{"id":45804,"date":"2025-07-01T12:06:27","date_gmt":"2025-07-01T12:06:27","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45804"},"modified":"2025-07-01T12:06:29","modified_gmt":"2025-07-01T12:06:29","slug":"sketch-and-label-a-qualitative-molecular-orbital-mo-energy-level-diagram-for-nitrogen-monofluoride-nf","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/sketch-and-label-a-qualitative-molecular-orbital-mo-energy-level-diagram-for-nitrogen-monofluoride-nf\/","title":{"rendered":"Sketch and label a qualitative molecular orbital (MO) energy level diagram for nitrogen monofluoride, NF."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Sketch and label a qualitative molecular orbital (MO) energy level diagram for nitrogen monofluoride, NF. (ii) Use your diagram to determine the bond order of NF, [NF]*, and [NF]. (iii) Use your diagram to determine whether each of NF, [NF]+, and [NF] is diamagnetic or paramagnetic. (b) (10 marks) The triboron cation (cyclo-B3+) has a triangular structure with each BB-B angle of 120 degrees. Construct a qualitative MO energy level diagram for the bonding in cyclo-B3+. Illustrate your answer by referencing clearly labeled sketches showing how the atomic orbitals interact to form the molecular orbitals.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) <strong>Molecular Orbital (MO) Diagram for NF (Nitrogen Monofluoride)<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">(i) <strong>Qualitative MO Energy Level Diagram for NF:<\/strong><\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Atomic Orbitals (AOs):<\/strong>\n<ul class=\"wp-block-list\">\n<li>Nitrogen (N) has the electronic configuration 1s22s22p31s^2 2s^2 2p^31s22s22p3.<\/li>\n\n\n\n<li>Fluorine (F) has the electronic configuration 1s22s22p51s^2 2s^2 2p^51s22s22p5.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>MO Diagram:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The 2s2s2s orbitals combine to form a bonding \u03c32s\\sigma_{2s}\u03c32s\u200b and an antibonding \u03c32s\u2217\\sigma^*_{2s}\u03c32s\u2217\u200b.<\/li>\n\n\n\n<li>The 2p2p2p orbitals combine in two ways:\n<ul class=\"wp-block-list\">\n<li>\u03c32pz\\sigma_{2p_z}\u03c32pz\u200b\u200b (bonding) and \u03c32pz\u2217\\sigma^*_{2p_z}\u03c32pz\u200b\u2217\u200b (antibonding).<\/li>\n\n\n\n<li>\u03c02px\\pi_{2p_x}\u03c02px\u200b\u200b, \u03c02py\\pi_{2p_y}\u03c02py\u200b\u200b (bonding), and \u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b, \u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b (antibonding).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Electron Placement:<\/strong><ul><li>From the atomic orbitals of N and F, there are 10 electrons to place in the molecular orbitals.<\/li><li>Electrons fill from the lowest energy orbital upward, following the Pauli exclusion principle and Hund\u2019s rule.<\/li><\/ul>The filling order for the molecular orbitals is as follows:\n<ul class=\"wp-block-list\">\n<li>2 electrons in \u03c32s\\sigma_{2s}\u03c32s\u200b<\/li>\n\n\n\n<li>2 electrons in \u03c32s\u2217\\sigma^*_{2s}\u03c32s\u2217\u200b<\/li>\n\n\n\n<li>2 electrons in \u03c02px\\pi_{2p_x}\u03c02px\u200b\u200b and \u03c02py\\pi_{2p_y}\u03c02py\u200b\u200b<\/li>\n\n\n\n<li>2 electrons in \u03c32pz\\sigma_{2p_z}\u03c32pz\u200b\u200b<\/li>\n\n\n\n<li>2 electrons in \u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b and \u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<h4 class=\"wp-block-heading\">(ii) <strong>Bond Order Calculation for NF:<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">The bond order is determined by the formula:Bond&nbsp;order=12(Number&nbsp;of&nbsp;electrons&nbsp;in&nbsp;bonding&nbsp;orbitals\u2212Number&nbsp;of&nbsp;electrons&nbsp;in&nbsp;antibonding&nbsp;orbitals)\\text{Bond order} = \\frac{1}{2} \\left( \\text{Number of electrons in bonding orbitals} &#8211; \\text{Number of electrons in antibonding orbitals} \\right)Bond&nbsp;order=21\u200b(Number&nbsp;of&nbsp;electrons&nbsp;in&nbsp;bonding&nbsp;orbitals\u2212Number&nbsp;of&nbsp;electrons&nbsp;in&nbsp;antibonding&nbsp;orbitals)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bonding electrons: 2 (\u03c32s\\sigma_{2s}\u03c32s\u200b) + 2 (\u03c02px\\pi_{2p_x}\u03c02px\u200b\u200b) + 2 (\u03c02py\\pi_{2p_y}\u03c02py\u200b\u200b) + 2 (\u03c32pz\\sigma_{2p_z}\u03c32pz\u200b\u200b) = 8 electrons.<\/li>\n\n\n\n<li>Antibonding electrons: 2 (\u03c32s\u2217\\sigma^*_{2s}\u03c32s\u2217\u200b) + 2 (\u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b) + 2 (\u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b) = 6 electrons.