{"id":27993,"date":"2025-06-20T08:39:18","date_gmt":"2025-06-20T08:39:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27993"},"modified":"2025-06-20T08:39:20","modified_gmt":"2025-06-20T08:39:20","slug":"determine-the-number-of-valence-electrons-in-snfand-then-draw-the-corresponding-lewis-structure","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-number-of-valence-electrons-in-snfand-then-draw-the-corresponding-lewis-structure\/","title":{"rendered":"Determine the number of valence electrons in SnFand then draw the corresponding Lewis structure."},"content":{"rendered":"\n<p>A) 51 B) 45 C) 47 D) 46 E) 48 Determine the number of valence electrons in SnF<br>and then draw the corresponding Lewis structure.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-red-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the number of valence electrons in <strong>SnF\u2086\u00b2\u207b<\/strong> (the hexafluorostannate(IV) ion), we begin by adding up the valence electrons from each atom and considering the extra electrons due to the negative charge.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count valence electrons<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sn (tin)<\/strong> is in Group 14 of the periodic table, so it contributes <strong>4 valence electrons<\/strong>.<\/li>\n\n\n\n<li>Each <strong>F (fluorine)<\/strong> atom is in Group 17 and contributes <strong>7 valence electrons<\/strong>. Since there are 6 fluorine atoms:<br>6 \u00d7 7 = <strong>42 valence electrons<\/strong><\/li>\n\n\n\n<li>The <strong>2- charge<\/strong> means we add <strong>2 more electrons<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Now we sum them up:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4 (Sn)<\/li>\n\n\n\n<li>42 (F)<\/li>\n\n\n\n<li>2 (charge)<br>= <strong>48 valence electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>So, the correct answer is:<br><strong>E) 48<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Draw the Lewis Structure<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Place Sn at the center<\/strong> since it is less electronegative and can form more than one bond.<\/li>\n\n\n\n<li>Surround it with six fluorine atoms.<\/li>\n\n\n\n<li>Create a <strong>single bond<\/strong> between Sn and each F atom. This uses <strong>6 bonds \u00d7 2 electrons = 12 electrons<\/strong>.<\/li>\n\n\n\n<li>Subtracting the bonding electrons from the total gives:<br>48 &#8211; 12 = <strong>36 electrons<\/strong> left.<\/li>\n\n\n\n<li>Distribute the remaining 36 electrons as lone pairs on the fluorine atoms. Each F atom needs 6 electrons to complete its octet (in addition to the 2 shared in the bond), so:<br>6 F atoms \u00d7 6 electrons = <strong>36 electrons<\/strong><\/li>\n<\/ol>\n\n\n\n<p>All electrons are now accounted for, and each atom has a complete octet.<\/p>\n\n\n\n<p>The <strong>Sn atom<\/strong> is surrounded by <strong>6 bonding pairs<\/strong>, making it <strong>hypervalent<\/strong>, which is acceptable for elements in period 5 or higher like tin.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total valence electrons: <strong>48<\/strong><\/li>\n\n\n\n<li>Central atom: <strong>Sn<\/strong><\/li>\n\n\n\n<li>Each F forms a single bond with Sn<\/li>\n\n\n\n<li>All F atoms have complete octets<\/li>\n\n\n\n<li>The structure obeys the octet rule for fluorine and expanded octet for tin<\/li>\n<\/ul>\n\n\n\n<p>Correct answer: <strong>E) 48<\/strong>.<\/p>\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\/06\/learnexams-banner5-83.jpeg\" alt=\"\" class=\"wp-image-27999\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-83.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-83-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-83-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-83-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A) 51 B) 45 C) 47 D) 46 E) 48 Determine the number of valence electrons in SnFand then draw the corresponding Lewis structure. The Correct Answer and Explanation is: To determine the number of valence electrons in SnF\u2086\u00b2\u207b (the hexafluorostannate(IV) ion), we begin by adding up the valence electrons from each atom and considering [&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-27993","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27993","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=27993"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27993\/revisions"}],"predecessor-version":[{"id":28000,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27993\/revisions\/28000"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27993"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27993"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}