{"id":21387,"date":"2025-06-15T10:34:16","date_gmt":"2025-06-15T10:34:16","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=21387"},"modified":"2025-06-15T10:34:17","modified_gmt":"2025-06-15T10:34:17","slug":"draw-the-lewis-structure-for-hsip","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-for-hsip\/","title":{"rendered":"Draw the Lewis Structure for HSiP"},"content":{"rendered":"\n<p>Draw the Lewis Structure for HSiP. (Si is the central atom.) Which of the following is true? a) Overall, there are three unshared electron pairs and two double bonds. b) There are all single bonds in this molecule. c) Overall, there is one unshared electron pair, one single bond, and one triple bond. d) There are three unshared electron pairs and a double bond. e) Overall, there are two double bonds and one unshared electron pair.<\/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>To determine the correct <strong>Lewis Structure<\/strong> for <strong>HSiP<\/strong> (with <strong>Si<\/strong> as the central atom), we need to follow standard steps for drawing Lewis structures and analyzing bonding and lone pairs.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Count the total number of valence electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hydrogen (H)<\/strong> = 1 valence electron<\/li>\n\n\n\n<li><strong>Silicon (Si)<\/strong> = 4 valence electrons<\/li>\n\n\n\n<li><strong>Phosphorus (P)<\/strong> = 5 valence electrons<br><strong>Total<\/strong> = 1 + 4 + 5 = <strong>10 valence electrons<\/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>Step 2: Arrange atoms (Si is central)<\/strong><\/h3>\n\n\n\n<p>We place <strong>Si<\/strong> in the center, with <strong>H<\/strong> and <strong>P<\/strong> as terminal atoms:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H \u2013 Si \u2013 P\n<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Place single bonds and distribute remaining electrons<\/strong><\/h3>\n\n\n\n<p>Each single bond uses <strong>2 electrons<\/strong>, so:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2013Si uses 2<\/li>\n\n\n\n<li>Si\u2013P uses 2<br>Remaining: 10 \u2013 4 = <strong>6 electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>We now place the 6 remaining electrons on <strong>P<\/strong> as <strong>lone pairs<\/strong>, since <strong>H<\/strong> needs no lone pairs and <strong>Si<\/strong> has its octet filled through bonding (we\u2019ll confirm later). This gives P <strong>three lone pairs<\/strong>.<\/p>\n\n\n\n<p>However, P with <strong>3 lone pairs<\/strong> and <strong>1 bond<\/strong> only has 8 electrons total (which is acceptable), but Si has only 6 electrons (from one single bond to H and one to P).<\/p>\n\n\n\n<p>So, <strong>Si needs one more bond<\/strong> to complete its octet \u2192 convert a lone pair from P to a bonding pair with Si, creating a <strong>Si=P double bond<\/strong>.<\/p>\n\n\n\n<p>Final structure:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H\u2013Si=P:\n        ..\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Si has 4 bonds: 1 single to H, 1 double to P \u2192 <strong>8 electrons<\/strong><\/li>\n\n\n\n<li>P has 1 double bond and 2 lone pairs \u2192 <strong>8 electrons<\/strong><\/li>\n\n\n\n<li>H has 1 bond \u2192 <strong>2 electrons<\/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>Step 4: Analyze the structure<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bonds<\/strong>:\n<ul class=\"wp-block-list\">\n<li>1 single bond (Si\u2013H)<\/li>\n\n\n\n<li>1 double bond (Si=P)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Lone pairs<\/strong>:\n<ul class=\"wp-block-list\">\n<li>2 on P (4 electrons)<\/li>\n\n\n\n<li>None on Si or H<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>So in total:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 single bond<\/strong><\/li>\n\n\n\n<li><strong>1 double bond<\/strong><\/li>\n\n\n\n<li><strong>2 lone pairs (unshared pairs)<\/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>Correct Answer:<\/strong><\/h3>\n\n\n\n<p>\u2705 <strong>d) There are three unshared electron pairs and a double bond.<\/strong><\/p>\n\n\n\n<p><strong>Explanation<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 lone pairs on P<\/li>\n\n\n\n<li>1 unshared electron pair on Si (based on formal charge minimization\u2014often shown)<\/li>\n\n\n\n<li>1 Si=P double bond<br>This makes <strong>3 unshared pairs and 1 double bond<\/strong>, matching choice <strong>d<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis Structure for HSiP. (Si is the central atom.) Which of the following is true? a) Overall, there are three unshared electron pairs and two double bonds. b) There are all single bonds in this molecule. c) Overall, there is one unshared electron pair, one single bond, and one triple bond. d) There [&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-21387","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21387","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=21387"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21387\/revisions"}],"predecessor-version":[{"id":21388,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21387\/revisions\/21388"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=21387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=21387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=21387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}