{"id":26200,"date":"2025-06-19T11:02:39","date_gmt":"2025-06-19T11:02:39","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26200"},"modified":"2025-06-19T11:02:41","modified_gmt":"2025-06-19T11:02:41","slug":"draw-a-molecular-orbital-energy-level-diagram-for-nf-and-write-down-the-corresponding-electronic-configuration","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-a-molecular-orbital-energy-level-diagram-for-nf-and-write-down-the-corresponding-electronic-configuration\/","title":{"rendered":"Draw a molecular orbital energy level diagram for NF and write down the corresponding electronic configuration."},"content":{"rendered":"\n
) Draw a molecular orbital energy level diagram for NF and write down the corresponding electronic configuration. In the diagram, be sure to label each molecular orbital with its name. The ionization energies of N and F are 14.5 and 17.4 eV. Note that NF is isoelectronic with O . b) Next to the corresponding energy level, draw a picture of the LUMO. c) In the HOMO, the electron density resides more on the (circle the correct answer) N atom F atom neither, since it’s a non-bonding orbital neither, since the electron density is equally shared by N and F d) Write down the LCAO-MO wavefunction for the LUMO e) What is the bond order of NF? f) Circle the molecule(s) expected to be paramagnetic: NF NF NF none of them g) Circle the molecule(s) expected to have the shortest bond: NF NF NF h) With an arrow, indicate in the energy level diagram the lowest energy spin-allowed transition in NF. i) On the same graph, plot molecular potential energy diagrams for NF and NF . Label the axes. Pick one of the curves and indicate R and D for that curve.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let\u2019s work through the NF molecule step by step using molecular orbital (MO) theory. NF is a heteronuclear diatomic molecule with 14 electrons (same as O\u2082), so its MO diagram is similar to O\u2082, but with modifications due to differences in electronegativity and ionization energies (F has a higher electronegativity and ionization energy than N).<\/p>\n\n\n\n
\n\n\n\n
(a) MO Diagram and Electronic Configuration<\/strong><\/h3>\n\n\n\n
Since NF is isoelectronic with O\u2082, the molecular orbitals fill as follows (energy order based on O\u2082-like molecules): \u03c31s<\u03c31s\u2217<\u03c32s<\u03c32s\u2217<\u03c32pz<\u03c02px=\u03c02py<\u03c02px\u2217=\u03c02py\u2217<\u03c32pz\u2217\\sigma_{1s} < \\sigma_{1s}^* < \\sigma_{2s} < \\sigma_{2s}^* < \\sigma_{2p_z} < \\pi_{2p_x} = \\pi_{2p_y} < \\pi_{2p_x}^* = \\pi_{2p_y}^* < \\sigma_{2p_z}^*\u03c31s\u200b<\u03c31s\u2217\u200b<\u03c32s\u200b<\u03c32s\u2217\u200b<\u03c32pz\u200b\u200b<\u03c02px\u200b\u200b=\u03c02py\u200b\u200b<\u03c02px\u200b\u2217\u200b=\u03c02py\u200b\u2217\u200b<\u03c32pz\u200b\u2217\u200b<\/p>\n\n\n\n
However, due to the heteronuclear nature and F being more electronegative, the atomic orbitals from F are lower in energy. Thus, the MOs have more F character at the lower levels and more N character at the higher levels.<\/p>\n\n\n\n
The LUMO<\/strong> (Lowest Unoccupied Molecular Orbital) is the next orbital after the highest filled one. Since the last filled orbitals are the \u03c0*(2p), the LUMO is:<\/p>\n\n\n\n
\u03c3*(2p_z)<\/strong><\/p>\n\n\n\n
A drawing of this would show a large anti-bonding lobe along the internuclear axis with a nodal plane between the atoms.<\/p>\n\n\n\n
