Let’s walk through how to draw the Lewis structures for the species you provided. I’ll first go over the general approach and then provide the Lewis structure for each molecule with formal charges where necessary.<\/p>\n\n\n\n
Step-by-Step Guide for Drawing Lewis Structures:<\/h3>\n\n\n\n\n- Determine the total number of valence electrons<\/strong>: This is calculated by adding up the valence electrons of each atom in the molecule.<\/li>\n\n\n\n
- Identify the central atom<\/strong>: Usually, it’s the least electronegative atom (excluding hydrogen) or the atom that can form the most bonds.<\/li>\n\n\n\n
- Connect atoms with single bonds<\/strong>: Draw bonds between atoms, and account for the electrons used in these bonds.<\/li>\n\n\n\n
- Distribute remaining electrons as lone pairs<\/strong>: After bonding, place remaining electrons on the atoms in pairs to satisfy the octet rule (or duet rule for hydrogen).<\/li>\n\n\n\n
- Adjust for formal charges<\/strong>: The goal is to minimize formal charges (charges on atoms) in the structure. Formal charge = (valence electrons) – (lone pair electrons) – (1\/2 bonding electrons).<\/li>\n<\/ol>\n\n\n\n
Now, let\u2019s go through each of the chemical species in your list:<\/p>\n\n\n\n
A. HBrO\u2082 (Hypobromous acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×2) = 1 + 7 + 12 = 20.<\/li>\n\n\n\n
- Central atom: Bromine (Br) since it’s the least electronegative.<\/li>\n\n\n\n
- Structure: Br forms single bonds with both O atoms and a single bond to H. Each oxygen atom will carry lone pairs to complete their octet.<\/li>\n<\/ul>\n\n\n\n
B. BrO\u2083\u207b (Bromate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×3) + 1 (for the negative charge) = 7 + 18 + 1 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), forms double bonds with oxygen atoms. The negative charge is likely to reside on the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
C. ClO\u2082 (Chlorine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Cl (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Chlorine is the central atom, and it forms a single bond to each oxygen, with one oxygen having a lone pair and the other being doubly bonded to chlorine.<\/li>\n<\/ul>\n\n\n\n
D. HPO\u2084\u00b2\u207b (Hydrogen phosphate)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), P (5), O (6×4) + 2 (for the 2 negative charges) = 1 + 5 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Phosphorus (P), which bonds to four oxygens. One oxygen will be bonded to a hydrogen atom, and the others will carry lone pairs.<\/li>\n<\/ul>\n\n\n\n
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-597.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Now, let\u2019s go through each of the chemical species in your list:<\/p>\n\n\n\n
A. HBrO\u2082 (Hypobromous acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×2) = 1 + 7 + 12 = 20.<\/li>\n\n\n\n
- Central atom: Bromine (Br) since it’s the least electronegative.<\/li>\n\n\n\n
- Structure: Br forms single bonds with both O atoms and a single bond to H. Each oxygen atom will carry lone pairs to complete their octet.<\/li>\n<\/ul>\n\n\n\n
B. BrO\u2083\u207b (Bromate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×3) + 1 (for the negative charge) = 7 + 18 + 1 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), forms double bonds with oxygen atoms. The negative charge is likely to reside on the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
C. ClO\u2082 (Chlorine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Cl (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Chlorine is the central atom, and it forms a single bond to each oxygen, with one oxygen having a lone pair and the other being doubly bonded to chlorine.<\/li>\n<\/ul>\n\n\n\n
D. HPO\u2084\u00b2\u207b (Hydrogen phosphate)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), P (5), O (6×4) + 2 (for the 2 negative charges) = 1 + 5 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Phosphorus (P), which bonds to four oxygens. One oxygen will be bonded to a hydrogen atom, and the others will carry lone pairs.<\/li>\n<\/ul>\n\n\n\n
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
B. BrO\u2083\u207b (Bromate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×3) + 1 (for the negative charge) = 7 + 18 + 1 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), forms double bonds with oxygen atoms. The negative charge is likely to reside on the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