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the bond order for NF is:Bond&nbsp;order=12(8\u22126)=1\\text{Bond order} = \\frac{1}{2} \\left( 8 &#8211; 6 \\right) = 1Bond&nbsp;order=21\u200b(8\u22126)=1<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">(iii) <strong>Magnetic Behavior of NF, [NF]*, and [NF]+:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>NF:<\/strong> Since there are no unpaired electrons in the molecular orbitals, NF is <strong>diamagnetic<\/strong>.<\/li>\n\n\n\n<li><strong>[NF]*:<\/strong>\n<ul class=\"wp-block-list\">\n<li>This ion has one additional electron, which will go into the \u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b or \u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b antibonding orbital.<\/li>\n\n\n\n<li>With one unpaired electron, [NF]* is <strong>paramagnetic<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>[NF]+:<\/strong>\n<ul class=\"wp-block-list\">\n<li>This ion has one fewer electron, which means an electron will be removed from one of the antibonding orbitals (likely \u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b or \u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b).<\/li>\n\n\n\n<li>With all electrons paired, [NF]+ is <strong>diamagnetic<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">(b) <strong>Qualitative MO Energy Level Diagram for Cyclo-B3+ (Triboron Cation)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Atomic Orbitals (AOs):<\/strong>\n<ul class=\"wp-block-list\">\n<li>Each boron atom has the electronic configuration 1s22s22p11s^2 2s^2 2p^11s22s22p1, so there are three 2p12p^12p1 orbitals from each boron atom.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>MO Diagram:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Since there are three boron atoms in cyclo-B3+, the three 2p12p^12p1 orbitals combine to form three molecular orbitals:\n<ul class=\"wp-block-list\">\n<li>One bonding \u03c02p\\pi_{2p}\u03c02p\u200b orbital (lower energy).<\/li>\n\n\n\n<li>One nonbonding \u03c02p\u2217\\pi^*_{2p}\u03c02p\u2217\u200b orbital (higher energy).<\/li>\n\n\n\n<li>One bonding \u03c32p\\sigma_{2p}\u03c32p\u200b orbital.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Electron Placement:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The total number of electrons in cyclo-B3+ is 5 (since each boron contributes 1 electron and there is a +1 charge).<\/li>\n\n\n\n<li>The 5 electrons fill the bonding \u03c02p\\pi_{2p}\u03c02p\u200b and \u03c32p\\sigma_{2p}\u03c32p\u200b molecular orbitals. Since the nonbonding orbital is empty, this makes the structure relatively stable.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Bonding and Geometry:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The bonding electrons form a molecular orbital that is symmetric about the center of the triangle.<\/li>\n\n\n\n<li>This results in a triangular molecular geometry with equal BB\u2212BBB-BBB\u2212B bond angles of 120 degrees.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">In summary, the bonding in cyclo-B3+ involves the overlap of the boron 2p2p2p orbitals to form both bonding and nonbonding molecular orbitals, with the structure adopting a triangular geometry due to the arrangement of the atoms in the molecule.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-80.jpeg\" alt=\"\" class=\"wp-image-45805\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-80.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-80-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-80-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-80-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Sketch and label a qualitative molecular orbital (MO) energy level diagram for nitrogen monofluoride, NF. (ii) Use your diagram to determine the bond order of NF, [NF]*, and [NF]. (iii) Use your diagram to determine whether each of NF, [NF]+, and [NF] is diamagnetic or paramagnetic. (b) (10 marks) The triboron cation (cyclo-B3+) has a [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-45804","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45804","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=45804"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45804\/revisions"}],"predecessor-version":[{"id":45806,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45804\/revisions\/45806"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45804"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45804"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45804"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}