{"id":199292,"date":"2025-03-11T04:24:37","date_gmt":"2025-03-11T04:24:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199292"},"modified":"2025-03-11T04:24:40","modified_gmt":"2025-03-11T04:24:40","slug":"the-complex-nicn4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/11\/the-complex-nicn4\/","title":{"rendered":"The complex [Ni(CN)4]"},"content":{"rendered":"\n

The complex [Ni(CN)4]
The complex [Ni(CN)4]2- is diamagnetic but [NiCl4]2- is paramagnetic (both d8), with two unpaired electrons. Similarly, [Fe(CN)6]3- has only one unpaired electron, but [Fe(OH2)6]3+ has 5. Explain these experimental observations using:<\/p>\n\n\n\n

a) the Valencia Bond Theory.<\/p>\n\n\n\n

b) the Crystalline Field Theory.<\/p>\n\n\n\n

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

Explanation of Magnetic Properties Using VBT and CFT<\/h3>\n\n\n\n

The complexes [Ni(CN)4]\u00b2\u207b<\/strong> and [NiCl4]\u00b2\u207b<\/strong> exhibit different magnetic properties despite both having d\u2078<\/strong> electronic configurations. Similarly, [Fe(CN)6]\u00b3\u207b<\/strong> and [Fe(OH2)6]\u00b3\u207a<\/strong> have different numbers of unpaired electrons despite both containing Fe\u00b3\u207a (d\u2075<\/strong>).<\/p>\n\n\n\n

\n\n\n\n

(a) Valence Bond Theory (VBT) Explanation<\/strong><\/h4>\n\n\n\n

VBT explains the magnetic properties based on hybridization and electron pairing in the metal\u2019s orbitals.<\/p>\n\n\n\n

    \n
  1. [Ni(CN)4]\u00b2\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Nickel is in the +2 oxidation state<\/strong>, meaning it has a d\u2078<\/strong> configuration.<\/li>\n\n\n\n
    • The strong field ligand CN\u207b<\/strong> forces the low-spin<\/strong> configuration.<\/li>\n\n\n\n
    • The Ni\u00b2\u207a<\/strong> ion undergoes dsp\u00b2 hybridization<\/strong>, resulting in a square planar<\/strong> structure.<\/li>\n\n\n\n
    • All electrons pair up, making the complex diamagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. [NiCl4]\u00b2\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Cl\u207b is a weak field ligand<\/strong>, leading to a high-spin<\/strong> configuration.<\/li>\n\n\n\n
        • The Ni\u00b2\u207a ion undergoes sp\u00b3 hybridization<\/strong>, resulting in a tetrahedral<\/strong> structure.<\/li>\n\n\n\n
        • Two unpaired electrons remain, making it paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. [Fe(CN)6]\u00b3\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • Fe\u00b3\u207a has a d\u2075<\/strong> configuration.<\/li>\n\n\n\n
            • CN\u207b is a strong field ligand<\/strong>, causing a low-spin<\/strong> state.<\/li>\n\n\n\n
            • Electrons pair up, leaving only one unpaired electron<\/strong> (t\u2082g\u2075 in an octahedral field).<\/li>\n<\/ul>\n\n\n\n
                \n
              1. [Fe(OH2)6]\u00b3\u207a<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • H\u2082O is a weak field ligand<\/strong>, leading to a high-spin<\/strong> configuration.<\/li>\n\n\n\n
                • The Fe\u00b3\u207a ion retains five unpaired electrons<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                  \n\n\n\n

                  (b) Crystal Field Theory (CFT) Explanation<\/strong><\/h4>\n\n\n\n

                  CFT explains magnetism by considering the splitting of d-orbitals<\/strong> due to ligand interactions.<\/p>\n\n\n\n

                    \n
                  1. [Ni(CN)4]\u00b2\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                      \n
                    • Square planar<\/strong> structure: The strong field CN\u207b<\/strong> ligand causes a large d-orbital splitting<\/strong>.<\/li>\n\n\n\n
                    • Electrons pair in the lower energy levels<\/strong>, leading to a low-spin, diamagnetic<\/strong> complex.<\/li>\n<\/ul>\n\n\n\n
                        \n
                      1. [NiCl4]\u00b2\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                          \n
                        • Tetrahedral structure<\/strong>: The weak field Cl\u207b<\/strong> ligand results in smaller d-orbital splitting<\/strong>.<\/li>\n\n\n\n
                        • High-spin state occurs, leaving two unpaired electrons<\/strong>, making it paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                            \n
                          1. [Fe(CN)6]\u00b3\u207b<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                              \n
                            • Octahedral field<\/strong>: Strong field CN\u207b<\/strong> ligand causes a large splitting<\/strong>.<\/li>\n\n\n\n
                            • A low-spin<\/strong> configuration (t\u2082g\u2075 eg\u2070) forms, leaving only one unpaired electron<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                                \n
                              1. [Fe(OH2)6]\u00b3\u207a<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                                  \n
                                • Octahedral field<\/strong>: Weak field H\u2082O<\/strong> ligand results in a small splitting<\/strong>.<\/li>\n\n\n\n
                                • A high-spin<\/strong> configuration (t\u2082g\u00b3 eg\u00b2) remains, keeping five unpaired electrons<\/strong>.<\/li>\n<\/ul>\n\n\n\n

                                  Conclusion<\/strong><\/h3>\n\n\n\n
                                    \n
                                  • Strong field ligands<\/strong> (CN\u207b) induce low-spin states<\/strong>, reducing the number of unpaired electrons.<\/li>\n\n\n\n
                                  • Weak field ligands<\/strong> (Cl\u207b, H\u2082O) lead to high-spin states<\/strong>, increasing unpaired electrons.<\/li>\n\n\n\n
                                  • Square planar (dsp\u00b2) hybridization results in diamagnetic behavior<\/strong>, while tetrahedral (sp\u00b3) hybridization allows unpaired electrons<\/strong>, leading to paramagnetic behavior.<\/li>\n<\/ul>\n\n\n\n

                                    Diagram for CFT Splitting in Different Complexes<\/strong><\/h4>\n\n\n\n
                                    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                                    The complex [Ni(CN)4]The complex [Ni(CN)4]2- is diamagnetic but [NiCl4]2- is paramagnetic (both d8), with two unpaired electrons. Similarly, [Fe(CN)6]3- has only one unpaired electron, but [Fe(OH2)6]3+ has 5. Explain these experimental observations using: a) the Valencia Bond Theory. b) the Crystalline Field Theory. The correct answer and explanation is : Explanation of Magnetic Properties Using […]<\/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-199292","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199292","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=199292"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199292\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199292"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199292"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}