{"id":204758,"date":"2025-03-22T14:43:44","date_gmt":"2025-03-22T14:43:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=204758"},"modified":"2025-03-22T14:43:46","modified_gmt":"2025-03-22T14:43:46","slug":"you-are-investigating-the-reactivity-of-oxygen-with-other-non-metals","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/22\/you-are-investigating-the-reactivity-of-oxygen-with-other-non-metals\/","title":{"rendered":"You are investigating the reactivity of oxygen with other non-metals"},"content":{"rendered":"\n
You are investigating the reactivity of oxygen with other non-metals.<\/p>\n\n\n\n
a) Draw a box diagram to show the electron configuration of oxygen.<\/p>\n\n\n\n
b) Give the symbol and charge of an oxygen ion.<\/p>\n\n\n\n
c) Draw the box diagram to show the electron configuration of the oxygen ion.<\/p>\n\n\n\n
d) Silicon dioxide can form a beautiful crystal known as quartz. Draw the Lewis dot structure. Show all steps and charges.<\/p>\n\n\n\n
e) Determine the geometry of silicon dioxide.<\/p>\n\n\n\n
f) What is the hybridisation of the central atom in silicon dioxide?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n
Here are the answers to your questions:<\/p>\n\n\n\n
a) Box Diagram of Oxygen Electron Configuration<\/strong><\/h3>\n\n\n\n
Oxygen has an atomic number of 8<\/strong>, meaning it has 8 electrons<\/strong>. The electron configuration is 1s\u00b2 2s\u00b2 2p\u2074<\/strong>. Below is the box diagram representation:<\/p>\n\n\n\n
Each box represents an orbital, and the arrows represent electrons (\u2191\u2193 for paired electrons and \u2191 for unpaired).<\/p>\n\n\n\n
\n\n\n\n
b) Symbol and Charge of an Oxygen Ion<\/strong><\/h3>\n\n\n\n
When oxygen gains two electrons to complete its octet, it forms an oxide ion (O\u00b2\u207b)<\/strong>. Symbol:<\/strong> O\u00b2\u207b Charge:<\/strong> -2<\/p>\n\n\n\n\n\n\n\n
c) Box Diagram of the Oxygen Ion (O\u00b2\u207b)<\/strong><\/h3>\n\n\n\n
The oxygen ion has two additional electrons<\/strong>, making its electron configuration 1s\u00b2 2s\u00b2 2p\u2076<\/strong>, the same as neon (Ne):<\/p>\n\n\n\n
d) Lewis Dot Structure of Silicon Dioxide (SiO\u2082)<\/strong><\/h3>\n\n\n\n
Silicon dioxide consists of one silicon (Si) atom covalently bonded to two oxygen (O) atoms. Since oxygen needs two bonds<\/strong> to complete its octet, each oxygen forms a double bond<\/strong> with silicon.<\/p>\n\n\n\n
Silicon forms two double bonds<\/strong> with oxygen to satisfy the octet rule.<\/li>\n<\/ul>\n\n\n\n\n
Final Structure:<\/strong><\/li>\n<\/ol>\n\n\n\n
O\u2550Si\u2550O<\/code><\/pre>\n\n\n\n
Each oxygen has four lone electrons (..)<\/strong>, and Si has a full octet.<\/p>\n\n\n\n\n\n\n\n
e) Geometry of Silicon Dioxide<\/strong><\/h3>\n\n\n\n
SiO\u2082 is a linear molecule<\/strong> due to the double bonds and sp hybridization<\/strong> of the silicon atom. The bond angle is 180\u00b0<\/strong>.<\/p>\n\n\n\n\n\n\n\n
f) Hybridization of Silicon in SiO\u2082<\/strong><\/h3>\n\n\n\n
The central Si atom is sp hybridized<\/strong>.<\/p>\n\n\n\n
Explanation (300 words)<\/strong><\/h4>\n\n\n\n
Silicon dioxide (SiO\u2082) has a linear molecular geometry<\/strong> due to the bonding between silicon and oxygen. To determine the hybridization<\/strong>, we analyze the valence electrons and bonding.<\/p>\n\n\n\n
Silicon belongs to group 14<\/strong> and has four valence electrons (3s\u00b23p\u00b2)<\/strong>. Oxygen belongs to group 16<\/strong> with six valence electrons<\/strong>. Each oxygen forms a double bond<\/strong> with silicon to satisfy the octet rule.<\/p>\n\n\n\n
In bonding, silicon needs to form two sigma (\u03c3) bonds<\/strong> with oxygen. The number of sigma bonds determines hybridization. Since silicon forms two \u03c3 bonds, its steric number is 2<\/strong> (number of sigma bonds + lone pairs).<\/p>\n\n\n\n
Using the steric number rule:<\/p>\n\n\n\n
\n
Steric Number 2<\/strong> \u2192 sp hybridization<\/strong><\/li>\n\n\n\n
Steric Number 3<\/strong> \u2192 sp\u00b2 hybridization<\/strong><\/li>\n\n\n\n
Steric Number 4<\/strong> \u2192 sp\u00b3 hybridization<\/strong><\/li>\n<\/ul>\n\n\n\n
Since SiO\u2082 has a steric number of 2<\/strong>, silicon undergoes sp hybridization<\/strong>. This means one s orbital<\/strong> and one p orbital<\/strong> mix to form two sp hybrid orbitals<\/strong>, aligning in a linear shape<\/strong> (180\u00b0 bond angle). The remaining two p orbitals<\/strong> remain unhybridized and participate in \u03c0 bonds<\/strong> with oxygen.<\/p>\n\n\n\n
Thus, SiO\u2082 has an sp hybridized silicon atom<\/strong>. The linear structure is due to sp hybridization<\/strong> and the double bonds between Si and O<\/strong>.<\/p>\n\n\n\n\n\n\n\n
Image Representation<\/strong><\/h3>\n\n\n\n
Here is an image of the SiO\u2082 Lewis structure and molecular geometry<\/strong>:<\/p>\n\n\n\n<\/figure>\n\n\n\n
Here is the diagram illustrating the Lewis dot structure<\/strong>, molecular geometry<\/strong>, and hybridization<\/strong> of silicon dioxide (SiO\u2082). Let me know if you need any modifications or further explanations!<\/p>\n","protected":false},"excerpt":{"rendered":"
You are investigating the reactivity of oxygen with other non-metals. a) Draw a box diagram to show the electron configuration of oxygen. b) Give the symbol and charge of an oxygen ion. c) Draw the box diagram to show the electron configuration of the oxygen ion. d) Silicon dioxide can form a beautiful crystal known […]<\/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-204758","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204758","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=204758"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204758\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204758"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204758"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204758"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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