{"id":47020,"date":"2025-07-02T09:48:40","date_gmt":"2025-07-02T09:48:40","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47020"},"modified":"2025-07-02T09:48:42","modified_gmt":"2025-07-02T09:48:42","slug":"in-the-space-below-draw-the-correct-lewis-structure-for-the-molecule-triiodide-i3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/in-the-space-below-draw-the-correct-lewis-structure-for-the-molecule-triiodide-i3\/","title":{"rendered":"In the space below, draw the correct Lewis Structure for the molecule, triiodide, I3-"},"content":{"rendered":"\n<p>In the space below, draw the correct Lewis Structure for the molecule, triiodide, I3-. Fill in the blanks on the right as well: Show both your initial trial structure and your modified final structure: Initial ne should reflect your initial trial structure: ae initial ne Hybridization electron geometry molecular geometry<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"700\" height=\"404\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-106.png\" alt=\"\" class=\"wp-image-47021\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-106.png 700w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-106-300x173.png 300w\" sizes=\"auto, (max-width: 700px) 100vw, 700px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the completed answers for the question about the triiodide ion (I\u2083\u207b).<\/p>\n\n\n\n<p><strong>Drawings:<\/strong><\/p>\n\n\n\n<p><strong>Initial Trial Structure:<\/strong><br>This structure represents the initial connectivity of the atoms before the distribution of all valence electrons.<br>I \u2014 I \u2014 I<\/p>\n\n\n\n<p><strong>Modified Final Structure:<\/strong><br>This is the correct and most stable Lewis structure, showing all valence electrons as lone pairs or bonding pairs.Generated code<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>..       ..       ..\n&#91; :\u00cf \u2014  \u00cf  \u2014 :\u00cf: ]\u207b\n  ..     |     ..\n         ..<\/code><\/pre>\n\n\n\n<p><em>In the final structure, each terminal iodine atom has 3 lone pairs, and the central iodine atom has 3 lone pairs.<\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Filled Blanks:<\/strong><\/p>\n\n\n\n<p><strong>ae =<\/strong>&nbsp;22<\/p>\n\n\n\n<p><strong>initial ne =<\/strong>&nbsp;24<\/p>\n\n\n\n<p><strong>Hybridization =<\/strong>&nbsp;sp\u00b3d<\/p>\n\n\n\n<p><strong>electron geometry =<\/strong>&nbsp;trigonal bipyramidal<\/p>\n\n\n\n<p><strong>molecular geometry =<\/strong>&nbsp;linear<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The correct Lewis structure and properties of the triiodide ion (I\u2083\u207b) are determined through a systematic process.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Available Electrons (ae):<\/strong>\u00a0First, we calculate the total number of valence electrons available for bonding. Each iodine (I) atom, being in Group 17, has 7 valence electrons. Since there are three iodine atoms, this gives us 3 x 7 = 21 electrons. The ion has a negative one charge, which means we add one extra electron. Therefore, the total available electrons (ae) are 21 + 1 = 22.<\/li>\n\n\n\n<li><strong>Initial Needed Electrons (initial ne):<\/strong>\u00a0For the initial trial structure, we assume that all atoms will obey the octet rule to achieve stability. Each of the three iodine atoms would need 8 electrons. Thus, the total needed electrons (initial ne) would be 3 x 8 = 24. The discrepancy between the needed (24) and available (22) electrons suggests that the simple octet rule will be violated, and one atom must have an expanded octet.<\/li>\n\n\n\n<li><strong>Lewis Structure Construction:<\/strong>\u00a0We start by connecting the three iodine atoms with single bonds (I-I-I), using 4 of the 22 available electrons. We then distribute the remaining 18 electrons as lone pairs, starting with the outer (terminal) atoms. Each terminal iodine receives 6 electrons (3 lone pairs) to complete its octet. This uses 12 more electrons. The final 6 electrons are placed on the central iodine atom as 3 lone pairs. In this final structure, the central iodine has 2 bonding pairs and 3 lone pairs, for a total of 10 valence electrons (an expanded octet, which is permissible for elements in period 3 and below). The formal charge on the central atom is -1, while the terminal atoms are neutral, which is the most stable arrangement.<\/li>\n\n\n\n<li><strong>VSEPR Theory and Geometry:<\/strong>\u00a0The central iodine atom has 5 electron domains (2 bonding pairs + 3 lone pairs), giving it a steric number of 5. This corresponds to a\u00a0<strong>trigonal bipyramidal electron geometry<\/strong>. According to VSEPR theory, to minimize repulsion, the three lone pairs occupy the equatorial positions, while the two bonding pairs occupy the axial positions. This arrangement results in a\u00a0<strong>linear molecular geometry<\/strong>, with a bond angle of 180\u00b0.<\/li>\n\n\n\n<li><strong>Hybridization:<\/strong>\u00a0The hybridization of the central atom is determined by its number of electron domains. For 5 electron domains, the atom utilizes one s, three p, and one d orbital, resulting in\u00a0<strong>sp\u00b3d<\/strong>\u00a0hybridization.<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-207.jpeg\" alt=\"\" class=\"wp-image-47026\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-207.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-207-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-207-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-207-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>In the space below, draw the correct Lewis Structure for the molecule, triiodide, I3-. Fill in the blanks on the right as well: Show both your initial trial structure and your modified final structure: Initial ne should reflect your initial trial structure: ae initial ne Hybridization electron geometry molecular geometry The Correct Answer and Explanation [&hellip;]<\/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-47020","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47020","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=47020"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47020\/revisions"}],"predecessor-version":[{"id":47027,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47020\/revisions\/47027"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}