Write down the electron configuration of neodymium (for Nd Z60). Rewrite it in terms of a noble gas configuration. What family does this element belong to? a) Write down the four quantum numbers of the electrons that are not part of the noble gas configuration. Show your answer in a table. b) Calculate the magnetic moment of the atom. c) Calculate the spectrocopic ground state for the atom. 2. An unknown metal has a density of 21.45 g\/ml. It crystallizes with a compact cubic packing and one edge of the unit cell measures 3.9231 A. a) Calculate the atomic weight of the metal. b) Calculate the atomic radius of the metal.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Electron Configuration<\/strong>: In terms of a noble gas configuration: Family<\/strong>: Neodymium belongs to the Lanthanide family<\/strong>, which is characterized by partially filled (4f) orbitals.<\/p>\n\n\n\n a) Four Quantum Numbers for Electrons Outside the Noble Gas Core<\/strong><\/p>\n\n\n\n The electrons outside the [Xe] core are (4f^4) and (6s^2). For these, the quantum numbers ((n, l, m_l, m_s)) are:<\/p>\n\n\n\n b) Magnetic Moment<\/strong> c) Spectroscopic Ground State<\/strong> a) Calculate Atomic Weight<\/strong> Density ((\\rho)) = 21.45 g\/mL<\/strong>, edge length ((a)) = 3.9231 \u00c5 = 3.9231 \\times 10^{-8} cm<\/strong>.<\/p>\n\n\n\n Volume of the unit cell: Mass of unit cell: Mass of one atom: Atomic weight: The atomic weight is approximately 195 g\/mol<\/strong>, corresponding to platinum (Pt).<\/p>\n\n\n\n b) Calculate Atomic Radius<\/strong> The atomic radius is approximately 1.39 \u00c5<\/strong>.<\/p>\n\n\n\n Neodymium (Nd) is a lanthanide with Z = 60. Its electron configuration highlights the filling of the (4f) orbitals, a characteristic of the lanthanides. The configuration simplifies to ([Xe] 4f^4 6s^2). Lanthanides exhibit unpaired (4f) electrons, contributing to magnetic properties. For Nd, with 4 unpaired electrons, the magnetic moment is (4.90 \\, \\mu_B). Hund\u2019s rules predict a ground state of ({}^5I_4), reflecting high spin multiplicity and orbital angular momentum.<\/p>\n\n\n\n For the unknown metal, its density and FCC structure allow us to calculate its atomic weight. FCC crystals have 4 atoms per unit cell, and using the unit cell’s dimensions and density, the atomic mass is determined as ~195 g\/mol, likely platinum. The atomic radius, derived from FCC geometry, is about 1.39 \u00c5, aligning with typical values for platinum. These calculations showcase the interplay of crystallography, density, and atomic properties in identifying metals.<\/p>\n","protected":false},"excerpt":{"rendered":" Write down the electron configuration of neodymium (for Nd Z60). Rewrite it in terms of a noble gas configuration. What family does this element belong to? a) Write down the four quantum numbers of the electrons that are not part of the noble gas configuration. Show your answer in a table. b) Calculate the magnetic […]<\/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-183892","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183892","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=183892"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183892\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183892"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183892"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183892"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Neodymium (Nd, Z = 60)<\/strong><\/h3>\n\n\n\n
The full electron configuration for neodymium is:
1s\u00b2 2s\u00b2 2p\u2076 3s\u00b2 3p\u2076 3d\u00b9\u2070 4s\u00b2 4p\u2076 4d\u00b9\u2070 5s\u00b2 5p\u2076 6s\u00b2 4f\u2074.<\/p>\n\n\n\n
[Xe] 4f\u2074 6s\u00b2<\/strong>.<\/p>\n\n\n\n
\n\n\n\nElectron<\/th> (n)<\/th> (l)<\/th> (m_l)<\/th> (m_s)<\/th><\/tr><\/thead> (4f^4)<\/td> 4<\/td> 3<\/td> -3 to 3<\/td> \u00b1\u00bd<\/td><\/tr> (6s^2)<\/td> 6<\/td> 0<\/td> 0<\/td> \u00b1\u00bd<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n
The formula for the magnetic moment is:
[
\\mu = \\sqrt{n(n+2)} \\, \\mu_B
]
where (n) is the number of unpaired electrons. Neodymium has 4 unpaired electrons in the (4f) subshell:
[
\\mu = \\sqrt{4(4+2)} = \\sqrt{24} \\approx 4.90 \\, \\mu_B
]
(\\mu_B) is the Bohr magneton.<\/p>\n\n\n\n
\n\n\n\n
The ground state of (4f^4) is determined using Hund\u2019s rules. For (4f^4):<\/p>\n\n\n\n\n
The term symbol is:
[
{}^5I_4
]<\/li>\n<\/ul>\n\n\n\n
\n\n\n\n2. Unknown Metal<\/strong><\/h3>\n\n\n\n
The metal crystallizes in a face-centered cubic (FCC) structure, where 4 atoms occupy each unit cell.<\/p>\n\n\n\n
[
V = a^3 = (3.9231 \\times 10^{-8})^3 \\approx 6.03 \\times 10^{-23} \\, \\text{cm}^3
]<\/p>\n\n\n\n
[
m_{\\text{cell}} = \\rho \\cdot V = 21.45 \\cdot 6.03 \\times 10^{-23} \\approx 1.29 \\times 10^{-21} \\, \\text{g}
]<\/p>\n\n\n\n
[
m_{\\text{atom}} = \\frac{m_{\\text{cell}}}{4} = \\frac{1.29 \\times 10^{-21}}{4} \\approx 3.23 \\times 10^{-22} \\, \\text{g}
]<\/p>\n\n\n\n
[
A = m_{\\text{atom}} \\cdot N_A = 3.23 \\times 10^{-22} \\cdot 6.022 \\times 10^{23} \\approx 195 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\n\n\n\n
For FCC, the relationship between edge length ((a)) and atomic radius ((r)) is:
[
a = 2\\sqrt{2}r \\implies r = \\frac{a}{2\\sqrt{2}} = \\frac{3.9231}{2\\sqrt{2}} \\approx 1.39 \\, \\text{\u00c5}
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/h3>\n\n\n\n