{"id":221274,"date":"2025-05-29T11:14:51","date_gmt":"2025-05-29T11:14:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221274"},"modified":"2025-05-29T11:14:54","modified_gmt":"2025-05-29T11:14:54","slug":"draw-the-structure-of-p4s3-and-describe-an-appropriate-bonding-scheme-for-this-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/draw-the-structure-of-p4s3-and-describe-an-appropriate-bonding-scheme-for-this-molecule\/","title":{"rendered":"Draw the structure of P4S3 and describe an appropriate bonding scheme for this molecule."},"content":{"rendered":"\n

(a) Draw the structure of P4S3 and describe an appropriate bonding scheme for this molecule. Compare the structures of P4S10, P4S3 and P4, and comment on the formal oxidation states of the P atoms in these species. (b) The electrical resistivity of Bi at 273 K is 1.07 \u00d7 10^(-6) m. How do you expect this property to change as the temperature increases? On what grounds have you drawn your conclusion? (c) Hydrated iron(III) nitrate was dissolved in hot HNO3 (100%), and the solution was placed in a desiccator with P2O5 until the sample had become a solid residue. The pure Fe(III) product (an ionic salt [NO2][X]) was collected by sublimation; crystals were extremely deliquescent. Suggest an identity for the product, clearly stating the charges on the ions. The Fe(III) centre has a coordination number of 8. Suggest how this is achieved.<\/p>\n\n\n\n

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

(a) Structure and Bonding of P\u2084S\u2083, Comparison with P\u2084 and P\u2084S\u2081\u2080, and Oxidation States of P<\/strong><\/h3>\n\n\n\n

Structure and Bonding of P\u2084S\u2083:<\/strong>
The structure of P\u2084S\u2083<\/strong> consists of a phosphorus tetrahedron<\/strong> with three sulfur atoms bridging three of the six edges. It resembles a distorted tetrahedron with one P atom bonded to three other P atoms and each of those three bridging to a sulfur atom. Each sulfur atom is thus bonded to two phosphorus atoms.<\/p>\n\n\n\n

Bonding scheme:<\/strong>
Each P atom forms three bonds \u2014 two to other P atoms and one to S. Sulfur bridges use lone pairs to form two bonds with phosphorus atoms. The bonding is mostly covalent, with dative interactions playing a minor role due to P and S having comparable electronegativities.<\/p>\n\n\n\n

Comparison with P\u2084 and P\u2084S\u2081\u2080:<\/strong><\/p>\n\n\n\n

    \n
  • P\u2084<\/strong>: A regular tetrahedron where each P is bonded to the other three P atoms. All P atoms are in the 0 oxidation state.<\/li>\n\n\n\n
  • P\u2084S\u2081\u2080<\/strong>: Each P is tetrahedrally coordinated to four S atoms. The molecule has a cage-like structure, and P atoms are in the +5 oxidation state.<\/li>\n\n\n\n
  • P\u2084S\u2083<\/strong>: Intermediate between P\u2084 and P\u2084S\u2081\u2080. The oxidation state of P is +3 on average, considering that each S typically contributes \u20132.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    (b) Electrical Resistivity of Bismuth with Temperature<\/strong><\/h3>\n\n\n\n

    At 273 K<\/strong>, Bi has a resistivity of 1.07 \u00d7 10\u207b\u2076 \u03a9\u00b7m<\/strong>.
    As temperature increases<\/strong>, the electrical resistivity of Bi is expected to increase<\/strong>. This is because bismuth is a semimetal<\/strong>, and in metals and semimetals, increasing temperature causes greater lattice vibrations<\/strong>, which scatter conduction electrons more<\/strong>, reducing mobility and increasing resistivity<\/strong>. This is consistent with standard solid-state physics behavior in conductive elements.<\/p>\n\n\n\n

    \n\n\n\n

    (c) Identity of Product from Fe(NO\u2083)\u2083 in HNO\u2083 and P\u2082O\u2085<\/strong><\/h3>\n\n\n\n

    Upon dissolving hydrated Fe(NO\u2083)\u2083 in concentrated HNO\u2083<\/strong>, followed by drying with P\u2082O\u2085<\/strong>, a highly deliquescent solid [NO\u2082][Fe(NO\u2083)\u2084]<\/strong> sublimes.<\/p>\n\n\n\n

      \n
    • The product is [NO\u2082\u207a][Fe(NO\u2083)\u2084\u207b]<\/strong>.<\/li>\n\n\n\n
    • NO\u2082\u207a<\/strong> is the nitronium ion<\/strong>, formed in concentrated HNO\u2083 via: 2HNO3\u2192NO2++NO3\u2212+H2O2 HNO_3 \\rightarrow NO_2^+ + NO_3^- + H_2O<\/li>\n\n\n\n
    • P\u2082O\u2085 removes water, driving this equilibrium.<\/li>\n<\/ul>\n\n\n\n

      The Fe(III)<\/strong> center forms the [Fe(NO\u2083)\u2084]\u207b<\/strong> anion, with each nitrate acting as a bidentate ligand<\/strong>, achieving a coordination number of 8<\/strong> (4 \u00d7 2 coordination sites).
      This compound is extremely deliquescent<\/strong> due to the high ionic character and strong hydration enthalpy of both ions.<\/p>\n\n\n\n

      \n\n\n\n

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

      (a) P\u2084S\u2083 has a cage-like structure similar to P\u2084, with three S atoms bridging three edges of a phosphorus tetrahedron. Each P atom is bonded to three atoms, with mixed P\u2013P and P\u2013S bonding. P\u2084 is a tetrahedron of only P atoms, with each at oxidation state 0. P\u2084S\u2081\u2080 has a more complex structure with each P atom bonded tetrahedrally to four S atoms; the P atoms are in the +5 oxidation state. In P\u2084S\u2083, the average oxidation state of P is +3.
      (b) Bismuth\u2019s electrical resistivity increases with temperature due to enhanced phonon scattering, which reduces electron mobility. As a semimetal, Bi follows the general trend of metals where thermal agitation increases resistivity.
      (c) Reacting hydrated Fe(NO\u2083)\u2083 with hot HNO\u2083 and drying with P\u2082O\u2085 removes water and forms nitronium ions (NO\u2082\u207a). These combine with Fe(III) to form the salt [NO\u2082][Fe(NO\u2083)\u2084], where Fe(III) is coordinated by four bidentate nitrate ions, reaching a coordination number of 8. The product is extremely deliquescent due to its highly ionic nature and affinity for water.<\/p>\n\n\n\n

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

      (a) Draw the structure of P4S3 and describe an appropriate bonding scheme for this molecule. Compare the structures of P4S10, P4S3 and P4, and comment on the formal oxidation states of the P atoms in these species. (b) The electrical resistivity of Bi at 273 K is 1.07 \u00d7 10^(-6) m. How do you expect […]<\/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-221274","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221274","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=221274"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221274\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221274"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221274"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221274"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}