{"id":24248,"date":"2025-06-18T11:55:56","date_gmt":"2025-06-18T11:55:56","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=24248"},"modified":"2025-06-18T11:55:58","modified_gmt":"2025-06-18T11:55:58","slug":"give-the-number-of-lone-pairs-around-the-central-atom-and-the-molecular-geometry-of-ibr5","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/give-the-number-of-lone-pairs-around-the-central-atom-and-the-molecular-geometry-of-ibr5\/","title":{"rendered":"Give the number of lone pairs around the central atom and the molecular geometry of IBr5"},"content":{"rendered":"\n<p>Give the number of lone pairs around the central atom and the molecular geometry of IBr5.<\/p>\n\n\n\n<p><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<p><strong>Correct Answer:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of lone pairs around the central atom (Iodine):<\/strong> 1 lone pair<\/li>\n\n\n\n<li><strong>Molecular geometry of IBr\u2085:<\/strong> Square pyramidal<\/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:<\/strong><\/h3>\n\n\n\n<p><strong>Step 1: Determine the total number of valence electrons<\/strong><\/p>\n\n\n\n<p>Iodine (I) belongs to group 17 and has <strong>7 valence electrons<\/strong>.<br>Each bromine (Br) atom also has 7 valence electrons.<br>IBr\u2085 has one iodine atom and five bromine atoms: Total&nbsp;valence&nbsp;electrons=7&nbsp;(from&nbsp;I)+5\u00d77&nbsp;(from&nbsp;Br)=7+35=42&nbsp;valence&nbsp;electrons\\text{Total valence electrons} = 7\\ (\\text{from I}) + 5 \\times 7\\ (\\text{from Br}) = 7 + 35 = 42\\ \\text{valence electrons}<\/p>\n\n\n\n<p><strong>Step 2: Choose the central atom<\/strong><\/p>\n\n\n\n<p>Iodine is less electronegative than bromine and can expand its octet (since it\u2019s in period 5), so it will be the central atom.<\/p>\n\n\n\n<p><strong>Step 3: Form bonds<\/strong><\/p>\n\n\n\n<p>Five Br atoms form single bonds with iodine, using <strong>2 electrons per bond<\/strong>, for a total of: 5\u00d72=10&nbsp;electrons5 \\times 2 = 10\\ \\text{electrons}<\/p>\n\n\n\n<p>Subtract these from the total: 42\u221210=32&nbsp;electrons&nbsp;left42 &#8211; 10 = 32\\ \\text{electrons left}<\/p>\n\n\n\n<p><strong>Step 4: Distribute remaining electrons<\/strong><\/p>\n\n\n\n<p>Each Br needs 6 more electrons (to complete their octets), and with 5 Br atoms: 5\u00d76=30&nbsp;electrons&nbsp;used&nbsp;for&nbsp;Br&nbsp;lone&nbsp;pairs5 \\times 6 = 30\\ \\text{electrons used for Br lone pairs}<\/p>\n\n\n\n<p>Now: 32\u221230=2&nbsp;electrons&nbsp;left32 &#8211; 30 = 2\\ \\text{electrons left}<\/p>\n\n\n\n<p>These remaining 2 electrons (1 lone pair) go on the central iodine atom.<\/p>\n\n\n\n<p><strong>Step 5: Determine electron geometry and molecular geometry<\/strong><\/p>\n\n\n\n<p>With <strong>5 bonding pairs<\/strong> and <strong>1 lone pair<\/strong> on iodine (6 total regions of electron density), the <strong>electron geometry<\/strong> is <strong>octahedral<\/strong>.<\/p>\n\n\n\n<p>However, the <strong>molecular geometry<\/strong> (which considers only atoms, not lone pairs) is <strong>square pyramidal<\/strong> \u2014 4 atoms form a square base around iodine, 1 atom is above the central atom, and the lone pair is opposite to that.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lone pairs on iodine:<\/strong> 1<\/li>\n\n\n\n<li><strong>Molecular geometry:<\/strong> Square pyramidal (AX\u2085E, based on VSEPR theory)<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Give the number of lone pairs around the central atom and the molecular geometry of IBr5. The correct answer and explanation is: Correct Answer: Explanation: Step 1: Determine the total number of valence electrons Iodine (I) belongs to group 17 and has 7 valence electrons.Each bromine (Br) atom also has 7 valence electrons.IBr\u2085 has one [&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-24248","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24248","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=24248"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24248\/revisions"}],"predecessor-version":[{"id":24249,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24248\/revisions\/24249"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=24248"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=24248"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=24248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}