Let’s address each question step-by-step, with correct answers followed by a detailed 300-word explanation.<\/p>\n\n\n\n
\n\n\n\n
1. Determine the Bond Order of BN.<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Bond Order = 2<\/p>\n\n\n\n2. Is BN\u00b2\u207a paramagnetic or diamagnetic?<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Diamagnetic<\/strong><\/p>\n\n\n\n3. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO\u207a?<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Stronger<\/strong><\/p>\n\n\n\n4. Compare and contrast VBT versus MOT for B\u2082 (Boron molecule).<\/strong><\/h3>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\n- Valence Bond Theory (VBT):<\/strong> Predicts no bond<\/strong> in B\u2082.<\/li>\n\n\n\n
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/strong><\/h3>\n\n\n\nTo determine the bond order<\/strong> of BN (boron nitride), we use Molecular Orbital Theory (MOT)<\/strong>. The total number of electrons in BN is 12 (B = 5, N = 7). Following the MOT for diatomic molecules with total electrons \u2264 14 (like N\u2082), the molecular orbital filling order is:<\/p>\n\n\n\n\u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2p_x)\u00b2 = \u03c0(2p_y)\u00b2<\/strong>, \u03c3(2p_z)\u00b2.<\/p>\n\n\n\nCounting bonding (8) and antibonding (4) electrons:<\/p>\n\n\n\n
Bond Order = (Bonding – Antibonding)\/2 = (8 – 4)\/2 = 2<\/strong>.<\/p>\n\n\n\nFor BN\u00b2\u207a<\/strong>, removing two electrons typically occurs from the highest-energy orbitals, likely the \u03c0(2p) or \u03c3(2p_z). If both are removed from \u03c0(2p), we now have paired electrons only, making BN\u00b2\u207a diamagnetic<\/strong> (no unpaired electrons).<\/p>\n\n\n\nNext, the bond strength in BO vs BO\u207a<\/strong>: BO has 11 electrons. BO\u207a has 10 electrons, meaning one electron is removed from an antibonding orbital. Fewer antibonding electrons lead to a higher bond order<\/strong> and a stronger bond<\/strong> in BO\u207a than in BO. Therefore, BO\u207a has a stronger bond<\/strong>.<\/p>\n\n\n\nComparing Valence Bond Theory (VBT)<\/strong> and Molecular Orbital Theory (MOT)<\/strong> for B\u2082:<\/p>\n\n\n\n\n- VBT struggles with delocalized bonding and cannot explain the magnetic behavior of molecules like B\u2082. VBT incorrectly predicts no bond in B\u2082 since it assumes each boron atom lacks unpaired electrons to overlap effectively.<\/li>\n\n\n\n
- MOT accurately predicts that B\u2082 has a bond order of 1<\/strong> and is paramagnetic<\/strong>, due to two unpaired electrons in the degenerate \u03c0(2p) orbitals.<\/li>\n<\/ul>\n\n\n\n
In conclusion, MOT provides better predictions for molecular properties such as bond order, magnetism, and stability, especially for homonuclear diatomics like B\u2082, BN, and BO.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-14.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Determine the Bond Order of BN. 6. Is BN2+ paramagnetic or diamagnetic? 7. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO? 8. Compare and contrast VBT verus MOT for B2 MOT Worksheet-Dr. Kim-Ver. The Correct Answer and Explanation is: Let’s address each question step-by-step, with correct […]<\/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-217598","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598","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=217598"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217598"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217598"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
2. Is BN\u00b2\u207a paramagnetic or diamagnetic?<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Diamagnetic<\/strong><\/p>\n\n\n\n3. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO\u207a?<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Stronger<\/strong><\/p>\n\n\n\n4. Compare and contrast VBT versus MOT for B\u2082 (Boron molecule).<\/strong><\/h3>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\n- Valence Bond Theory (VBT):<\/strong> Predicts no bond<\/strong> in B\u2082.<\/li>\n\n\n\n
