To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps:<\/p>\n\n\n\n
Step 1: Count the total number of valence electrons.<\/h3>\n\n\n\n\n- Xenon (Xe)<\/strong> is in group 18, so it has 8 valence electrons.<\/li>\n\n\n\n
- Fluorine (F)<\/strong> is in group 17, and each fluorine atom has 7 valence electrons.<\/li>\n\n\n\n
- Since there are 4 fluorine atoms, the total number of valence electrons from fluorine is 4\u00d77=284 \\times 7 = 284\u00d77=28.<\/li>\n<\/ul>\n\n\n\n
So, the total number of valence electrons is:8 (from Xe)+28 (from F)=36 valence electrons.8 \\text{ (from Xe)} + 28 \\text{ (from F)} = 36 \\text{ valence electrons.}8 (from Xe)+28 (from F)=36 valence electrons.<\/p>\n\n\n\n
Step 2: Create the skeletal structure.<\/h3>\n\n\n\n\n- Xenon (Xe) will be the central atom because it can expand its octet, and fluorine atoms will be placed around it.<\/li>\n<\/ul>\n\n\n\n
Step 3: Distribute electrons.<\/h3>\n\n\n\n\n- Each F atom will form a single bond with Xe, using 2 electrons per bond. This accounts for 4\u00d72=84 \\times 2 = 84\u00d72=8 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 4: Complete the octets of the fluorine atoms.<\/h3>\n\n\n\n\n- Each F atom now has 6 remaining valence electrons (since it already shares 2 electrons with Xe). So, place 3 lone pairs on each F atom, using 4\u00d76=244 \\times 6 = 244\u00d76=24 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 5: Distribute the remaining electrons around Xe.<\/h3>\n\n\n\n\n- Xenon still needs electrons to complete its octet, but since it’s in period 5, it can have more than 8 electrons. After the 4 bonds (8 electrons) to fluorine, Xenon will have 4 electrons left to form 2 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-314.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
So, the total number of valence electrons is:8 (from Xe)+28 (from F)=36 valence electrons.8 \\text{ (from Xe)} + 28 \\text{ (from F)} = 36 \\text{ valence electrons.}8 (from Xe)+28 (from F)=36 valence electrons.<\/p>\n\n\n\n
Step 2: Create the skeletal structure.<\/h3>\n\n\n\n\n- Xenon (Xe) will be the central atom because it can expand its octet, and fluorine atoms will be placed around it.<\/li>\n<\/ul>\n\n\n\n
Step 3: Distribute electrons.<\/h3>\n\n\n\n\n- Each F atom will form a single bond with Xe, using 2 electrons per bond. This accounts for 4\u00d72=84 \\times 2 = 84\u00d72=8 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 4: Complete the octets of the fluorine atoms.<\/h3>\n\n\n\n\n- Each F atom now has 6 remaining valence electrons (since it already shares 2 electrons with Xe). So, place 3 lone pairs on each F atom, using 4\u00d76=244 \\times 6 = 244\u00d76=24 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 5: Distribute the remaining electrons around Xe.<\/h3>\n\n\n\n\n- Xenon still needs electrons to complete its octet, but since it’s in period 5, it can have more than 8 electrons. After the 4 bonds (8 electrons) to fluorine, Xenon will have 4 electrons left to form 2 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 3: Distribute electrons.<\/h3>\n\n\n\n\n- Each F atom will form a single bond with Xe, using 2 electrons per bond. This accounts for 4\u00d72=84 \\times 2 = 84\u00d72=8 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 4: Complete the octets of the fluorine atoms.<\/h3>\n\n\n\n\n- Each F atom now has 6 remaining valence electrons (since it already shares 2 electrons with Xe). So, place 3 lone pairs on each F atom, using 4\u00d76=244 \\times 6 = 244\u00d76=24 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 5: Distribute the remaining electrons around Xe.<\/h3>\n\n\n\n\n- Xenon still needs electrons to complete its octet, but since it’s in period 5, it can have more than 8 electrons. After the 4 bonds (8 electrons) to fluorine, Xenon will have 4 electrons left to form 2 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 4: Complete the octets of the fluorine atoms.<\/h3>\n\n\n\n\n- Each F atom now has 6 remaining valence electrons (since it already shares 2 electrons with Xe). So, place 3 lone pairs on each F atom, using 4\u00d76=244 \\times 6 = 244\u00d76=24 electrons.<\/li>\n<\/ul>\n\n\n\n
Step 5: Distribute the remaining electrons around Xe.<\/h3>\n\n\n\n\n- Xenon still needs electrons to complete its octet, but since it’s in period 5, it can have more than 8 electrons. After the 4 bonds (8 electrons) to fluorine, Xenon will have 4 electrons left to form 2 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 5: Distribute the remaining electrons around Xe.<\/h3>\n\n\n\n\n- Xenon still needs electrons to complete its octet, but since it’s in period 5, it can have more than 8 electrons. After the 4 bonds (8 electrons) to fluorine, Xenon will have 4 electrons left to form 2 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Step 6: Check the final structure.<\/h3>\n\n\n\n\n- Xenon has 2 lone pairs of electrons and 4 bonds with fluorine atoms. Each fluorine atom has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Final Structure:<\/h3>\n\n\n\nrCopyEdit F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
F\n |\n F--Xe--F\n |\n F\n<\/code><\/pre>\n\n\n\nHow many lone pairs are around Xe?<\/h3>\n\n\n\n
There are 2 lone pairs<\/strong> of electrons around Xe.<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
Xenon in XeF\u2084 has an expanded octet because it is a noble gas in period 5. Xenon can accommodate more than 8 electrons due to the availability of d-orbitals. After bonding with the four fluorine atoms, which each contribute one bonding electron pair, the remaining electrons are placed as lone pairs on xenon. This gives Xenon 2 lone pairs of electrons. The fluorine atoms each have 3 lone pairs to complete their octet.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the Lewis structure for XeFand then answer the following question: How many lone pairs of electrons are around Xe? 0 0 0 1 0 2 0 3 0 4 The Correct Answer and Explanation is: To draw the Lewis structure for xenon fluoride (XeF\u2084), follow these steps: Step 1: Count the total number of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}