{"id":248369,"date":"2025-07-08T10:33:12","date_gmt":"2025-07-08T10:33:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248369"},"modified":"2025-07-08T10:33:14","modified_gmt":"2025-07-08T10:33:14","slug":"determine-the-correct-oxidation-numbers-for-all-three-elements-in-na2c2o4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/determine-the-correct-oxidation-numbers-for-all-three-elements-in-na2c2o4\/","title":{"rendered":"Determine the correct oxidation numbers for all three elements in Na2C2O4."},"content":{"rendered":"\n

Determine the correct oxidation numbers for all three elements in Na2C2O4. (Na 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 = +1, Group II = +2, Group III = -3 3) Fluorine = -1 (always) 4) Halogens (except in compounds with oxygen and other halogens) = -1 5) Hydrogen = +1 (except in metal hydrides e.g. LiH, Rule 2) 6) Oxygen = -2 (except in compounds with fluorine, Rule 3, and compounds with O-O bonds, Rules 2 & 4) A. Na: +1, C: +3, O: -2 B. Na: +1, C: +4, O: -2 C. Na: +1, C: +2, O: -2 D. Na: +1, C: +6, O: -2 E. Na: +2, C: +4, O: -2<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The correct answer is A. Na: +1, C: +3, O: -2<\/strong>.<\/p>\n\n\n\n

Here\u2019s how the oxidation states are determined:<\/p>\n\n\n\n

Step 1: Assign Oxidation States Based on Rules<\/h3>\n\n\n\n
    \n
  1. Sodium (Na)<\/strong>:\n
      \n
    • Sodium is a Group I element, and according to Rule 2, elements in Group I have an oxidation number of +1<\/strong>. Therefore, Na is assigned an oxidation state of +1<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Oxygen (O)<\/strong>:\n
        \n
      • Oxygen typically has an oxidation number of -2<\/strong>, as per Rule 6, unless it is part of a peroxide or compound with fluorine. Since oxygen is in the oxalate ion (C\u2082O\u2084\u00b2\u207b), we apply the rule and assign -2<\/strong> to each oxygen atom.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Carbon (C)<\/strong>:\n
          \n
        • The molecule is sodium oxalate (Na\u2082C\u2082O\u2084)<\/strong>, which contains two carbon atoms. To find the oxidation number of carbon, we use the fact that the sum of all oxidation numbers in a neutral compound must equal zero.<\/li>\n\n\n\n
        • We have 2 sodium atoms with an oxidation number of +1<\/strong> each, and 4 oxygen atoms with an oxidation number of -2<\/strong> each. So, we can set up an equation to solve for the oxidation state of carbon (C): 2(Na)+2(C)+4(O)=02(\\text{Na}) + 2(\\text{C}) + 4(\\text{O}) = 02(Na)+2(C)+4(O)=0 Substituting known values: 2(+1)+2(C)+4(\u22122)=02(+1) + 2(\\text{C}) + 4(-2) = 02(+1)+2(C)+4(\u22122)=0 2+2(C)\u22128=02 + 2(\\text{C}) – 8 = 02+2(C)\u22128=0 2(C)\u22126=02(\\text{C}) – 6 = 02(C)\u22126=0 2(C)=62(\\text{C}) = 62(C)=6 C=+3\\text{C} = +3C=+3 Thus, the oxidation state of carbon in sodium oxalate is +3<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

          Step 2: Verify the Sum of Oxidation Numbers<\/h3>\n\n\n\n
            \n
          • The sum of the oxidation states should be zero for a neutral compound:\n
              \n
            • Sodium (Na): 2 \u00d7 (+1) = +2<\/li>\n\n\n\n
            • Carbon (C): 2 \u00d7 (+3) = +6<\/li>\n\n\n\n
            • Oxygen (O): 4 \u00d7 (-2) = -8<\/li>\n\n\n\n
            • Total: +2 + 6 + (-8) = 0<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

              Therefore, the oxidation numbers are:<\/p>\n\n\n\n

                \n
              • Na: +1<\/strong><\/li>\n\n\n\n
              • C: +3<\/strong><\/li>\n\n\n\n
              • O: -2<\/strong><\/li>\n<\/ul>\n\n\n\n

                Thus, the correct answer is A<\/strong>.<\/p>\n\n\n\n

                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                Determine the correct oxidation numbers for all three elements in Na2C2O4. (Na 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 = +1, Group II = +2, Group III = […]<\/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-248369","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248369","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=248369"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248369\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}