{"id":221704,"date":"2025-05-30T12:44:08","date_gmt":"2025-05-30T12:44:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221704"},"modified":"2025-05-30T12:44:11","modified_gmt":"2025-05-30T12:44:11","slug":"molecular-models-data-sheet-for-each-formula-given-perform-the-following-in-the-chart-provided","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/molecular-models-data-sheet-for-each-formula-given-perform-the-following-in-the-chart-provided\/","title":{"rendered":"MOLECULAR MODELS DATA SHEET For each formula given, perform the following in the chart provided"},"content":{"rendered":"\n

MOLECULAR MODELS DATA SHEET For each formula given, perform the following in the chart provided: QUESTIONS ARE VERTICAL-WORK \u201cDOWN\u201d A COLUMN Formula DRAW THE LEWIS DOT STRUCTURE FOR EACH FORMULA INDICATED AND WRITE ANY RESONANCE STRUCTURES, IF APPLICABLE CHEMISTRY 121: CALCULATE FORMAL CHARGE FOR EACH ATOM IN YOUR STRUCTURE INDICATE THE ELECTRON PAIR GEOMETRY (AS PREDICTED BY VSEPR) ABOUT THE ATOM INDICATED AND INDICATE THE PREDICTED BOND ANGLES (AS PREDICTED BY VSEPR) ABOUT THE ATOM INDICATED INDICATE WHETHER THE MOLECULAR GEOMETRY IS THE SAME AS THE ELECTRON PAIR GEOMETRY OR DIFFERENT (JUST WRITE “SAME” OR “DIFFERENT”) Lewis Structure with any resonance structures VSEPR: Electron pair geometry AND bond angle about: Formal charge calculation only for Chemistry 121<\/em> Molecular geometry and bond angle about atom SAME or DIFFERENT than VSEPR? CS2 C CN C H3O+ O O CO3\u00b2 C Page 2 of 4<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here’s a completed Molecular Models Data Sheet<\/strong> for the compounds you listed (CS\u2082, CN\u207b, CH\u2083O\u207a, and CO\u2083\u00b2\u207b<\/strong>), followed by a 300-word explanation:<\/p>\n\n\n\n

\n\n\n\n

1. CS\u2082 (Carbon Disulfide)<\/strong><\/h3>\n\n\n\n
    \n
  • Lewis Structure & Resonance<\/strong>:
    :S\u2261C\u2261S: (linear, no resonance)<\/li>\n\n\n\n
  • VSEPR Geometry (central atom = C)<\/strong>:
    Electron pair geometry<\/strong>: Linear
    Bond angle<\/strong>: 180\u00b0<\/li>\n\n\n\n
  • Formal Charges<\/strong>:
    C = 0; Each S = 0<\/li>\n\n\n\n
  • Molecular Geometry<\/strong>: Linear
    Bond angle<\/strong>: 180\u00b0<\/li>\n\n\n\n
  • Same or Different?<\/strong> SAME<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    2. CN\u207b (Cyanide Ion)<\/strong><\/h3>\n\n\n\n
      \n
    • Lewis Structure & Resonance<\/strong>:
      \u207b:C\u2261N: (no resonance)<\/li>\n\n\n\n
    • VSEPR Geometry (central atom = C)<\/strong>:
      Electron pair geometry<\/strong>: Linear
      Bond angle<\/strong>: 180\u00b0<\/li>\n\n\n\n
    • Formal Charges<\/strong>:
      C = 0; N = \u20131<\/li>\n\n\n\n
    • Molecular Geometry<\/strong>: Linear
      Bond angle<\/strong>: 180\u00b0<\/li>\n\n\n\n
    • Same or Different?<\/strong> SAME<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      3. CH\u2083O\u207a (Methyloxonium Ion)<\/strong><\/h3>\n\n\n\n
        \n
      • Lewis Structure & Resonance<\/strong>:
        H
        |
        H\u2013C\u2013O\u207a\u2013H<\/li>\n\n\n\n
      • VSEPR Geometry (central atom = O)<\/strong>:
        Electron pair geometry<\/strong>: Tetrahedral
        Bond angle<\/strong>: ~109.5\u00b0<\/li>\n\n\n\n
      • Formal Charges<\/strong>:
        C = 0; O = +1; H = 0<\/li>\n\n\n\n
      • Molecular Geometry<\/strong>: Trigonal pyramidal
        Bond angle<\/strong>: ~107\u00b0<\/li>\n\n\n\n
      • Same or Different?<\/strong> DIFFERENT<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        4. CO\u2083\u00b2\u207b (Carbonate Ion)<\/strong><\/h3>\n\n\n\n
          \n
        • Lewis Structure & Resonance<\/strong>:
          O
          ||
          C \u2013 O\u207b \u2194 O\u207b \u2013 C = O \u2194 O = C \u2013 O\u207b (3 resonance structures)<\/li>\n\n\n\n
        • VSEPR Geometry (central atom = C)<\/strong>:
          Electron pair geometry<\/strong>: Trigonal planar
          Bond angle<\/strong>: 120\u00b0<\/li>\n\n\n\n
        • Formal Charges<\/strong>:
          Double-bond O = 0; Single-bond O = \u20131; C = 0<\/li>\n\n\n\n
        • Molecular Geometry<\/strong>: Trigonal planar
          Bond angle<\/strong>: 120\u00b0<\/li>\n\n\n\n
        • Same or Different?<\/strong> SAME<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Explanation<\/strong><\/h3>\n\n\n\n

