{"id":248073,"date":"2025-07-08T06:17:57","date_gmt":"2025-07-08T06:17:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248073"},"modified":"2025-07-08T06:17:59","modified_gmt":"2025-07-08T06:17:59","slug":"lewis-structures-of-two-chemical-species-are-shown-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/lewis-structures-of-two-chemical-species-are-shown-below\/","title":{"rendered":"Lewis structures of two chemical species are shown below."},"content":{"rendered":"\n<p>Lewis structures of two chemical species are shown below. For each, give the formal charge on each atom and (ii) the overall charge on each species. Structure a: Atom 1: C Atom 2: I Atom 3: O Atom 4: F Atom 5: S Structure b: Atom 1: O Atom 2: b Atom 3: O Atom 4: c Which one of the following molecules or ions has one lone pair on the central atom? (If there are 3 elements in the molecule, the first atom is the central atom and the others are bonded to it). Br2 IO2F SF IF3 CIO Draw two equivalent resonance forms for bicarbonate ion, HCO3-. How many sigma bonds are there? How many pi bonds?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-289.png\" alt=\"\" class=\"wp-image-248074\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the correct answers to the chemistry questions.<\/p>\n\n\n\n<p><strong>3. Formal Charges and Overall Charge<\/strong><\/p>\n\n\n\n<p>The formal charge on an atom in a Lewis structure is calculated using the formula: Formal Charge = (Valence Electrons) &#8211; (Non-bonding Electrons) &#8211; (\u00bd \u00d7 Bonding Electrons). The overall charge of the species is the sum of the formal charges on all its atoms.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Structure a (N\u2082O\u2083)<\/strong>: The structure shown is O\u2090=N\u2090\u2212N\u266d(=O\u266d)\u2212O\ua700.\n<ul class=\"wp-block-list\">\n<li><strong>N\u2090<\/strong>: Has 5 valence electrons. It has 1 lone pair (2 non-bonding electrons) and 3 bonds (6 bonding electrons). Formal Charge = 5 \u2212 2 \u2212 (\u00bd \u00d7 6) = 0.<\/li>\n\n\n\n<li><strong>N\u266d<\/strong>: Has 5 valence electrons. It has no lone pairs (0 non-bonding electrons) and 4 bonds (8 bonding electrons). Formal Charge = 5 \u2212 0 \u2212 (\u00bd \u00d7 8) = +1.<\/li>\n\n\n\n<li><strong>O\u2090<\/strong>: Has 6 valence electrons. It has 2 lone pairs (4 non-bonding electrons) and a double bond (4 bonding electrons). Formal Charge = 6 \u2212 4 \u2212 (\u00bd \u00d7 4) = 0.<\/li>\n\n\n\n<li><strong>O\u266d<\/strong>: Has 6 valence electrons. It has 2 lone pairs (4 non-bonding electrons) and a double bond (4 bonding electrons). Formal Charge = 6 \u2212 4 \u2212 (\u00bd \u00d7 4) = 0.<\/li>\n\n\n\n<li><strong>O\ua700<\/strong>: Has 6 valence electrons. It has 3 lone pairs (6 non-bonding electrons) and a single bond (2 bonding electrons). Formal Charge = 6 \u2212 6 \u2212 (\u00bd \u00d7 2) = -1.<\/li>\n\n\n\n<li><strong>(ii) Overall Charge on Structure a<\/strong>: The sum of the formal charges is 0 + (+1) + 0 + 0 + (-1) = 0.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Structure b ([ClO\u2082F\u2082]\u207a)<\/strong>: Assuming the central atom is Cl as listed, not C as drawn.\n<ul class=\"wp-block-list\">\n<li><strong>Cl<\/strong>: Has 7 valence electrons. It has no lone pairs (0 non-bonding electrons) and 4 single bonds (8 bonding electrons). Formal Charge = 7 \u2212 0 \u2212 (\u00bd \u00d7 8) = +3.<\/li>\n\n\n\n<li><strong>O\u2090 and O\u266d<\/strong>: Each has 6 valence electrons, 3 lone pairs (6 non-bonding electrons), and 1 single bond (2 bonding electrons). Formal Charge = 6 \u2212 6 \u2212 (\u00bd \u00d7 2) = -1.<\/li>\n\n\n\n<li><strong>F\u2090 and F\u266d<\/strong>: Each has 7 valence electrons, 3 lone pairs (6 non-bonding electrons), and 1 single bond (2 bonding electrons). Formal Charge = 7 \u2212 6 \u2212 (\u00bd \u00d7 2) = 0.