{"id":251254,"date":"2025-07-10T15:58:49","date_gmt":"2025-07-10T15:58:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=251254"},"modified":"2025-07-10T15:58:51","modified_gmt":"2025-07-10T15:58:51","slug":"write-orbital-diagrams-for-each-ion-and-determine-if-the-ion-is-diamagnetic-or-paramagnetic-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/write-orbital-diagrams-for-each-ion-and-determine-if-the-ion-is-diamagnetic-or-paramagnetic-2\/","title":{"rendered":"Write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic"},"content":{"rendered":"\n<p>Write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic. a. N2+ b. V2+ c. Pd2+ d. Ru?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-382.png\" alt=\"\" class=\"wp-image-251255\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the orbital diagrams and magnetic properties for each ion.<\/p>\n\n\n\n<p><strong>a. Niobium(V) ion, Nb\u2075\u207a<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Configuration:<\/strong>\u00a0The neutral Niobium atom (Nb, atomic number 41) has the electron configuration [Kr] 5s\u00b9 4d\u2074. To form the Nb\u2075\u207a ion, five electrons are removed. These are the one 5s electron and the four 4d electrons. This leaves the electron configuration of the noble gas Krypton, [Kr].<\/li>\n\n\n\n<li><strong>Orbital Diagram:<\/strong>\u00a0The electron configuration [Kr] or [Ar] 3d\u00b9\u2070 4s\u00b2 4p\u2076 means all orbitals in the n=1, 2, 3, and 4 shells are completely filled. For the outermost shell (n=4):\n<ul class=\"wp-block-list\">\n<li>4s: [\u2191\u2193]<\/li>\n\n\n\n<li>4p: [\u2191\u2193] [\u2191\u2193] [\u2191\u2193]<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Magnetic Property:<\/strong>\u00a0Since all electrons are paired,\u00a0<strong>Nb\u2075\u207a is diamagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>b. Vanadium(II) ion, V\u00b2\u207a<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Configuration:<\/strong>\u00a0The neutral Vanadium atom (V, atomic number 23) has the electron configuration [Ar] 4s\u00b2 3d\u00b3. To form the V\u00b2\u207a ion, two electrons are removed from the outermost shell, which is the 4s orbital. The resulting configuration is [Ar] 3d\u00b3.<\/li>\n\n\n\n<li><strong>Orbital Diagram (for 3d electrons):<\/strong>\u00a0According to Hund&#8217;s rule, electrons will occupy separate orbitals before pairing up.\n<ul class=\"wp-block-list\">\n<li>3d: [\u2191] [\u2191] [\u2191] [ ] [ ]<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Magnetic Property:<\/strong>\u00a0The ion has three unpaired electrons, making\u00a0<strong>V\u00b2\u207a paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>c. Palladium(II) ion, Pd\u00b2\u207a<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Configuration:<\/strong>\u00a0The neutral Palladium atom (Pd, atomic number 46) has an exceptional electron configuration of [Kr] 4d\u00b9\u2070. To form the Pd\u00b2\u207a ion, two electrons are removed from the 4d orbital, resulting in the configuration [Kr] 4d\u2078.<\/li>\n\n\n\n<li><strong>Orbital Diagram (for 4d electrons):<\/strong>\u00a0The eight electrons fill the five d-orbitals.\n<ul class=\"wp-block-list\">\n<li>4d: [\u2191\u2193] [\u2191\u2193] [\u2191\u2193] [\u2191] [\u2191]<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Magnetic Property:<\/strong>\u00a0The ion has two unpaired electrons, which means\u00a0<strong>Pd\u00b2\u207a is paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>d. Ruthenium(III) ion, Ru\u00b3\u207a<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Configuration:<\/strong>\u00a0The neutral Ruthenium atom (Ru, atomic number 44) has an exceptional electron configuration of [Kr] 5s\u00b9 4d\u2077. To form the Ru\u00b3\u207a ion, three electrons are removed: the one 5s electron first, followed by two 4d electrons. This leaves a configuration of [Kr] 4d\u2075.<\/li>\n\n\n\n<li><strong>Orbital Diagram (for 4d electrons):<\/strong>\u00a0A half-filled d-subshell has one electron in each orbital.\n<ul class=\"wp-block-list\">\n<li>4d: [\u2191] [\u2191] [\u2191] [\u2191] [\u2191]<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Magnetic Property:<\/strong>\u00a0The ion has five unpaired electrons, so\u00a0<strong>Ru\u00b3\u207a is paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>To determine if an ion is paramagnetic or diamagnetic, we must first find its electron configuration and create an orbital diagram for its valence electrons. Paramagnetic species possess one or more unpaired electrons and are attracted to an external magnetic field. Diamagnetic species have all their electrons paired and are weakly repelled by a magnetic field.<\/p>\n\n\n\n<p>For transition metal cations, electrons are first removed from the s-orbital of the highest principal quantum number (n), and then from the (n-1)d orbital.<\/p>\n\n\n\n<p>The Niobium(V) ion, Nb\u2075\u207a, is formed from a neutral Nb atom ([Kr] 5s\u00b9 4d\u2074) by removing all five of its valence electrons. This results in the stable electron configuration of Krypton ([Kr]), where all electron orbitals are completely filled. With no unpaired electrons, Nb\u2075\u207a is diamagnetic.<\/p>\n\n\n\n<p>The Vanadium(II) ion, V\u00b2\u207a, is formed from a neutral V atom ([Ar] 4s\u00b2 3d\u00b3) by removing the two 4s electrons. This leaves a configuration of [Ar] 3d\u00b3. The orbital diagram for the 3d subshell shows three unpaired electrons, each in a separate orbital as dictated by Hund\u2019s rule. Therefore, V\u00b2\u207a is paramagnetic.<\/p>\n\n\n\n<p>The Palladium(II) ion, Pd\u00b2\u207a, originates from a neutral Pd atom, which has an exceptional configuration of [Kr] 4d\u00b9\u2070. Removing two electrons gives the ion a [Kr] 4d\u2078 configuration. Its orbital diagram shows three filled d-orbitals and two d-orbitals with a single, unpaired electron each. The presence of these two unpaired electrons makes Pd\u00b2\u207a paramagnetic.<\/p>\n\n\n\n<p>Finally, the Ruthenium(III) ion, Ru\u00b3\u207a, comes from a neutral Ru atom ([Kr] 5s\u00b9 4d\u2077). Removing three electrons (one from 5s, two from 4d) yields a [Kr] 4d\u2075 configuration. This corresponds to a half-filled d-subshell, with five unpaired electrons. Consequently, Ru\u00b3\u207a is strongly paramagnetic.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-40.jpeg\" alt=\"\" class=\"wp-image-251259\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic. a. N2+ b. V2+ c. Pd2+ d. Ru? The Correct Answer and Explanation is: Here are the orbital diagrams and magnetic properties for each ion. a. Niobium(V) ion, Nb\u2075\u207a b. Vanadium(II) ion, V\u00b2\u207a c. Palladium(II) ion, Pd\u00b2\u207a d. Ruthenium(III) ion, [&hellip;]<\/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-251254","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251254","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=251254"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251254\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=251254"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=251254"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=251254"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}