{"id":166393,"date":"2024-11-13T16:37:20","date_gmt":"2024-11-13T16:37:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=166393"},"modified":"2024-11-13T16:37:22","modified_gmt":"2024-11-13T16:37:22","slug":"what-is-the-oxidation-state-of-each-element-in-coh2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/13\/what-is-the-oxidation-state-of-each-element-in-coh2\/","title":{"rendered":"what is the oxidation state of each element in coh2"},"content":{"rendered":"\n<p>what is the oxidation state of each element in coh2? c o h what is the oxidation state of each element in febr3? fe br<\/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>To determine the oxidation states of each element in ( \\text{COH}_2 ) and ( \\text{FeBr}_3 ), let&#8217;s go through the process step by step for each compound.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Oxidation States in ( \\text{COH}_2 ):<\/h3>\n\n\n\n<p>This is a molecular compound containing carbon (C), oxygen (O), and hydrogen (H).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hydrogen (H)<\/strong>: In most compounds, hydrogen has an oxidation state of +1. Since there are two hydrogen atoms in ( \\text{COH}_2 ), the total oxidation state of hydrogen is ( 2 \\times (+1) = +2 ).<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong>: Oxygen typically has an oxidation state of -2 in most compounds, including this one.<\/li>\n\n\n\n<li><strong>Carbon (C)<\/strong>: The oxidation state of carbon is what we need to determine.<\/li>\n<\/ul>\n\n\n\n<p>Now, let\u2019s balance the total oxidation state of the molecule. The sum of all oxidation states in a neutral molecule must equal 0.<\/p>\n\n\n\n<p>Let the oxidation state of carbon be ( x ).<\/p>\n\n\n\n<p>The equation for the sum of oxidation states is:<br>[<br>x + (-2) + 2(+1) = 0<br>]<br>Simplifying:<br>[<br>x &#8211; 2 + 2 = 0 \\quad \\Rightarrow \\quad x = 0<br>]<br>So, the oxidation states are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong>: 0<\/li>\n\n\n\n<li><strong>O<\/strong>: -2<\/li>\n\n\n\n<li><strong>H<\/strong>: +1 (for each hydrogen atom)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2. Oxidation States in ( \\text{FeBr}_3 ):<\/h3>\n\n\n\n<p>This is an ionic compound consisting of iron (Fe) and bromine (Br).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bromine (Br)<\/strong>: Bromine is a halogen and typically has an oxidation state of -1 in its compounds, such as ( \\text{FeBr}_3 ). Since there are three bromine atoms, the total oxidation state of bromine is ( 3 \\times (-1) = -3 ).<\/li>\n\n\n\n<li><strong>Iron (Fe)<\/strong>: The oxidation state of iron can be determined by considering the total charge of the molecule. ( \\text{FeBr}_3 ) is neutral, so the oxidation state of iron must balance the -3 from the bromine atoms.<\/li>\n<\/ul>\n\n\n\n<p>Let the oxidation state of iron be ( x ).<\/p>\n\n\n\n<p>The equation for the sum of oxidation states is:<br>[<br>x + 3(-1) = 0<br>]<br>Simplifying:<br>[<br>x &#8211; 3 = 0 \\quad \\Rightarrow \\quad x = +3<br>]<br>So, the oxidation states are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Fe<\/strong>: +3<\/li>\n\n\n\n<li><strong>Br<\/strong>: -1 (for each bromine atom)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>In ( \\text{COH}_2 )<\/strong>:<\/li>\n\n\n\n<li>C: 0<\/li>\n\n\n\n<li>O: -2<\/li>\n\n\n\n<li>H: +1 (for each hydrogen atom)<\/li>\n\n\n\n<li><strong>In ( \\text{FeBr}_3 )<\/strong>:<\/li>\n\n\n\n<li>Fe: +3<\/li>\n\n\n\n<li>Br: -1 (for each bromine atom)<\/li>\n<\/ul>\n\n\n\n<p>In general, the oxidation state of an element in a compound reflects how electrons are distributed between the atoms involved. The sum of oxidation states in a neutral compound is zero, which we used to determine the unknown oxidation states.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>what is the oxidation state of each element in coh2? c o h what is the oxidation state of each element in febr3? fe br The Correct Answer and Explanation is : To determine the oxidation states of each element in ( \\text{COH}_2 ) and ( \\text{FeBr}_3 ), let&#8217;s go through the process step by [&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-166393","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/166393","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=166393"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/166393\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=166393"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=166393"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=166393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}