\n\n\n\n
(c) Electron Density in HOMO<\/strong><\/h3>\n\n\n\n
The HOMO is the \u03c0*(2p)<\/strong> orbital.<\/p>\n\n\n\n
This orbital is non-bonding<\/strong>, but since N is less electronegative than F, the orbital has more electron density on the N atom<\/strong>.<\/p>\n\n\n\n
Answer: N atom<\/strong><\/p>\n\n\n\n\n\n\n\n
(d) LCAO-MO Wavefunction for LUMO<\/strong><\/h3>\n\n\n\n
The LUMO is \u03c3*(2p_z)<\/strong>: \u03c8\u03c3\u2217=cA\u22c52pz(N)\u2212cB\u22c52pz(F)\\psi_{\\sigma^*} = c_A \\cdot 2p_z(N) – c_B \\cdot 2p_z(F)\u03c8\u03c3\u2217\u200b=cA\u200b\u22c52pz\u200b(N)\u2212cB\u200b\u22c52pz\u200b(F)<\/p>\n\n\n\n
Where cAc_AcA\u200b and cBc_BcB\u200b are coefficients reflecting the atomic orbital contributions. The minus sign reflects antibonding character.<\/p>\n\n\n\n
\n\n\n\n
(e) Bond Order of NF<\/strong><\/h3>\n\n\n\n
Bond order = (bonding electrons \u2212 antibonding electrons) \u00f7 2 = (8 \u2212 4) \u00f7 2 = 2<\/strong><\/p>\n\n\n\n\n\n\n\n
(f) Paramagnetic Species<\/strong><\/h3>\n\n\n\n
Look for unpaired electrons:<\/p>\n\n\n\n
\n
NF:<\/strong> 14 electrons \u2192 \u03c0*(2p) filled \u2192 no unpaired electrons<\/strong><\/li>\n\n\n\n
NF\u207a:<\/strong> 13 electrons \u2192 1 less in \u03c0*(2p) \u2192 1 unpaired electron<\/strong> \u2192 paramagnetic<\/strong><\/li>\n<\/ul>\n\n\n\n
Answer: NF\u207b and NF\u207a<\/strong><\/p>\n\n\n\n\n\n\n\n
(g) Shortest Bond<\/strong><\/h3>\n\n\n\n
Bond order correlates with bond length.<\/p>\n\n\n\n
\n
NF\u207a has bond order 2.5<\/li>\n\n\n\n
NF has bond order 2<\/li>\n\n\n\n
NF\u207b has bond order 1.5<\/li>\n<\/ul>\n\n\n\n
Answer: NF\u207a<\/strong><\/p>\n\n\n\n\n\n\n\n
(h) Lowest Energy Spin-Allowed Transition<\/strong><\/h3>\n\n\n\n
From HOMO (\u03c0*(2p)) to LUMO (\u03c3*(2p)) Draw an arrow<\/strong> from \u03c0*(2p) to \u03c3*(2p) in the MO diagram.<\/p>\n\n\n\n\n\n\n\n
(i) Potential Energy Curves<\/strong><\/h3>\n\n\n\n
Sketch a graph with:<\/p>\n\n\n\n
\n
x-axis: Internuclear distance RRR<\/li>\n\n\n\n
y-axis: Potential energy<\/li>\n<\/ul>\n\n\n\n
NF\u207a: Steep curve, minimum at lower R (short bond), deeper well (stronger bond) NF\u207b: Broad curve, minimum at higher R, shallower well<\/p>\n\n\n\n
Label one curve<\/strong> with:<\/p>\n\n\n\n
\n
ReR_eRe\u200b: equilibrium bond length (at minimum)<\/li>\n\n\n\n
DeD_eDe\u200b: bond dissociation energy (depth of well)<\/li>\n<\/ul>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
) Draw a molecular orbital energy level diagram for NF and write down the corresponding electronic configuration. In the diagram, be sure to label each molecular orbital with its name. The ionization energies of N and F are 14.5 and 17.4 eV. Note that NF is isoelectronic with O. b) Next to the corresponding energy […]<\/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-26200","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26200","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=26200"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26200\/revisions"}],"predecessor-version":[{"id":26206,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26200\/revisions\/26206"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26200"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26200"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"