Determine the Lewis structure of O2? oxygen gas. Step 1: Complete the Lewis dot symbols for the oxygen atoms below 00 Step 2: Determine the number of valence electrons available in the O2? molecule. total valence electrons = Use two of these electrons to make a bond between the 2 oxygen atoms. 0:0 Step 3: Determine the number of unused valence electrons. Step 4: Place the remaining electrons as pairs of electrons around the O2? structure. 0:0 Does each of the 0 atoms have an octet of electrons?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Each oxygen atom has 6 valence electrons<\/strong> (since oxygen is in Group 16 of the periodic table). The Lewis dot symbols for two oxygen atoms are:<\/p>\n\n\n\n [ [ Each oxygen atom has 6 valence electrons<\/strong>, so for the O\u2082 molecule:<\/p>\n\n\n\n [ Two of these electrons will be used to form a bond between the oxygen atoms.<\/p>\n\n\n\n After forming a single bond<\/strong>, we subtract 2 electrons<\/strong>, leaving:<\/p>\n\n\n\n [ The remaining 10 electrons are placed around both oxygen atoms. However, when placing electrons, we must ensure that each atom satisfies the octet rule<\/strong> (having 8 electrons in its valence shell). Placing the electrons as lone pairs<\/strong> leads to the realization that a single bond is insufficient<\/strong> to complete both octets.<\/p>\n\n\n\n To satisfy the octet rule, a second bond must form<\/strong>, creating a double bond between the oxygen atoms:<\/p>\n\n\n\n [ Each oxygen atom now has:<\/p>\n\n\n\n This results in each oxygen atom having 8 valence electrons<\/strong>, satisfying the octet rule.<\/p>\n\n\n\n [ Oxygen gas (O\u2082) is composed of two oxygen atoms, each with 6 valence electrons. To determine its Lewis structure, we begin by considering the total valence electrons, which is 12<\/strong> (6 from each oxygen atom). Initially, a single bond (O\u2013O)<\/strong> would leave each oxygen atom with only 7 electrons, violating the octet rule.<\/p>\n\n\n\n To resolve this, a double bond (O=O)<\/strong> is formed, using 4 electrons<\/strong> (2 pairs) to create a stronger bond between the atoms. This leaves 4 electrons per oxygen atom as lone pairs<\/strong>, satisfying the octet rule.<\/p>\n\n\n\n The O\u2082 molecule exhibits paramagnetism<\/strong>, meaning it has unpaired electrons in its molecular orbital theory, which is not evident in the simple Lewis structure. However, in Lewis structures, the double bond representation is the most accepted form.<\/p>\n\n\n\n Thus, the correct Lewis structure of O\u2082 is O=O<\/strong>, where each oxygen has 8 valence electrons (4 from the bond and 4 from lone pairs), satisfying the octet rule.<\/p>\n","protected":false},"excerpt":{"rendered":" Determine the Lewis structure of O2? oxygen gas. Step 1: Complete the Lewis dot symbols for the oxygen atoms below 00 Step 2: Determine the number of valence electrons available in the O2? molecule. total valence electrons = Use two of these electrons to make a bond between the 2 oxygen atoms. 0:0 Step 3: […]<\/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-187692","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187692","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=187692"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187692\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187692"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187692"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Lewis Structure of O\u2082 (Oxygen Gas)<\/strong><\/h3>\n\n\n\n
Step 1: Lewis Dot Symbols for Oxygen Atoms<\/strong><\/h4>\n\n\n\n
\\cdot \\cdot O \\cdot \\cdot
]<\/p>\n\n\n\n
\\cdot \\cdot O \\cdot \\cdot
]<\/p>\n\n\n\nStep 2: Total Valence Electrons in O\u2082<\/strong><\/h4>\n\n\n\n
6 + 6 = 12 \\text{ total valence electrons}
]<\/p>\n\n\n\nStep 3: Determine the Number of Unused Valence Electrons<\/strong><\/h4>\n\n\n\n
12 – 2 = 10 \\text{ electrons remaining}
]<\/p>\n\n\n\nStep 4: Distribute the Remaining Electrons<\/strong><\/h4>\n\n\n\n
\\cdot \\cdot O = O \\cdot \\cdot
]<\/p>\n\n\n\n\n
Final Lewis Structure of O\u2082<\/strong><\/h3>\n\n\n\n
\\cdot \\cdot O = O \\cdot \\cdot
]<\/p>\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n