{"id":196044,"date":"2025-03-02T08:26:01","date_gmt":"2025-03-02T08:26:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196044"},"modified":"2025-03-02T08:26:04","modified_gmt":"2025-03-02T08:26:04","slug":"bond-order-of-kr2-b2-c2-n2-respectively-are","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/02\/bond-order-of-kr2-b2-c2-n2-respectively-are\/","title":{"rendered":"Bond Order of Kr2, B2, C2, N2 respectively are"},"content":{"rendered":"\n

Bond Order of Kr2, B2, C2, N2 respectively are
A+0,+3,+1,+2
B+2,+3,+1,0
C+3,+2,+1,0
D+1, +3, +1,0
E+2, +2, +1,0
FQ,+1,+2,+3<\/p>\n\n\n\n

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

To calculate the bond order of diatomic molecules like ( \\text{Kr}_2, \\text{B}_2, \\text{C}_2, \\text{N}_2 ), we need to consider the molecular orbital theory. The bond order is given by the formula:<\/p>\n\n\n\n

[
\\text{Bond Order} = \\frac{1}{2} \\left( \\text{Number of bonding electrons} – \\text{Number of antibonding electrons} \\right)
]<\/p>\n\n\n\n

Let\u2019s go over the molecular orbital theory for each of the given molecules, calculating the bond order step by step:<\/p>\n\n\n\n

1. Krypton ((\\text{Kr}_2))<\/strong><\/h3>\n\n\n\n

Krypton has an atomic number of 36, so for the (\\text{Kr}_2) molecule, we have a total of (36 + 36 = 72) electrons. In this case, krypton is a noble gas, and the electron configuration will place electrons in both bonding and antibonding molecular orbitals. However, the bond order for (\\text{Kr}_2) is typically found to be 0<\/strong>, since the number of bonding electrons is equal to the number of antibonding electrons.<\/p>\n\n\n\n

Bond Order of (\\text{Kr}_2 = 0)<\/strong>.<\/p>\n\n\n\n

2. Boron ((\\text{B}_2))<\/strong><\/h3>\n\n\n\n

Boron has an atomic number of 5, so for (\\text{B}_2), we have a total of (5 + 5 = 10) electrons. The molecular orbital filling for (\\text{B}_2) will have electrons in both bonding and antibonding orbitals, and the bond order comes out to be 1<\/strong>.<\/p>\n\n\n\n

Bond Order of (\\text{B}_2 = 1)<\/strong>.<\/p>\n\n\n\n

3. Carbon ((\\text{C}_2))<\/strong><\/h3>\n\n\n\n

Carbon has an atomic number of 6, so for (\\text{C}_2), we have a total of (6 + 6 = 12) electrons. Carbon follows similar molecular orbital filling principles as boron, but with slightly more bonding electrons. The bond order comes out to be 2<\/strong>.<\/p>\n\n\n\n

Bond Order of (\\text{C}_2 = 2)<\/strong>.<\/p>\n\n\n\n

4. Nitrogen ((\\text{N}_2))<\/strong><\/h3>\n\n\n\n

Nitrogen has an atomic number of 7, so for (\\text{N}_2), we have a total of (7 + 7 = 14) electrons. The molecular orbital filling of (\\text{N}_2) results in a bond order of 3<\/strong>.<\/p>\n\n\n\n

Bond Order of (\\text{N}_2 = 3)<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Answer Choices:<\/h3>\n\n\n\n
    \n
  • A<\/strong>: 0, 3, 1, 2<\/li>\n\n\n\n
  • B<\/strong>: 2, 3, 1, 0<\/li>\n\n\n\n
  • C<\/strong>: 3, 2, 1, 0<\/li>\n\n\n\n
  • D<\/strong>: 1, 3, 1, 0<\/li>\n\n\n\n
  • E<\/strong>: 2, 2, 1, 0<\/li>\n\n\n\n
  • F<\/strong>: 1, 2, 3<\/li>\n<\/ul>\n\n\n\n

    From the analysis, the correct answer is C<\/strong> (0, 3, 1, 2). This corresponds to the bond order of Krypton, Boron, Carbon, and Nitrogen molecules, respectively.<\/p>\n\n\n\n

    Conclusion:<\/h3>\n\n\n\n

    The bond order depends on the number of bonding and antibonding electrons in the molecular orbitals. For (\\text{Kr}_2), the bond order is 0 due to equal bonding and antibonding electrons. For (\\text{B}_2), (\\text{C}_2), and (\\text{N}_2), the bond orders are 1, 2, and 3, respectively, based on their electron configurations and molecular orbital theory.<\/p>\n\n\n\n

    Let me generate the image now.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Here is the diagram illustrating the molecular orbital theory bond order calculations for (\\text{Kr}_2, \\text{B}_2, \\text{C}_2,) and (\\text{N}_2). It shows the bonding and antibonding electron configurations, helping to visualize how bond order is derived.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Bond Order of Kr2, B2, C2, N2 respectively areA+0,+3,+1,+2B+2,+3,+1,0C+3,+2,+1,0D+1, +3, +1,0E+2, +2, +1,0FQ,+1,+2,+3 The Correct Answer and Explanation is : To calculate the bond order of diatomic molecules like ( \\text{Kr}_2, \\text{B}_2, \\text{C}_2, \\text{N}_2 ), we need to consider the molecular orbital theory. The bond order is given by the formula: [\\text{Bond Order} = \\frac{1}{2} […]<\/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-196044","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196044","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=196044"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196044\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196044"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196044"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196044"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}