Using noble gas notation, write electron configurations for atoms of the following elements. (Express your answer using the noble gas notation. For example, the electron configuration of Li would be entered as [He]2s+.)<\/p>\n\n\n\n
a. Titanium, Ti. Titanium was discovered in Cornwall, Great Britain, by William Gregor in 1791 and named by Martin Heinrich Klaproth for the Titans of Greek mythology. Electron configuration:<\/p>\n\n\n\n
b. Cobalt, Co. Cobalt-based blue pigments (cobalt blue) have been used since ancient times for jewelry and paints, and to impart a distinctive blue tint to glass. Electron configuration:<\/p>\n\n\n\n
c. Zinc, Zn. It is the 24th most abundant element in the Earth’s crust and has five stable isotopes. The most common zinc ore is sphalerite (zinc blende), a zinc sulfide mineral. Electron configuration:<\/p>\n\n\n\n
d. Tin, Sn. The metal was used in the ancient world. Alloys of tin (solder, bronze, and pewter) are important. Electron configuration:<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n a. Titanium (Ti):<\/strong> b. Cobalt (Co):<\/strong> c. Zinc (Zn):<\/strong> d. Tin (Sn):<\/strong> Electron configuration describes the distribution of electrons in an atom\u2019s orbitals. Using noble gas notation streamlines this process by replacing the electron configuration of the nearest previous noble gas with its symbol in brackets.<\/p>\n\n\n\n Titanium (Ti)<\/strong> has 22 electrons. The nearest noble gas before Ti is argon (Ar)<\/strong>, which has 18 electrons. The remaining 4 electrons go into the 4s and 3d orbitals: 2 in 4s and 2 in 3d, resulting in Cobalt (Co)<\/strong> has 27 electrons. Starting from [Ar]<\/strong>, which accounts for 18 electrons, we place the next 2 electrons in 4s and the remaining 7 in the 3d orbital, yielding Zinc (Zn)<\/strong>, with 30 electrons, follows the same noble gas core Tin (Sn)<\/strong>, with 50 electrons, uses krypton (Kr)<\/strong> as the core, which covers 36 electrons. The next 14 electrons fill 5s\u00b2 (2), 4d\u00b9\u2070 (10), and 5p\u00b2 (2). Thus, the configuration is These configurations reflect periodic trends and influence each element\u2019s chemical behavior, including bonding, ion formation, and reactivity.<\/p>\n","protected":false},"excerpt":{"rendered":" Using noble gas notation, write electron configurations for atoms of the following elements. (Express your answer using the noble gas notation. For example, the electron configuration of Li would be entered as [He]2s+.) a. Titanium, Ti. Titanium was discovered in Cornwall, Great Britain, by William Gregor in 1791 and named by Martin Heinrich Klaproth for […]<\/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-209588","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209588","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=209588"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209588\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Electron Configurations Using Noble Gas Notation:<\/h3>\n\n\n\n
Atomic number = 22
Electron configuration:<\/strong> [Ar] 4s\u00b2 3d\u00b2<\/code><\/p>\n\n\n\n
Atomic number = 27
Electron configuration:<\/strong> [Ar] 4s\u00b2 3d\u2077<\/code><\/p>\n\n\n\n
Atomic number = 30
Electron configuration:<\/strong> [Ar] 4s\u00b2 3d\u00b9\u2070<\/code><\/p>\n\n\n\n
Atomic number = 50
Electron configuration:<\/strong> [Kr] 5s\u00b2 4d\u00b9\u2070 5p\u00b2<\/code><\/p>\n\n\n\n
\n\n\n\nExplanation (Approx. 300 Words):<\/h3>\n\n\n\n
[Ar] 4s\u00b2 3d\u00b2<\/code>.<\/p>\n\n\n\n[Ar] 4s\u00b2 3d\u2077<\/code>. This configuration reflects cobalt\u2019s partial filling of the 3d subshell, typical of transition metals.<\/p>\n\n\n\n[Ar]<\/code>. After filling 4s with 2 electrons, the 3d subshell is filled with 10 electrons. So, its configuration is [Ar] 4s\u00b2 3d\u00b9\u2070<\/code>, showing a fully filled d-block, which contributes to zinc\u2019s chemical stability and lack of multiple oxidation states.<\/p>\n\n\n\n[Kr] 5s\u00b2 4d\u00b9\u2070 5p\u00b2<\/code>. Tin’s configuration shows it lies in the p-block, and the presence of two 5p electrons corresponds with its common +2 and +4 oxidation states.<\/p>\n\n\n\n