{"id":23766,"date":"2025-06-17T21:27:54","date_gmt":"2025-06-17T21:27:54","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=23766"},"modified":"2025-06-17T21:27:56","modified_gmt":"2025-06-17T21:27:56","slug":"draw-the-lewis-structure-for-icl3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-for-icl3\/","title":{"rendered":"Draw the Lewis structure for ICl3"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Draw the Lewis structure for ICl3. What is the hybridization on the I (iodine) atom?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Correct Answer:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure for ICl\u2083<\/strong>:<br>Iodine (I) is the central atom, bonded to three chlorine (Cl) atoms, with two lone pairs on iodine.<\/li>\n\n\n\n<li><strong>Hybridization of iodine in ICl\u2083<\/strong>:<br><strong>sp\u00b3d<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 words):<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 1: Count the total valence electrons<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Iodine (I) has 7 valence electrons.<\/li>\n\n\n\n<li>Each chlorine (Cl) also has 7 valence electrons.<\/li>\n\n\n\n<li>Total valence electrons:<br>7(I)+3\u00d77(Cl)=287 (I) + 3 \\times 7 (Cl) = 28 electrons<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 2: Draw the skeleton structure<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Place iodine in the center with three single bonds to chlorine atoms:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   Cl\n    |\nCl\u2013I\u2013Cl\n<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">This uses up 3\u00d72=63 \\times 2 = 6 electrons for bonding.<br>Remaining electrons: 28\u22126=2228 &#8211; 6 = 22<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 3: Complete octets of chlorine<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Each Cl atom needs 6 more electrons to complete an octet:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 Cl \u00d7 6 electrons = 18 electrons<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Remaining electrons: 22\u221218=422 &#8211; 18 = 4 electrons<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 4: Place remaining electrons on iodine<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Place the remaining 4 electrons (2 lone pairs) on iodine.<br>Now iodine has:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 bonding pairs (from Cl)<\/li>\n\n\n\n<li>2 lone pairs<br>This gives <strong>5 regions of electron density<\/strong>, meaning <strong>trigonal bipyramidal electron geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Step 5: Determine hybridization<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The number of electron regions (bonding + lone pairs) around iodine is 5 \u2192 hybridization is <strong>sp\u00b3d<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Molecular Shape:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Due to two lone pairs on iodine, the actual molecular shape of ICl\u2083 is <strong>T-shaped<\/strong>, not trigonal bipyramidal.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Summary:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: Iodine in center, three Cl atoms bonded, two lone pairs on I.<\/li>\n\n\n\n<li><strong>Electron regions<\/strong>: 5 (3 bonds + 2 lone pairs)<\/li>\n\n\n\n<li><strong>Hybridization<\/strong>: <strong>sp\u00b3d<\/strong><\/li>\n\n\n\n<li><strong>Molecular geometry<\/strong>: <strong>T-shaped<\/strong> due to lone pairs causing electron repulsion.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for ICl3. What is the hybridization on the I (iodine) atom? The correct answer and explanation is: Correct Answer: Explanation (300 words): Step 1: Count the total valence electrons Step 2: Draw the skeleton structure Place iodine in the center with three single bonds to chlorine atoms: This uses up 3\u00d72=63 [&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-23766","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23766","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=23766"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23766\/revisions"}],"predecessor-version":[{"id":23767,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23766\/revisions\/23767"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23766"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23766"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23766"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}