          This molecular modeling exercise involves analyzing various compounds by drawing Lewis structures, determining electron and molecular geometries via VSEPR theory, and calculating formal charges. This method helps visualize bonding and predict molecular shapes and behaviors.<\/p>\n\n\n\n

          CS\u2082<\/strong> is linear with a central carbon double bonded to two sulfurs. No lone pairs on carbon mean the electron pair geometry is linear<\/strong>, matching the molecular geometry<\/strong> with 180\u00b0<\/strong> bond angles. The formal charges are all zero, indicating a stable structure.<\/p>\n\n\n\n

          CN\u207b<\/strong> has a triple bond between carbon and nitrogen. As a diatomic molecule, it\u2019s linear<\/strong> with 180\u00b0<\/strong> bond angle. The negative charge resides on nitrogen, as it\u2019s more electronegative and stabilizes the extra electron, giving N a \u20131 formal charge.<\/p>\n\n\n\n

          CH\u2083O\u207a<\/strong> features a positively charged oxygen bonded to three atoms (two Hs and one carbon). The oxygen has one lone pair, making its electron geometry tetrahedral<\/strong>, but the molecular geometry is trigonal pyramidal<\/strong> due to the lone pair repelling bonding pairs slightly, reducing bond angles to about 107\u00b0<\/strong>. The difference<\/strong> between electron and molecular geometry arises due to lone pairs.<\/p>\n\n\n\n

          CO\u2083\u00b2\u207b<\/strong> is a classic example of resonance<\/strong>, where the negative charge is delocalized over three oxygen atoms, equalizing the C\u2013O bond lengths. The carbon is bonded to three regions of electron density, resulting in trigonal planar<\/strong> geometry with 120\u00b0<\/strong> angles. All resonance forms contribute equally, and formal charges are well-distributed (each single-bonded O carries a \u20131).<\/p>\n\n\n\n

          These models show how molecular shapes and electron arrangements influence polarity, reactivity, and intermolecular interactions\u2014crucial in predicting chemical behavior<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          .<\/p>\n","protected":false},"excerpt":{"rendered":"

          MOLECULAR MODELS DATA SHEET For each formula given, perform the following in the chart provided: QUESTIONS ARE VERTICAL-WORK \u201cDOWN\u201d A COLUMN Formula DRAW THE LEWIS DOT STRUCTURE FOR EACH FORMULA INDICATED AND WRITE ANY RESONANCE STRUCTURES, IF APPLICABLE CHEMISTRY 121: CALCULATE FORMAL CHARGE FOR EACH ATOM IN YOUR STRUCTURE INDICATE THE ELECTRON PAIR GEOMETRY (AS […]<\/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-221704","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221704","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=221704"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221704\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221704"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221704"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221704"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}