{"id":220801,"date":"2025-05-28T19:42:50","date_gmt":"2025-05-28T19:42:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220801"},"modified":"2025-05-28T19:42:55","modified_gmt":"2025-05-28T19:42:55","slug":"what-is-the-correct-oxidation-number-of-each-of-its-elements","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/what-is-the-correct-oxidation-number-of-each-of-its-elements\/","title":{"rendered":"What is the correct oxidation number of each of its elements"},"content":{"rendered":"\n<p>In the compound K2C1O4 , what is the correct oxidation number of each of its elements? (K is a Group I element) (The rules for assigning Oxidation States are given as below) 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 +1 ; Group II \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 +2 ; Group I \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 +3 3) Fluorine \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 -1 (always) Halogens &#8211; -1 (except in compounds with O and other halogens) 4) Hydrogen \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 +1 (except in metal hydrides, e.g. LiH, Rule 2 and H has -1) 5) Oxygen -2 (except in compounds with F, Rule 3) and compounds with 0-0 bonds (Rules 2 &amp; 4) OA K:+2 , Cr:+2 ,O: -2 O B. K:+1 , Cr:+4 ,O: -2 ocK:+1 , Cr:+2 ,O: -2 O D.K:+2, Cr:+6 ,O: -2 K:+1 , Cr:+6 ,O: -2 O E. None In the compound K\u00e2\u201a\u201aCrO4, what is the correct oxidation number of each of its elements? (K is a Group I element) (The rules for assigning Oxidation States are given as below) 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I \u00e2\u2020\u2019 +1; Group II \u00e2\u2020\u2019 +2 ; Group III \u00e2\u2020\u2019 +3 3) Fluorine \u00e2\u2020\u2019\u00e2\u2020\u2019-1 (always) Halogens -1 (except in compounds with O and other halogens) 4) Hydrogen +1 (except in metal hydrides, e.g. LiH, Rule 2 and H has -1) 5) Oxygen \u00e2\u2020\u2019 -2 (except in compounds with F, Rule 3) and compounds with O-O bonds (Rules 2 &amp; 4) \u00e2\u2020\u2019 OAK:+2. Cr:+2.0: -2 B. K:+1, Cr:+4,0:-2 Oc. K:+1, Cr:+2,0:-2 OD. K:+2, Cr:+6,0:-2 K:+1, Cr:+6,0: -2 O E.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is:<br><strong>E. K: +1, Cr: +6, O: \u20132<\/strong><\/p>\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>To determine the oxidation numbers in the compound <strong>K\u2082CrO\u2084 (potassium chromate)<\/strong>, we apply standard oxidation number rules:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. Total Oxidation Number = 0<\/strong><\/h4>\n\n\n\n<p>Since K\u2082CrO\u2084 is a <strong>neutral compound<\/strong>, the <strong>sum of the oxidation numbers<\/strong> of all atoms must equal <strong>zero<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. Potassium (K):<\/strong><\/h4>\n\n\n\n<p>Potassium is a <strong>Group I element<\/strong>, and <strong>Rule 2<\/strong> tells us Group I elements always have an oxidation number of <strong>+1<\/strong> in compounds.<br>There are <strong>two K atoms<\/strong>, so the total contribution from potassium is:<br><strong>2 \u00d7 (+1) = +2<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>3. Oxygen (O):<\/strong><\/h4>\n\n\n\n<p>Oxygen typically has an oxidation number of <strong>\u20132<\/strong>, as per <strong>Rule 5<\/strong>.<br>There are <strong>four O atoms<\/strong>, so the total contribution from oxygen is:<br><strong>4 \u00d7 (\u20132) = \u20138<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>4. Chromium (Cr):<\/strong><\/h4>\n\n\n\n<p>Let <strong>x<\/strong> be the oxidation number of Cr.<br>Now apply the rule that the sum of oxidation numbers = 0: 2(+1)+x+4(\u22122)=0\u21d22+x\u22128=0\u21d2x=+62(+1) + x + 4(-2) = 0 \\Rightarrow 2 + x &#8211; 8 = 0 \\Rightarrow x = +6<\/p>\n\n\n\n<p>So, <strong>Cr has an oxidation number of +6<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Summary of Oxidation Numbers in K\u2082CrO\u2084:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>K = +1<\/strong><\/li>\n\n\n\n<li><strong>Cr = +6<\/strong><\/li>\n\n\n\n<li><strong>O = \u20132<\/strong><\/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>Conclusion:<\/strong><\/h3>\n\n\n\n<p>Therefore, the correct oxidation states are:<br><strong>K: +1, Cr: +6, O: \u20132<\/strong>, which matches <strong>option E<\/strong>.<\/p>\n\n\n\n<p>This structure is consistent with the known chemistry of <strong>chromates<\/strong>, where chromium is commonly in the +6 oxidation state.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner7-73.jpeg\" alt=\"\" class=\"wp-image-220802\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>In the compound K2C1O4 , what is the correct oxidation number of each of its elements? (K is a Group I element) (The rules for assigning Oxidation States are given as below) 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I \u00c3\u00a2\u00e2\u20ac \u00e2\u20ac\u2122 +1 ; [&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-220801","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220801","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=220801"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220801\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220801"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220801"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}