C. ClO\u2082 (Chlorine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Cl (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Chlorine is the central atom, and it forms a single bond to each oxygen, with one oxygen having a lone pair and the other being doubly bonded to chlorine.<\/li>\n<\/ul>\n\n\n\n
D. HPO\u2084\u00b2\u207b (Hydrogen phosphate)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), P (5), O (6×4) + 2 (for the 2 negative charges) = 1 + 5 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Phosphorus (P), which bonds to four oxygens. One oxygen will be bonded to a hydrogen atom, and the others will carry lone pairs.<\/li>\n<\/ul>\n\n\n\n
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
C. ClO\u2082 (Chlorine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Cl (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Chlorine is the central atom, and it forms a single bond to each oxygen, with one oxygen having a lone pair and the other being doubly bonded to chlorine.<\/li>\n<\/ul>\n\n\n\n
D. HPO\u2084\u00b2\u207b (Hydrogen phosphate)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), P (5), O (6×4) + 2 (for the 2 negative charges) = 1 + 5 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Phosphorus (P), which bonds to four oxygens. One oxygen will be bonded to a hydrogen atom, and the others will carry lone pairs.<\/li>\n<\/ul>\n\n\n\n
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
D. HPO\u2084\u00b2\u207b (Hydrogen phosphate)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), P (5), O (6×4) + 2 (for the 2 negative charges) = 1 + 5 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Phosphorus (P), which bonds to four oxygens. One oxygen will be bonded to a hydrogen atom, and the others will carry lone pairs.<\/li>\n<\/ul>\n\n\n\n
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
E. SeO\u2084\u00b2\u207b (Selenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Se (6), O (6×4) + 2 (for 2 negative charges) = 6 + 24 + 2 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms double bonds with the four oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
F. HBrO\u2083 (Bromic acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Br (7), O (6×3) = 1 + 7 + 18 = 26.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms bonds with three oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
G. BrO\u2082 (Bromine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: Br (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Bromine (Br), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
H. IO\u2082 (Iodine dioxide)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: I (7), O (6×2) = 7 + 12 = 19.<\/li>\n\n\n\n
- Central atom: Iodine (I), which forms single bonds to two oxygen atoms.<\/li>\n<\/ul>\n\n\n\n
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
I. HSeO\u2084\u207b (Hydroselenate ion)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1), Se (6), O (6×4) + 1 (for the negative charge) = 1 + 6 + 24 + 1 = 32.<\/li>\n\n\n\n
- Central atom: Selenium (Se), which forms bonds with four oxygens. One oxygen will be bonded to hydrogen.<\/li>\n<\/ul>\n\n\n\n
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
J. H\u2082SO\u2084 (Sulfuric acid)<\/strong>:<\/h3>\n\n\n\n\n- Total valence electrons: H (1×2), S (6), O (6×4) = 2 + 6 + 24 = 32.<\/li>\n\n\n\n
- Central atom: Sulfur (S), which forms double bonds with each oxygen, two oxygens will also carry hydrogen atoms.<\/li>\n<\/ul>\n\n\n\n
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Let me create and display the Lewis structure for one or a couple of these species, and I’ll send you the images.<\/p>\n\n\n\n
Give me a moment!<\/p>\n\n\n\n
Here is the Lewis structure for HBrO\u2082<\/strong> (Hypobromous acid). This structure includes the formal charges on the atoms where appropriate. Would you like me to generate more structures for the other species you mentioned?<\/p>\n\n\n\n Draw the BEST Lewis structure (with formal charges) of the indicated chemical species. A. HBrO2 B.BrO3 E. CIO2 C. Se042- H. 103 D. HPO42- I. HSeO4 F. HBrO3 G. BrO2 HIO2 : IBV (1)(7): 7 30 (3) (4) 18 11 Charge 26e4 -4 20\u20ac’s 0: FC-6-[(3) + 1\/22 (6)] 10 (C-7-[(4) \u00b9 \u00bd (6) 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-191590","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590","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=191590"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191590\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}<\/figure>\n","protected":false},"excerpt":{"rendered":"