{"id":191271,"date":"2025-02-14T11:19:10","date_gmt":"2025-02-14T11:19:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191271"},"modified":"2025-02-14T11:19:12","modified_gmt":"2025-02-14T11:19:12","slug":"indicate-the-hybridization-of-the-central-atom-bcl3-alcl4-cs2-krf2-pf6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/14\/indicate-the-hybridization-of-the-central-atom-bcl3-alcl4-cs2-krf2-pf6\/","title":{"rendered":"Indicate the hybridization of the central atom BCl3 AlCl4 – CS2 KrF2 PF6 –"},"content":{"rendered":"\n

Indicate the hybridization of the central atom BCl3 AlCl4 – CS2 KrF2 PF6 –<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Here\u2019s a breakdown of the hybridization for the central atoms of the compounds:<\/p>\n\n\n\n

    \n
  1. BCl\u2083 (Boron Trichloride)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • The central atom is boron (B).<\/li>\n\n\n\n
    • Boron in BCl\u2083 forms three covalent bonds with three chlorine (Cl) atoms.<\/li>\n\n\n\n
    • Since boron has three valence electrons, it uses three orbitals (one from the 2s and two from the 2p orbitals) to form bonds with chlorine.<\/li>\n\n\n\n
    • Therefore, the hybridization of the boron atom in BCl\u2083 is sp\u00b2<\/strong>. This is because boron forms three sigma bonds in a trigonal planar geometry, and there are no lone pairs on the boron atom.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. AlCl\u2084\u207b (Tetrachoridoaluminate ion)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • The central atom is aluminum (Al).<\/li>\n\n\n\n
        • In AlCl\u2084\u207b, aluminum is surrounded by four chloride ions (Cl\u207b), forming a tetrahedral shape.<\/li>\n\n\n\n
        • Aluminum has three valence electrons and receives an extra electron from the negative charge, making a total of four valence electrons.<\/li>\n\n\n\n
        • To accommodate four bonding pairs, aluminum undergoes sp\u00b3 hybridization<\/strong>. This creates four equivalent orbitals that overlap with the chlorine atoms to form four sigma bonds.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. CS\u2082 (Carbon Disulfide)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • The central atom is carbon (C).<\/li>\n\n\n\n
            • Carbon in CS\u2082 forms two double bonds with sulfur (S) atoms.<\/li>\n\n\n\n
            • Carbon has four valence electrons and forms two sigma bonds and two pi bonds, one with each sulfur atom.<\/li>\n\n\n\n
            • Since carbon forms two sigma bonds and two pi bonds, it uses two sp hybridized orbitals for bonding, leaving two unhybridized p orbitals for pi bonding.<\/li>\n\n\n\n
            • Therefore, the hybridization of carbon in CS\u2082 is sp<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                \n
              1. KrF\u2082 (Krypton Difluoride)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • The central atom is krypton (Kr).<\/li>\n\n\n\n
                • Krypton is a noble gas and has 8 valence electrons. In KrF\u2082, it forms two bonds with two fluorine (F) atoms, leaving three lone pairs of electrons.<\/li>\n\n\n\n
                • Given the presence of lone pairs, krypton in KrF\u2082 undergoes sp\u00b3d hybridization<\/strong> to accommodate five electron pairs (two bonding pairs and three lone pairs) in a linear geometry.<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. PF\u2086\u207b (Hexafluorophosphate ion)<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                      \n
                    • The central atom is phosphorus (P).<\/li>\n\n\n\n
                    • Phosphorus in PF\u2086\u207b is surrounded by six fluorine (F) atoms, and it carries a negative charge.<\/li>\n\n\n\n
                    • Phosphorus has five valence electrons, and the negative charge adds one more electron, making a total of six valence electrons.<\/li>\n\n\n\n
                    • To form six bonds with fluorine atoms, phosphorus undergoes sp\u00b3d\u00b2 hybridization<\/strong>, resulting in an octahedral geometry.<\/li>\n<\/ul>\n\n\n\n

                      Image:<\/h3>\n\n\n\n

                      I’ll generate an image representing the hybridization and geometries of these molecules.<\/p>\n\n\n\n

                      Here is the diagram showing the hybridization and geometry of the molecules you asked about: BCl\u2083, AlCl\u2084\u207b, CS\u2082, KrF\u2082, and PF\u2086\u207b. Each molecule is labeled with its corresponding hybridization type and molecular geometry. Let me know if you’d like more details or explanations!<\/p>\n\n\n\n

                      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                      Indicate the hybridization of the central atom BCl3 AlCl4 – CS2 KrF2 PF6 – The Correct Answer and Explanation is : Here\u2019s a breakdown of the hybridization for the central atoms of the compounds: Image: I’ll generate an image representing the hybridization and geometries of these molecules. Here is the diagram showing the hybridization and […]<\/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-191271","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191271","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=191271"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191271\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191271"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191271"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}