Draw the Lewis structure for carbonyl fluoride, COF2. Describe this molecule’s electron-pair geometry and molecular geometry. What is the hybridisation of the carbon atom? What orbitals overlap to form o and a bonds? (20) 2p 7 2p 2p 2p 2p 2p 2p 2p 2p 2p ENERGY TV TIHTI Tap Tap The highest occupied molecular orbital (the HOMO). 11 025 25 11 25 carbon 025 oxygen orbitals carbon monoxide, CO orbitals Figure 1: The molecular orbitals of carbon monoxide and the HOMO of the molecule. (c) Assume that the energy level diagram for carbon monoxide shown in Figure 1 above can also be applied to the heteronuclear diatomic molecule chlorine monoxide, CIO. (i) Write the electron configuration for CIO (ii) What is the highest energy, occupied molecular orbital (HOMO)? (iii) is the molecule diamagnetic or paramagnetic? (iv) What is the net number of o and a bonds and what is the CIO bond order? (10) (d) Which of the homonuclear, diatomic molecules of second period elements (from Li2 to Ne2) are paramagnetic? Which have a bond order of 1? Which have a bond order of 2? Which diatomic molecule has the highest bond order? (10)<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n Let’s address each part of the question with correct answers followed by a 300-word detailed explanation.<\/p>\n\n\n\n (i) Electron configuration<\/strong>: (ii) HOMO<\/strong>: (iii) Magnetic character<\/strong>: (iv) Net \u03c3 and \u03c0 bonds<\/strong>: Carbonyl fluoride (COF\u2082) is a triatomic molecule composed of one carbon (C), one oxygen (O), and two fluorine (F) atoms. Carbon is the central atom. To draw the Lewis structure, we begin by counting valence electrons: carbon contributes 4, oxygen 6, and each fluorine 7, totaling 24 electrons. Carbon forms a double bond with oxygen and single bonds with each fluorine, completing the octet of all atoms. This arrangement gives COF\u2082 a trigonal planar<\/strong> shape due to three regions of electron density around the carbon atom. The bond angles are approximately 120\u00b0, and carbon is sp\u00b2 hybridized<\/strong>, forming sigma bonds using its sp\u00b2 orbitals and a \u03c0 bond via unhybridized 2p orbitals overlapping with oxygen’s 2p orbitals.<\/p>\n\n\n\n In part (c), chlorine monoxide (ClO) has 25 total valence electrons. Its molecular orbital diagram resembles that of CO. Electrons fill the molecular orbitals following Hund\u2019s rule and the Aufbau principle. The 25th electron enters a \u03c0*2p orbital, making it the HOMO<\/strong>. Because it contains an unpaired electron, ClO is paramagnetic<\/strong>. The bond order is calculated as: Finally, in part (d), among second-period diatomic molecules:<\/p>\n\n\n\n Draw the Lewis structure for carbonyl fluoride, COF2. Describe this molecule’s electron-pair geometry and molecular geometry. What is the hybridisation of the carbon atom? What orbitals overlap to form o and a bonds? (20) 2p 7 2p 2p 2p 2p 2p 2p 2p 2p 2p ENERGY TV TIHTI Tap Tap The highest occupied molecular orbital […]<\/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-215318","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215318","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=215318"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215318\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215318"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215318"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215318"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n(b) Carbonyl Fluoride, COF\u2082<\/strong><\/h3>\n\n\n\n
\n
..\n :F\u2013C(=O)\u2013F:\n ..<\/code><\/pre>\n\n\n\n\n
\n\n\n\n(c) Chlorine Monoxide, ClO<\/strong><\/h3>\n\n\n\n
Total electrons = Cl (17) + O (8) = 25 electrons<\/strong><\/p>\n\n\n\n
\u03c0*2p (degenerate antibonding orbitals)<\/p>\n\n\n\n
Paramagnetic<\/strong> (due to one unpaired electron)<\/p>\n\n\n\n
1 \u03c3 bond, 1 \u03c0 bond (but reduced by antibonding)
Bond order<\/strong> = \u00bd \u00d7 (bonding e\u207b \u2013 antibonding e\u207b) = 1.5<\/strong><\/p>\n\n\n\n
\n\n\n\n(d) Diatomic Molecules from Li\u2082 to Ne\u2082<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\n300-Word Explanation:<\/h3>\n\n\n\n
$\\text{Bond Order} = \\frac{1}{2}(13 – 10) = 1.5$<\/p>\n\n\n\n\n