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/strong><\/h3>\n\n\n\nTo determine the bond order<\/strong> of BN (boron nitride), we use Molecular Orbital Theory (MOT)<\/strong>. The total number of electrons in BN is 12 (B = 5, N = 7). Following the MOT for diatomic molecules with total electrons \u2264 14 (like N\u2082), the molecular orbital filling order is:<\/p>\n\n\n\n\u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2p_x)\u00b2 = \u03c0(2p_y)\u00b2<\/strong>, \u03c3(2p_z)\u00b2.<\/p>\n\n\n\nCounting bonding (8) and antibonding (4) electrons:<\/p>\n\n\n\n
Bond Order = (Bonding – Antibonding)\/2 = (8 – 4)\/2 = 2<\/strong>.<\/p>\n\n\n\nFor BN\u00b2\u207a<\/strong>, removing two electrons typically occurs from the highest-energy orbitals, likely the \u03c0(2p) or \u03c3(2p_z). If both are removed from \u03c0(2p), we now have paired electrons only, making BN\u00b2\u207a diamagnetic<\/strong> (no unpaired electrons).<\/p>\n\n\n\nNext, the bond strength in BO vs BO\u207a<\/strong>: BO has 11 electrons. BO\u207a has 10 electrons, meaning one electron is removed from an antibonding orbital. Fewer antibonding electrons lead to a higher bond order<\/strong> and a stronger bond<\/strong> in BO\u207a than in BO. Therefore, BO\u207a has a stronger bond<\/strong>.<\/p>\n\n\n\nComparing Valence Bond Theory (VBT)<\/strong> and Molecular Orbital Theory (MOT)<\/strong> for B\u2082:<\/p>\n\n\n\n\n- VBT struggles with delocalized bonding and cannot explain the magnetic behavior of molecules like B\u2082. VBT incorrectly predicts no bond in B\u2082 since it assumes each boron atom lacks unpaired electrons to overlap effectively.<\/li>\n\n\n\n
- MOT accurately predicts that B\u2082 has a bond order of 1<\/strong> and is paramagnetic<\/strong>, due to two unpaired electrons in the degenerate \u03c0(2p) orbitals.<\/li>\n<\/ul>\n\n\n\n
In conclusion, MOT provides better predictions for molecular properties such as bond order, magnetism, and stability, especially for homonuclear diatomics like B\u2082, BN, and BO.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Determine the Bond Order of BN. 6. Is BN2+ paramagnetic or diamagnetic? 7. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO? 8. Compare and contrast VBT verus MOT for B2 MOT Worksheet-Dr. Kim-Ver. The Correct Answer and Explanation is: Let’s address each question step-by-step, with correct […]<\/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-217598","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598","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=217598"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217598"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217598"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
3. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO\u207a?<\/strong><\/h3>\n\n\n\nAnswer:<\/strong> Stronger<\/strong><\/p>\n\n\n\n4. Compare and contrast VBT versus MOT for B\u2082 (Boron molecule).<\/strong><\/h3>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\n- Valence Bond Theory (VBT):<\/strong> Predicts no bond<\/strong> in B\u2082.<\/li>\n\n\n\n
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/strong><\/h3>\n\n\n\nTo determine the bond order<\/strong> of BN (boron nitride), we use Molecular Orbital Theory (MOT)<\/strong>. The total number of electrons in BN is 12 (B = 5, N = 7). Following the MOT for diatomic molecules with total electrons \u2264 14 (like N\u2082), the molecular orbital filling order is:<\/p>\n\n\n\n\u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2p_x)\u00b2 = \u03c0(2p_y)\u00b2<\/strong>, \u03c3(2p_z)\u00b2.<\/p>\n\n\n\nCounting bonding (8) and antibonding (4) electrons:<\/p>\n\n\n\n
