{"id":36157,"date":"2025-06-25T03:50:22","date_gmt":"2025-06-25T03:50:22","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36157"},"modified":"2025-06-25T03:50:25","modified_gmt":"2025-06-25T03:50:25","slug":"consider-each-of-the-following-molecules-and-or-ions","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/consider-each-of-the-following-molecules-and-or-ions\/","title":{"rendered":"Consider each of the following molecules and\/or ions."},"content":{"rendered":"\n

Consider each of the following molecules and\/or ions. Give the hybridization for the central atom in each case, and also give the molecular geometry. Ignore any distortions in the geometry due to non-equivalent terminal atoms. molecule or ion central atom hybridization for central atom molecular geometry (a) NOF N (b)
C (c)
I (d)
C (e) COS C<\/p>\n\n\n\n

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

Here are the hybridizations and molecular geometries for each of the given molecules and ions:<\/p>\n\n\n\n

Molecule or Ion<\/th>Central Atom<\/th>Hybridization<\/th>Molecular Geometry<\/th><\/tr><\/thead>
(a) NOF<\/td>N<\/td>sp\u00b2<\/td>Bent<\/td><\/tr>
(b) CH\u2083\u207b<\/td>C<\/td>sp\u00b3<\/td>Trigonal pyramidal<\/td><\/tr>
(c) IO\u2084\u207b<\/td>I<\/td>sp\u00b3<\/td>Tetrahedral<\/td><\/tr>
(d) CH\u2083\u207a<\/td>C<\/td>sp\u00b2<\/td>Trigonal planar<\/td><\/tr>
(e) COS<\/td>C<\/td>sp<\/td>Linear<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

To determine the hybridization and molecular geometry of a molecule, we look at the number of electron domains (bonding and lone pairs) around the central atom. This leads us to the electron domain geometry, from which we derive the molecular geometry. Hybridization corresponds to the number of electron domains:<\/p>\n\n\n\n

    \n
  • 2 domains<\/strong> \u2192 sp<\/li>\n\n\n\n
  • 3 domains<\/strong> \u2192 sp\u00b2<\/li>\n\n\n\n
  • 4 domains<\/strong> \u2192 sp\u00b3<\/li>\n<\/ul>\n\n\n\n

    (a) NOF<\/strong>: Nitrogen is bonded to oxygen and fluorine and has one lone pair, giving it three electron domains. Hence, hybridization is sp\u00b2<\/strong>, and the molecular geometry is bent<\/strong>.<\/p>\n\n\n\n

    (b) CH\u2083\u207b<\/strong>: The carbon has three single bonds with hydrogen and one lone pair, resulting in four electron domains. This leads to sp\u00b3<\/strong> hybridization and a trigonal pyramidal<\/strong> geometry.<\/p>\n\n\n\n

    (c) IO\u2084\u207b<\/strong>: Iodine is bonded to four oxygen atoms and has no lone pairs, giving it four electron domains. This results in sp\u00b3<\/strong> hybridization and a tetrahedral<\/strong> geometry.<\/p>\n\n\n\n

    (d) CH\u2083\u207a<\/strong>: Carbon forms three bonds with hydrogen atoms and has no lone pair, totaling three electron domains. Therefore, the hybridization is sp\u00b2<\/strong>, and the geometry is trigonal planar<\/strong>.<\/p>\n\n\n\n

    (e) COS<\/strong>: Carbon is doubly bonded to both oxygen and sulfur, forming two regions of electron density. This leads to sp<\/strong> hybridization and a linear<\/strong> molecular geometry.<\/p>\n\n\n\n

    By following the number of electron regions and applying VSEPR theory, one can systematically deduce both the hybridization and geometry for each case.<\/p>\n\n\n\n

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

    Consider each of the following molecules and\/or ions. Give the hybridization for the central atom in each case, and also give the molecular geometry. Ignore any distortions in the geometry due to non-equivalent terminal atoms. molecule or ion central atom hybridization for central atom molecular geometry (a) NOF N (b)C (c)I (d)C (e) COS C […]<\/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-36157","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36157","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=36157"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36157\/revisions"}],"predecessor-version":[{"id":36159,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36157\/revisions\/36159"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}