{"id":225677,"date":"2025-06-04T11:33:16","date_gmt":"2025-06-04T11:33:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225677"},"modified":"2025-06-04T11:33:18","modified_gmt":"2025-06-04T11:33:18","slug":"fluoroaluminate-anions-alf4-and-alf6-3-have-been-known-as-the-structure-pentafluoroaluminate-ion-alf5-2-1","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/fluoroaluminate-anions-alf4-and-alf6-3-have-been-known-as-the-structure-pentafluoroaluminate-ion-alf5-2-1\/","title":{"rendered":"Fluoroaluminate anions AlF4- and AlF6 3- have been known as the structure pentafluoroaluminate ion, AlF5 2-. 1"},"content":{"rendered":"\n

Fluoroaluminate anions AlF4- and AlF6 3- have been known as the structure pentafluoroaluminate ion, AlF5 2-. 1. Draw the Lewis structures for AlF3, AlF4-, AlF5 2-, and AlF6 3-. 2. What is the molecular geometry of AlF3, AlF4-, AlF5 2-, and AlF6 3-? 3. What are the bonds in AlF3, AlF4-, AlF5 2-, and AlF6 3- using VSEPR.<\/p>\n\n\n\n

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

\n\n\n\n

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

a. AlF\u2083 (Aluminum trifluoride)<\/strong><\/h4>\n\n\n\n
    \n
  • Al has 3 valence electrons, F has 7 each.<\/li>\n\n\n\n
  • Al forms three single bonds<\/strong> with three fluorine atoms.<\/li>\n\n\n\n
  • Each F gets 3 lone pairs; Al has no lone pairs.<\/li>\n\n\n\n
  • Al has only 6 electrons<\/strong>, making it electron-deficient<\/strong>.<\/li>\n<\/ul>\n\n\n\n

    b. AlF\u2084\u207b (Tetrafluoroaluminate)<\/strong><\/h4>\n\n\n\n
      \n
    • Al has 3 valence electrons + 1 extra electron (due to charge) = 4 bonding electrons.<\/li>\n\n\n\n
    • Forms 4 single bonds<\/strong> with 4 F atoms.<\/li>\n\n\n\n
    • Al now has an octet<\/strong>, and each F has 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n

      c. AlF\u2085\u00b2\u207b (Pentafluoroaluminate)<\/strong><\/h4>\n\n\n\n
        \n
      • Al has 3 valence + 2 extra electrons = 5 bonding electrons.<\/li>\n\n\n\n
      • Forms 5 single bonds<\/strong> with 5 F atoms.<\/li>\n\n\n\n
      • Each F has 3 lone pairs.<\/li>\n\n\n\n
      • Al has 10 electrons<\/strong>\u2014this is hypervalent<\/strong> (expanded octet).<\/li>\n<\/ul>\n\n\n\n

        d. AlF\u2086\u00b3\u207b (Hexafluoroaluminate)<\/strong><\/h4>\n\n\n\n
          \n
        • Al has 3 + 3 extra electrons = 6 bonding electrons.<\/li>\n\n\n\n
        • Forms 6 single bonds<\/strong> with 6 F atoms.<\/li>\n\n\n\n
        • Al has 12 electrons<\/strong>, another hypervalent case.<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          2. Molecular Geometry (VSEPR Theory)<\/strong><\/h3>\n\n\n\n
          Species<\/th>Electron Domains<\/th>Shape<\/th>VSEPR Geometry<\/th><\/tr><\/thead>
          AlF\u2083<\/strong><\/td>3<\/td>Trigonal planar<\/td>AX\u2083 (sp\u00b2)<\/td><\/tr>
          AlF\u2084\u207b<\/strong><\/td>4<\/td>Tetrahedral<\/td>AX\u2084 (sp\u00b3)<\/td><\/tr>
          AlF\u2085\u00b2\u207b<\/strong><\/td>5<\/td>Trigonal bipyramidal<\/td>AX\u2085 (sp\u00b3d)<\/td><\/tr>
          AlF\u2086\u00b3\u207b<\/strong><\/td>6<\/td>Octahedral<\/td>AX\u2086 (sp\u00b3d\u00b2)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
          \n\n\n\n