Bond Order = (Bonding – Antibonding)\/2 = (8 – 4)\/2 = 2<\/strong>.<\/p>\n\n\n\nFor BN\u00b2\u207a<\/strong>, removing two electrons typically occurs from the highest-energy orbitals, likely the \u03c0(2p) or \u03c3(2p_z). If both are removed from \u03c0(2p), we now have paired electrons only, making BN\u00b2\u207a diamagnetic<\/strong> (no unpaired electrons).<\/p>\n\n\n\nNext, the bond strength in BO vs BO\u207a<\/strong>: BO has 11 electrons. BO\u207a has 10 electrons, meaning one electron is removed from an antibonding orbital. Fewer antibonding electrons lead to a higher bond order<\/strong> and a stronger bond<\/strong> in BO\u207a than in BO. Therefore, BO\u207a has a stronger bond<\/strong>.<\/p>\n\n\n\nComparing Valence Bond Theory (VBT)<\/strong> and Molecular Orbital Theory (MOT)<\/strong> for B\u2082:<\/p>\n\n\n\n\n- VBT struggles with delocalized bonding and cannot explain the magnetic behavior of molecules like B\u2082. VBT incorrectly predicts no bond in B\u2082 since it assumes each boron atom lacks unpaired electrons to overlap effectively.<\/li>\n\n\n\n
- MOT accurately predicts that B\u2082 has a bond order of 1<\/strong> and is paramagnetic<\/strong>, due to two unpaired electrons in the degenerate \u03c0(2p) orbitals.<\/li>\n<\/ul>\n\n\n\n
In conclusion, MOT provides better predictions for molecular properties such as bond order, magnetism, and stability, especially for homonuclear diatomics like B\u2082, BN, and BO.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Determine the Bond Order of BN. 6. Is BN2+ paramagnetic or diamagnetic? 7. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO? 8. Compare and contrast VBT verus MOT for B2 MOT Worksheet-Dr. Kim-Ver. The Correct Answer and Explanation is: Let’s address each question step-by-step, with correct […]<\/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-217598","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598","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=217598"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217598"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217598"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
4. Compare and contrast VBT versus MOT for B\u2082 (Boron molecule).<\/strong><\/h3>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\n- Valence Bond Theory (VBT):<\/strong> Predicts no bond<\/strong> in B\u2082.<\/li>\n\n\n\n
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/strong><\/h3>\n\n\n\nTo determine the bond order<\/strong> of BN (boron nitride), we use Molecular Orbital Theory (MOT)<\/strong>. The total number of electrons in BN is 12 (B = 5, N = 7). Following the MOT for diatomic molecules with total electrons \u2264 14 (like N\u2082), the molecular orbital filling order is:<\/p>\n\n\n\n\u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2p_x)\u00b2 = \u03c0(2p_y)\u00b2<\/strong>, \u03c3(2p_z)\u00b2.<\/p>\n\n\n\nCounting bonding (8) and antibonding (4) electrons:<\/p>\n\n\n\n
Bond Order = (Bonding – Antibonding)\/2 = (8 – 4)\/2 = 2<\/strong>.<\/p>\n\n\n\nFor BN\u00b2\u207a<\/strong>, removing two electrons typically occurs from the highest-energy orbitals, likely the \u03c0(2p) or \u03c3(2p_z). If both are removed from \u03c0(2p), we now have paired electrons only, making BN\u00b2\u207a diamagnetic<\/strong> (no unpaired electrons).<\/p>\n\n\n\nNext, the bond strength in BO vs BO\u207a<\/strong>: BO has 11 electrons. BO\u207a has 10 electrons, meaning one electron is removed from an antibonding orbital. Fewer antibonding electrons lead to a higher bond order<\/strong> and a stronger bond<\/strong> in BO\u207a than in BO. Therefore, BO\u207a has a stronger bond<\/strong>.<\/p>\n\n\n\nComparing Valence Bond Theory (VBT)<\/strong> and Molecular Orbital Theory (MOT)<\/strong> for B\u2082:<\/p>\n\n\n\n\n- VBT struggles with delocalized bonding and cannot explain the magnetic behavior of molecules like B\u2082. VBT incorrectly predicts no bond in B\u2082 since it assumes each boron atom lacks unpaired electrons to overlap effectively.<\/li>\n\n\n\n
- MOT accurately predicts that B\u2082 has a bond order of 1<\/strong> and is paramagnetic<\/strong>, due to two unpaired electrons in the degenerate \u03c0(2p) orbitals.<\/li>\n<\/ul>\n\n\n\n
