{"id":181525,"date":"2025-01-11T02:08:21","date_gmt":"2025-01-11T02:08:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181525"},"modified":"2025-01-11T02:08:24","modified_gmt":"2025-01-11T02:08:24","slug":"draw-the-lewis-formula-for-a-brf3-b-nh2-and-c-sic14-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/11\/draw-the-lewis-formula-for-a-brf3-b-nh2-and-c-sic14-2\/","title":{"rendered":"Draw the Lewis formula for a) BrF3; b) NH2- ; and c) SiC14 2"},"content":{"rendered":"\n
    \n
  1. Draw the Lewis formula for a) BrF3; b) NH2- ; and c) SiC14 2. What terms describe the shapes of the species mentioned in number 1? a) BrF3; b) NH2- ; and c) SiC14 3. Predict the geometry of the following species: a) SCO | b) BeCl2 c) PC15 4. Give the hybridization of the central atom in each of the following species? a) SiC14 b) NH2CI c) ONCI 5. How may electron domains are there in the following molecule? a. H2 b. NH3 c. XeF22- d. SF4 e. ICI4- None<\/li>\n<\/ol>\n\n\n\n

    the correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

    1. Lewis Structures<\/h3>\n\n\n\n

    a) BrF\u2083<\/strong>:
    Bromine (Br) is the central atom with three single bonds to fluorine atoms and two lone pairs.<\/p>\n\n\n\n

    b) NH\u2082\u207b<\/strong>:
    Nitrogen (N) is the central atom, bonded to two hydrogen atoms with one lone pair and an extra electron (giving it a negative charge).<\/p>\n\n\n\n

    c) SiCl\u2084<\/strong>:
    Silicon (Si) is the central atom, with four single bonds to chlorine atoms and no lone pairs.<\/p>\n\n\n\n

    \n\n\n\n

    2. Molecular Shapes<\/h3>\n\n\n\n

    a) BrF\u2083<\/strong>: T-shaped.
    The presence of two lone pairs on bromine causes a T-shaped molecular geometry.<\/p>\n\n\n\n

    b) NH\u2082\u207b<\/strong>: Bent.
    The lone pair on nitrogen creates a bent geometry.<\/p>\n\n\n\n

    c) SiCl\u2084<\/strong>: Tetrahedral.
    With no lone pairs on silicon, the geometry is tetrahedral.<\/p>\n\n\n\n

    \n\n\n\n

    3. Predicted Geometry<\/h3>\n\n\n\n

    a) SCO<\/strong>: Linear.
    SCO has a triple bond between C and O and a single bond between S and C, forming a linear shape.<\/p>\n\n\n\n

    b) BeCl\u2082<\/strong>: Linear.
    Beryllium chloride has no lone pairs on Be, leading to a linear geometry.<\/p>\n\n\n\n

    c) PCl\u2085<\/strong>: Trigonal bipyramidal.
    Phosphorus is surrounded by five bonding pairs, forming a trigonal bipyramidal shape.<\/p>\n\n\n\n

    \n\n\n\n

    4. Hybridization of the Central Atom<\/h3>\n\n\n\n

    a) SiCl\u2084<\/strong>: sp\u00b3.
    Silicon forms four sigma bonds, requiring sp\u00b3 hybridization.<\/p>\n\n\n\n

    b) NH\u2082Cl<\/strong>: sp\u00b3.
    Nitrogen is bonded to two hydrogens and one chlorine, with one lone pair, resulting in sp\u00b3 hybridization.<\/p>\n\n\n\n

    c) ONCl<\/strong>: sp\u00b2.
    Oxygen forms a double bond with nitrogen and a single bond with chlorine, requiring sp\u00b2 hybridization.<\/p>\n\n\n\n

    \n\n\n\n

    5. Electron Domains<\/h3>\n\n\n\n

    a) H\u2082<\/strong>: 1 electron domain (single bond).
    b) NH\u2083<\/strong>: 4 electron domains (3 bonds, 1 lone pair).
    c) XeF\u2082\u00b2\u207b<\/strong>: 5 electron domains (2 bonds, 3 lone pairs).
    d) SF\u2084<\/strong>: 5 electron domains (4 bonds, 1 lone pair).
    e) ICl\u2084\u207b<\/strong>: 6 electron domains (4 bonds, 2 lone pairs).<\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/h3>\n\n\n\n

    The Lewis structures of molecules are based on valence electrons around the central atom. Bromine in BrF\u2083 uses five valence electrons to bond with three fluorine atoms, leaving two lone pairs. The T-shaped geometry results from electron-pair repulsion. NH\u2082\u207b has nitrogen with two bonding pairs and a lone pair, forming a bent shape. SiCl\u2084 has silicon bonded to four chlorine atoms in a tetrahedral shape due to even electron distribution.<\/p>\n\n\n\n

    For predicting geometries, we apply the VSEPR theory. SCO is linear because of sp hybridization in carbon. BeCl\u2082 is linear as Be uses sp hybrid orbitals to minimize repulsion. PCl\u2085 adopts a trigonal bipyramidal shape, consistent with five bonding pairs around phosphorus.<\/p>\n\n\n\n

    Hybridization explains orbital mixing for sigma bonding. Si in SiCl\u2084 uses sp\u00b3 orbitals for its tetrahedral geometry. Nitrogen in NH\u2082Cl uses sp\u00b3 orbitals due to one lone pair. In ONCl, oxygen uses sp\u00b2 orbitals as it participates in a double bond.<\/p>\n\n\n\n

    Electron domains count lone pairs and bonding regions. H\u2082 has a single bonding domain. NH\u2083 includes three bonding domains and one lone pair. XeF\u2082\u00b2\u207b has five domains, including lone pairs on xenon. SF\u2084 has five domains with one lone pair, and ICl\u2084\u207b has six domains, with two lone pairs.<\/p>\n\n\n\n

    These principles demonstrate how molecular shape, hybridization, and electron domains are interrelated.<\/p>\n","protected":false},"excerpt":{"rendered":"

    the correct answer and explanation is: 1. Lewis Structures a) BrF\u2083:Bromine (Br) is the central atom with three single bonds to fluorine atoms and two lone pairs. b) NH\u2082\u207b:Nitrogen (N) is the central atom, bonded to two hydrogen atoms with one lone pair and an extra electron (giving it a negative charge). c) SiCl\u2084:Silicon (Si) […]<\/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-181525","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181525","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=181525"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181525\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181525"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181525"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}