{"id":261355,"date":"2025-07-20T05:41:57","date_gmt":"2025-07-20T05:41:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261355"},"modified":"2025-07-20T05:42:00","modified_gmt":"2025-07-20T05:42:00","slug":"consider-the-following-polyatomic-ion-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/consider-the-following-polyatomic-ion-2\/","title":{"rendered":"Consider the following polyatomic ion"},"content":{"rendered":"\n
Consider the following polyatomic ion: AsO2-. – Show the full electron configuration for As. – Draw the most correct Lewis structure for AsO2-. – Briefly explain why your Lewis structure is correct. – If the structure is stabilized by resonance, draw at least one of the possible resonance forms. If it is not stabilized by resonance, briefly explain why. – Consider the following molecule: GeBr2. Indicate the bond angles around the central atom. What is the electronic geometry of this molecule? What is its molecular shape? – Does GeBr2 have a dipole moment? Briefly justify your answer.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s break down the questions and answer them systematically:<\/p>\n\n\n\n
1. Electron Configuration for As<\/strong> (Arsenic)<\/h3>\n\n\n\n
Arsenic (As) has an atomic number of 33. Its full electron configuration is:<\/p>\n\n\n\n
3 electrons in the 4p orbital.<\/li>\n<\/ul>\n\n\n\n
2. Lewis Structure for AsO2- (Arsenite Ion)<\/strong><\/h3>\n\n\n\n
To draw the Lewis structure for AsO2-, follow these steps:<\/p>\n\n\n\n
\n
Step 1: Count the valence electrons.<\/strong>\n
\n
Arsenic (As) is in group 15, so it has 5 valence electrons.<\/li>\n\n\n\n
Each oxygen (O) is in group 16, so each oxygen has 6 valence electrons. There are 2 oxygens, so 2\u00d76=122 \\times 6 = 122\u00d76=12.<\/li>\n\n\n\n
The negative charge means an additional electron, so we add 1 more electron. Thus, the total number of valence electrons is: 5(As)+12(2\u00a0O\u2019s)+1(negative\u00a0charge)=18\u00a0electrons.5 (\\text{As}) + 12 (\\text{2 O’s}) + 1 (\\text{negative charge}) = 18 \\text{ electrons.}5(As)+12(2\u00a0O\u2019s)+1(negative\u00a0charge)=18\u00a0electrons.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Step 2: Place As in the center.<\/strong>\n
\n
As will be the central atom, surrounded by the two oxygen atoms.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Step 3: Draw single bonds between As and O.<\/strong>\n
\n
Each As-O bond is a single bond, consuming 4 electrons (2 electrons per bond).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Step 4: Distribute the remaining electrons.<\/strong>\n
\n
After placing the bonds, you have 18\u22124=1418 – 4 = 1418\u22124=14 electrons left. Distribute these electrons as lone pairs on oxygen atoms.<\/li>\n\n\n\n
Place 6 electrons (3 lone pairs) on each oxygen atom.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Step 5: Check the octet rule.<\/strong>\n
\n
Oxygen atoms satisfy the octet rule (each has 8 electrons). Arsenic has 6 electrons in its valence shell (counting the 2 bonds and 2 lone pairs), so it does not satisfy the octet rule. To fix this, arsenic can form a double bond with one of the oxygen atoms.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Step 6: Add a double bond to one oxygen atom.<\/strong>\n
\n
If one oxygen forms a double bond with arsenic, it will use 2 electrons instead of 1. The other oxygen remains with a single bond and 3 lone pairs.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
The most plausible Lewis structure for AsO2- would look like this:<\/p>\n\n\n\n
\n
One O atom double-bonded to As.<\/li>\n\n\n\n
The other O atom single-bonded to As and having 3 lone pairs.<\/li>\n\n\n\n
A negative charge on the single-bonded O.<\/li>\n<\/ul>\n\n\n\n
This structure satisfies the octet rule for oxygen and gives arsenic an expanded octet.<\/p>\n\n\n\n
3. Resonance in AsO2-<\/strong><\/h3>\n\n\n\n
Yes, this structure is stabilized by resonance. In this case, the position of the double bond can switch between the two oxygen atoms, leading to two equivalent resonance forms:<\/p>\n\n\n\n
\n
In one resonance form, one oxygen is double-bonded to arsenic, and the other is single-bonded with a lone pair and the negative charge.<\/li>\n\n\n\n
In the other resonance form, the roles of the oxygens are reversed.<\/li>\n<\/ul>\n\n\n\n
Both forms are of equal energy, and the actual structure is a hybrid of these forms.<\/p>\n\n\n\n
Bond Angles:<\/strong> The central atom in GeBr2 is germanium (Ge), which is in group 14. It forms two bonds with bromine (Br), and no lone pairs are on Ge. This results in a linear geometry<\/strong>.\n
\n
The bond angle between the two Ge-Br bonds is 180\u00b0<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Electronic Geometry:<\/strong> Since there are two bonding pairs and no lone pairs on germanium, the electronic geometry is linear<\/strong>.<\/li>\n\n\n\n
Molecular Shape:<\/strong> The molecular shape of GeBr2 is also linear<\/strong>, as the two bromine atoms are positioned opposite each other, minimizing repulsion.<\/li>\n\n\n\n
Dipole Moment:<\/strong> No, GeBr2 does not have a dipole moment.\n
\n
This is because the two bromine atoms are symmetrically arranged on either side of the central germanium atom, and their electronegativities are almost identical. As a result, the dipoles created by each Ge-Br bond cancel each other out, leading to no overall dipole moment.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
Conclusion<\/h3>\n\n\n\n
\n
AsO2-<\/strong>: The correct Lewis structure involves resonance and an expanded octet for arsenic. The ion’s structure is stabilized by resonance.<\/li>\n\n\n\n
GeBr2<\/strong>: The molecule has a linear electronic and molecular geometry, with no dipole moment due to the symmetry of the two Ge-Br bonds.<\/li>\n<\/ul>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Consider the following polyatomic ion: AsO2-. – Show the full electron configuration for As. – Draw the most correct Lewis structure for AsO2-. – Briefly explain why your Lewis structure is correct. – If the structure is stabilized by resonance, draw at least one of the possible resonance forms. If it is not stabilized by […]<\/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-261355","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261355","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=261355"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261355\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261355"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261355"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261355"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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