Lewis Dot Structure of Argon (Ar):<\/h3>\n\n\n\n
Argon (Ar) is a noble gas with an atomic number of 18. This means it has 18 electrons. In its Lewis dot structure, Argon has 8 electrons in its valence shell, which are arranged in pairs around the symbol Ar<\/strong> as follows:<\/p>\n\n\n\n Argon is chemically inert because its valence shell is complete with 8 electrons, fulfilling the octet rule.<\/p>\n\n\n\n Boron (B) has an atomic number of 5, meaning it has 5 electrons. The Lewis dot structure for Boron shows that it has 3 valence electrons:<\/p>\n\n\n\n Boron typically forms covalent bonds by sharing these 3 electrons to complete its octet when bonding with other atoms.<\/p>\n\n\n\n Sulfur (S) has an atomic number of 16, so it has 6 valence electrons. The sulfide ion (S\u00b2\u207b) has gained 2 electrons to form a stable 2\u207b charge, making its valence electron count 8. The Lewis dot structure for S\u00b2\u207b looks like:<\/p>\n\n\n\n The 2 extra electrons fill the valence shell, completing the octet and giving the ion a full, stable configuration.<\/p>\n\n\n\n Answer: (e) KF<\/strong> Answer: (a) F and Al<\/strong> Copper(II) refers to copper in the +2 oxidation state (Cu\u00b2\u207a), and phosphate is a polyatomic ion with the formula PO\u2084\u00b3\u207b<\/strong>. To balance the charges, we need 3 Cu\u00b2\u207a ions for every 2 PO\u2084\u00b3\u207b ions. The correct formula is:<\/p>\n\n\n\n Cu\u2083(PO\u2084)\u2082<\/strong><\/p>\n\n\n\n Iron(II) refers to iron in the +2 oxidation state (Fe\u00b2\u207a), and nitrate is a polyatomic ion with the formula NO\u2083\u207b<\/strong>. To balance the charges, we need 1 Fe\u00b2\u207a ion for every 2 NO\u2083\u207b ions. The correct formula is:<\/p>\n\n\n\n Fe(NO\u2083)\u2082<\/strong><\/p>\n\n\n\n In summary, the Lewis dot structures represent the valence electrons of elements and ions, while ionic compounds are formed through the transfer of electrons between a metal and a non-metal.<\/p>\n\n\n\n Write the Lewis dot structure of Argon. 2. Write Lewis dot structure for Boron. 3. Write Lewis dot structure for Sion. 4. Which of the following compounds is ionic? (circle one answer) (a) ICI (b) CS(c) CO(d) CCl(e) KF 5. Which pair of element would be most likely to form an ionic compound (a) F […]<\/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-253585","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253585","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=253585"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253585\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253585"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253585"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253585"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}makefileCopyEdit
Ar: \u2022 \u2022\n \u2022 \u2022\n \u2022 \u2022\n<\/code><\/pre>\n\n\n\n2. Lewis Dot Structure for Boron (B):<\/h3>\n\n\n\n
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B: \u2022 \u2022 \u2022\n<\/code><\/pre>\n\n\n\n3. Lewis Dot Structure for Sulfide Ion (S\u00b2\u207b):<\/h3>\n\n\n\n
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S: \u2022 \u2022 \u2022\n \u2022 \u2022\n \u2022 \u2022\n<\/code><\/pre>\n\n\n\n4. Which of the following compounds is ionic?<\/h3>\n\n\n\n
Explanation: Potassium (K) is a metal that easily loses its 1 electron to form a K\u207a ion, and Fluorine (F) is a non-metal that gains that electron to form an F\u207b ion. The opposite charges between K\u207a and F\u207b lead to an ionic bond, making KF an ionic compound.<\/p>\n\n\n\n5. Which pair of elements would be most likely to form an ionic compound?<\/h3>\n\n\n\n
Explanation: Fluorine (F) is a highly electronegative non-metal that readily gains 1 electron to form F\u207b, while Aluminum (Al) is a metal that can lose 3 electrons to form Al\u00b3\u207a. The large difference in electronegativity between these two elements makes them highly likely to form an ionic bond.<\/p>\n\n\n\n6. Correct Formula for Copper(II) Phosphate:<\/h3>\n\n\n\n
7. Chemical Formula for Iron(II) Nitrate:<\/h3>\n\n\n\n
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