In conclusion, MOT provides better predictions for molecular properties such as bond order, magnetism, and stability, especially for homonuclear diatomics like B\u2082, BN, and BO.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Determine the Bond Order of BN. 6. Is BN2+ paramagnetic or diamagnetic? 7. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO? 8. Compare and contrast VBT verus MOT for B2 MOT Worksheet-Dr. Kim-Ver. The Correct Answer and Explanation is: Let’s address each question step-by-step, with correct […]<\/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-217598","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598","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=217598"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217598"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217598"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- \n
- Valence Bond Theory (VBT):<\/strong> Predicts no bond<\/strong> in B\u2082.<\/li>\n\n\n\n
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nDetailed Explanation (300+ words):<\/strong><\/h3>\n\n\n\n
To determine the bond order<\/strong> of BN (boron nitride), we use Molecular Orbital Theory (MOT)<\/strong>. The total number of electrons in BN is 12 (B = 5, N = 7). Following the MOT for diatomic molecules with total electrons \u2264 14 (like N\u2082), the molecular orbital filling order is:<\/p>\n\n\n\n
\u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2p_x)\u00b2 = \u03c0(2p_y)\u00b2<\/strong>, \u03c3(2p_z)\u00b2.<\/p>\n\n\n\n
Counting bonding (8) and antibonding (4) electrons:<\/p>\n\n\n\n
Bond Order = (Bonding – Antibonding)\/2 = (8 – 4)\/2 = 2<\/strong>.<\/p>\n\n\n\n
For BN\u00b2\u207a<\/strong>, removing two electrons typically occurs from the highest-energy orbitals, likely the \u03c0(2p) or \u03c3(2p_z). If both are removed from \u03c0(2p), we now have paired electrons only, making BN\u00b2\u207a diamagnetic<\/strong> (no unpaired electrons).<\/p>\n\n\n\n
Next, the bond strength in BO vs BO\u207a<\/strong>: BO has 11 electrons. BO\u207a has 10 electrons, meaning one electron is removed from an antibonding orbital. Fewer antibonding electrons lead to a higher bond order<\/strong> and a stronger bond<\/strong> in BO\u207a than in BO. Therefore, BO\u207a has a stronger bond<\/strong>.<\/p>\n\n\n\n
Comparing Valence Bond Theory (VBT)<\/strong> and Molecular Orbital Theory (MOT)<\/strong> for B\u2082:<\/p>\n\n\n\n
- \n
- VBT struggles with delocalized bonding and cannot explain the magnetic behavior of molecules like B\u2082. VBT incorrectly predicts no bond in B\u2082 since it assumes each boron atom lacks unpaired electrons to overlap effectively.<\/li>\n\n\n\n
- MOT accurately predicts that B\u2082 has a bond order of 1<\/strong> and is paramagnetic<\/strong>, due to two unpaired electrons in the degenerate \u03c0(2p) orbitals.<\/li>\n<\/ul>\n\n\n\n
In conclusion, MOT provides better predictions for molecular properties such as bond order, magnetism, and stability, especially for homonuclear diatomics like B\u2082, BN, and BO.<\/p>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"Determine the Bond Order of BN. 6. Is BN2+ paramagnetic or diamagnetic? 7. Does the bond between boron and oxygen in BO become stronger or weaker when it becomes BO? 8. Compare and contrast VBT verus MOT for B2 MOT Worksheet-Dr. Kim-Ver. The Correct Answer and Explanation is: Let’s address each question step-by-step, with correct […]<\/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-217598","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598","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=217598"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217598\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217598"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217598"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217598"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Molecular Orbital Theory (MOT):<\/strong> Predicts a bond order of 1<\/strong> and paramagnetism<\/strong> due to unpaired electrons in \u03c0 orbitals.<\/li>\n<\/ul>\n\n\n\n