{"id":35470,"date":"2025-06-24T11:04:56","date_gmt":"2025-06-24T11:04:56","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35470"},"modified":"2025-06-24T11:04:58","modified_gmt":"2025-06-24T11:04:58","slug":"draw-on-paper-lewis-structures-for-tetrabromophosphonium-ion-and-the-nitrate-ion","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-on-paper-lewis-structures-for-tetrabromophosphonium-ion-and-the-nitrate-ion\/","title":{"rendered":"Draw (on paper) Lewis structures for tetrabromophosphonium ion and the nitrate ion."},"content":{"rendered":"\n<p>Draw (on paper) Lewis structures for tetrabromophosphonium ion and the nitrate ion. How many equivalent Lewis structures are necessary to describe the bonding in PBr<br>? How many equivalent Lewis structures are necessary to describe the bonding in NO<\/p>\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>To answer this question, we will draw the Lewis structures of the <strong>tetrabromophosphonium ion (PBr\u2084\u207a)<\/strong> and the <strong>nitrate ion (NO\u2083\u207b)<\/strong>, and then explain how many equivalent Lewis structures are necessary for each.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Tetrabromophosphonium Ion (PBr\u2084\u207a)<\/strong><\/h3>\n\n\n\n<p><strong>Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Central atom: Phosphorus (P)<\/li>\n\n\n\n<li>Four bonded atoms: Bromine (Br)<\/li>\n\n\n\n<li>Charge: +1<\/li>\n<\/ul>\n\n\n\n<p><strong>Drawing the structure:<\/strong><\/p>\n\n\n\n<p>Phosphorus has 5 valence electrons. Each bromine atom has 7 valence electrons. The +1 charge indicates a loss of one electron from the total count.<\/p>\n\n\n\n<p>Total valence electrons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P: 5<\/li>\n\n\n\n<li>4 Br atoms: 4 \u00d7 7 = 28<\/li>\n\n\n\n<li>Subtract 1 for the +1 charge:<br><strong>Total = 5 + 28 \u2013 1 = 32 electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Place P in the center and form single bonds with each Br (each bond uses 2 electrons): 4 \u00d7 2 = 8 electrons used.<\/p>\n\n\n\n<p>Remaining: 32 \u2013 8 = 24 electrons, used to complete octets of the Br atoms (6 each).<\/p>\n\n\n\n<p><strong>Final structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P has four single bonds to Br.<\/li>\n\n\n\n<li>Br atoms each have three lone pairs.<\/li>\n\n\n\n<li>The structure has a +1 formal charge mostly localized on the phosphorus.<\/li>\n<\/ul>\n\n\n\n<p><strong>Number of equivalent Lewis structures:<\/strong><br>Only <strong>1<\/strong> Lewis structure is needed. All Br atoms are in identical positions and single bonded; there are no resonance structures.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Nitrate Ion (NO\u2083\u207b)<\/strong><\/h3>\n\n\n\n<p><strong>Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Central atom: Nitrogen (N)<\/li>\n\n\n\n<li>Three bonded atoms: Oxygen (O)<\/li>\n\n\n\n<li>Charge: \u20131<\/li>\n<\/ul>\n\n\n\n<p><strong>Total valence electrons:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>N: 5<\/li>\n\n\n\n<li>3 O: 3 \u00d7 6 = 18<\/li>\n\n\n\n<li>Add 1 for the \u20131 charge:<br><strong>Total = 5 + 18 + 1 = 24 electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Place N in the center and connect each O with a single bond: 3 \u00d7 2 = 6 electrons.<\/p>\n\n\n\n<p>Remaining: 24 \u2013 6 = 18 electrons<\/p>\n\n\n\n<p>Assign 6 electrons (3 lone pairs) to each O atom: 3 \u00d7 6 = 18 electrons used.<\/p>\n\n\n\n<p>Now check octets: Each O has 8 electrons, N only has 6. To satisfy the octet rule, create a double bond with one O by converting a lone pair into a bonding pair.<\/p>\n\n\n\n<p><strong>Final structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One N=O double bond<\/li>\n\n\n\n<li>Two N\u2013O single bonds<\/li>\n\n\n\n<li>Formal charges: The two single-bonded O atoms carry a \u20131 charge each; N has a +1 charge.<\/li>\n<\/ul>\n\n\n\n<p><strong>Number of equivalent Lewis structures:<\/strong><br><strong>3<\/strong> resonance structures are needed. The double bond can be placed with any of the three O atoms. These are equivalent and collectively describe the delocalized bonding in the nitrate ion.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>PBr\u2084\u207a<\/strong>: 1 Lewis structure is sufficient; no resonance.<\/li>\n\n\n\n<li><strong>NO\u2083\u207b<\/strong>: 3 equivalent Lewis structures are required due to resonance among the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"722\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-310.jpeg\" alt=\"\" class=\"wp-image-35471\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-310.jpeg 722w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-310-212x300.jpeg 212w\" sizes=\"auto, (max-width: 722px) 100vw, 722px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw (on paper) Lewis structures for tetrabromophosphonium ion and the nitrate ion. How many equivalent Lewis structures are necessary to describe the bonding in PBr? How many equivalent Lewis structures are necessary to describe the bonding in NO The Correct Answer and Explanation is: To answer this question, we will draw the Lewis structures of [&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-35470","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35470","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=35470"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35470\/revisions"}],"predecessor-version":[{"id":35472,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35470\/revisions\/35472"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35470"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35470"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}