          3. Bonds and VSEPR Explanation<\/strong><\/h3>\n\n\n\n

          AlF\u2083<\/strong> has three bonding pairs and no lone pairs, giving a trigonal planar structure (AX\u2083). The bonding is limited due to Al’s electron deficiency.<\/p>\n\n\n\n

          AlF\u2084\u207b<\/strong> has four bonding pairs with no lone pairs\u2014ideal tetrahedral (AX\u2084). The negative charge stabilizes the octet.<\/p>\n\n\n\n

          AlF\u2085\u00b2\u207b<\/strong> uses expanded octet (10 electrons) to accommodate five F atoms. This gives a trigonal bipyramidal structure (AX\u2085), with 3 F in equatorial and 2 in axial positions.<\/p>\n\n\n\n

          AlF\u2086\u00b3\u207b<\/strong> is an octahedral ion (AX\u2086), with Al forming 6 bonds using its 3 valence and 3 extra electrons from charge. All F atoms are 90\u00b0 apart.<\/p>\n\n\n\n

          \n\n\n\n

          Summary<\/strong><\/h3>\n\n\n\n

          The fluoroaluminate species AlF\u2083, AlF\u2084\u207b, AlF\u2085\u00b2\u207b, and AlF\u2086\u00b3\u207b demonstrate progressive complexity in bonding and geometry, explained effectively by Lewis structures and VSEPR theory. AlF\u2083<\/strong> has a trigonal planar geometry (AX\u2083) because aluminum forms three covalent bonds with fluorine atoms and has no lone pairs. It is electron-deficient<\/strong>, holding only six electrons around the Al center.<\/p>\n\n\n\n

          AlF\u2084\u207b<\/strong>, on the other hand, has a tetrahedral geometry (AX\u2084). The additional electron (due to the -1 charge) allows aluminum to achieve an octet<\/strong>. The four F atoms bond symmetrically, minimizing repulsion.<\/p>\n\n\n\n

          As more fluorines are added, AlF\u2085\u00b2\u207b<\/strong> adopts a trigonal bipyramidal<\/strong> geometry (AX\u2085). This indicates hypervalency<\/strong>, where aluminum expands its octet to accommodate 10 electrons. Three F atoms lie in the equatorial plane (120\u00b0 apart), while two occupy axial positions (180\u00b0 apart).<\/p>\n\n\n\n

          Finally, AlF\u2086\u00b3\u207b<\/strong> adopts an octahedral<\/strong> geometry (AX\u2086), with six bonding pairs and 12 electrons around aluminum. This structure is highly symmetrical, with 90\u00b0 bond angles between adjacent fluorine atoms.<\/p>\n\n\n\n

          All structures consist of polar covalent bonds<\/strong> (Al\u2013F), but as the number of F atoms increases, the molecular symmetry increases, reducing overall polarity in some cases (e.g., AlF\u2086\u00b3\u207b is nonpolar due to symmetry). VSEPR theory explains these shapes as resulting from repulsion minimization among bonding pairs of electrons around the central aluminum atom, leading to predictable and distinct geometries for each fluoroaluminate species.<\/p>\n\n\n\n

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

          Fluoroaluminate anions AlF4- and AlF6 3- have been known as the structure pentafluoroaluminate ion, AlF5 2-. 1. Draw the Lewis structures for AlF3, AlF4-, AlF5 2-, and AlF6 3-. 2. What is the molecular geometry of AlF3, AlF4-, AlF5 2-, and AlF6 3-? 3. What are the bonds in AlF3, AlF4-, AlF5 2-, and AlF6 […]<\/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-225677","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225677","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=225677"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225677\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225677"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225677"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225677"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}