<\/li>\n\n\n\n<li><strong>(ii) Overall Charge on Structure b<\/strong>: The sum of the formal charges is (+3) + (-1) + (-1) + 0 + 0 = +1.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>4. Molecule or Ion with One Lone Pair on the Central Atom<\/strong><\/p>\n\n\n\n<p>The molecule with one lone pair on the central atom is&nbsp;<strong>IO\u2082F<\/strong>.<\/p>\n\n\n\n<p>To determine this, we find the number of lone pairs on the central atom for each species. The central atom for IO\u2082F is Iodine (I).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total valence electrons in IO\u2082F = 7 (from I) + 2\u00d76 (from O) + 7 (from F) = 26 electrons.<\/li>\n\n\n\n<li>Iodine is bonded to three other atoms (two O, one F). A stable Lewis structure that minimizes formal charges will have two double bonds to oxygen and one single bond to fluorine (O=I(=O)\u2212F).<\/li>\n\n\n\n<li>In this structure, Iodine forms 5 bonds (using 10 electrons) and has a complete valence shell of 12 electrons (an expanded octet, which is common for elements in period 3 and below).<\/li>\n\n\n\n<li>The number of electrons on the central iodine atom is 12. Ten are bonding electrons. The remaining 12 \u2212 10 = 2 electrons form\u00a0<strong>one lone pair<\/strong>.<\/li>\n\n\n\n<li>The other options do not have one lone pair on the central atom: Br\u2082O has two lone pairs on the central oxygen; SF\u2086 and ClO\u2084\u207b have zero lone pairs; and IF\u2083 has two lone pairs.<\/li>\n<\/ul>\n\n\n\n<p><strong>5. Resonance Forms and Bonding in Bicarbonate Ion (HCO\u2083\u207b)<\/strong><\/p>\n\n\n\n<p>The bicarbonate ion has two equivalent resonance forms. In both structures, the carbon atom is central. It is single-bonded to an OH group and has one single bond and one double bond to the other two oxygen atoms. The negative charge is delocalized between these two oxygen atoms.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Resonance Form 1<\/strong>: Carbon is double-bonded to the &#8220;top&#8221; oxygen and single-bonded to the &#8220;right&#8221; oxygen, which holds the negative charge.<\/li>\n\n\n\n<li><strong>Resonance Form 2<\/strong>: Carbon is single-bonded to the &#8220;top&#8221; oxygen (which now holds the negative charge) and double-bonded to the &#8220;right&#8221; oxygen.<\/li>\n\n\n\n<li><strong>Sigma (\u03c3) and Pi (\u03c0) Bonds<\/strong>: We can count the bonds in either resonance structure.\n<ul class=\"wp-block-list\">\n<li>There is one sigma bond between H and O.<\/li>\n\n\n\n<li>There is one sigma bond between O and C.<\/li>\n\n\n\n<li>There is one sigma bond in the C-O single bond.<\/li>\n\n\n\n<li>There is one sigma bond within the C=O double bond.<\/li>\n\n\n\n<li>This gives a total of\u00a0<strong>4 sigma bonds<\/strong>.<\/li>\n\n\n\n<li>There is one pi bond within the C=O double bond.<\/li>\n\n\n\n<li>This gives a total of\u00a0<strong>1 pi bond<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-142.jpeg\" alt=\"\" class=\"wp-image-248075\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Lewis structures of two chemical species are shown below. For each, give the formal charge on each atom and (ii) the overall charge on each species. Structure a: Atom 1: C Atom 2: I Atom 3: O Atom 4: F Atom 5: S Structure b: Atom 1: O Atom 2: b Atom 3: O Atom [&hellip;]<\/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-248073","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248073","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=248073"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248073\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248073"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248073"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248073"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}