)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>iii. CF\u2084 (Carbon Tetrafluoride):<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure:<\/strong>\u00a0Carbon is the central atom, bonded to four fluorine atoms. Each fluorine atom has three lone pairs of electrons.Generated code<code>:F: | :F - C - F: | :F:<\/code>content_copydownloadUse code\u00a0<a href=\"https:\/\/support.google.com\/legal\/answer\/13505487\" target=\"_blank\" rel=\"noreferrer noopener\">with caution<\/a>.<\/li>\n\n\n\n<li><strong>Molecular Geometry:<\/strong>\u00a0Tetrahedral<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>(b) Formation of Calcium Oxide (CaO):<\/strong><\/p>\n\n\n\n<p>The formation involves the transfer of two electrons from a calcium atom to an oxygen atom.Generated code<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>..        -&gt;       2+      .. 2-\n    Ca   +   :O:          -&gt;    &#91;Ca]   +  &#91;:O:]\n      ..                       ..<\/code><\/pre>\n\n\n\n<p>content_copydownloadUse code&nbsp;<a href=\"https:\/\/support.google.com\/legal\/answer\/13505487\" target=\"_blank\" rel=\"noreferrer noopener\">with caution<\/a>.<\/p>\n\n\n\n<p>Calcium (Ca) loses its two valence electrons to become a calcium ion (Ca\u00b2\u207a). Oxygen (O) gains these two electrons to complete its octet, becoming an oxide ion (O\u00b2\u207b).<\/p>\n\n\n\n<p><strong>(c) Geometry of NF\u2083:<\/strong><\/p>\n\n\n\n<p>The NF\u2083 molecule has a trigonal pyramidal geometry. The central nitrogen atom has three bonding pairs (with fluorine) and one lone pair of electrons.<\/p>\n\n\n\n<p><em>(A sketch showing a nitrogen atom at the apex of a pyramid, bonded to three fluorine atoms at the base, with a lone pair of electrons shown on the nitrogen atom.)<\/em><\/p>\n\n\n\n<p><strong>(d) Molecular Geometry of Atoms:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>i) Carbonyl Carbon:<\/strong>\u00a0Trigonal planar<\/li>\n\n\n\n<li><strong>ii) Methylene Carbon:<\/strong>\u00a0Tetrahedral<\/li>\n\n\n\n<li><strong>iii) Hydroxyl Oxygen:<\/strong>\u00a0Bent (or V-shaped)<\/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\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The answers are determined using Lewis structures and the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states that electron pairs in the valence shell of a central atom repel each other and will arrange themselves in space to be as far apart as possible, which minimizes repulsion and determines the molecule&#8217;s geometry.<\/p>\n\n\n\n<p>For the covalent molecules in part (a) and (c), we first determine the total number of valence electrons to draw the Lewis structure. For NH\u2083 and NF\u2083, the central nitrogen atom has four electron domains (three bonding pairs and one lone pair). This results in a tetrahedral electron geometry, but the molecular geometry, which only considers the positions of the atoms, is trigonal pyramidal. The lone pair exerts a stronger repulsion than bonding pairs, pushing the bonded atoms closer together. For H\u2082O, the central oxygen has four electron domains (two bonding pairs and two lone pairs), leading to a bent molecular geometry. In CF\u2084, the central carbon has four bonding pairs and no lone pairs (AX\u2084), so its electron and molecular geometries are both tetrahedral.<\/p>\n\n\n\n<p>Part (b) illustrates ionic bonding, which involves the complete transfer of electrons from a metal to a nonmetal. Calcium, a Group 2 metal, readily donates its two valence electrons to achieve a stable configuration. Oxygen, a Group 16 nonmetal, accepts these two electrons to complete its valence shell (octet). The resulting oppositely charged ions, Ca\u00b2\u207a and O\u00b2\u207b, are held together by strong electrostatic attraction, forming the ionic compound calcium oxide.<\/p>\n\n\n\n<p>For part (d), VSEPR theory is applied to individual atoms within the larger molecule. The carbon atom at (i) is bonded to three other atoms (three electron domains), resulting in a trigonal planar geometry. The carbon at (ii) is bonded to four other atoms (four electron domains), giving it a tetrahedral geometry. The oxygen at (iii) has two bonding pairs and two lone pairs (four electron domains), resulting in a bent geometry around that atom.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-176.jpeg\" alt=\"\" class=\"wp-image-241751\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>CF\u2084 (6) (b) Using Lewis dot structures, indicate the movement of electrons during the formation of the ionic compound calcium oxide (CaO) from calcium (Ca) and oxygen (O). (c) Sketch the geometry of the NF\u2083 molecule. Indicate lone pairs of electrons on nitrogen. (d) Give the molecular geometry of the atoms for i), ii), and [&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-241749","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241749","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=241749"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241749\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=241749"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=241749"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=241749"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}