{"id":226494,"date":"2025-06-05T10:15:00","date_gmt":"2025-06-05T10:15:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226494"},"modified":"2025-06-05T10:15:03","modified_gmt":"2025-06-05T10:15:03","slug":"when-lead-iv-combines-with-nitrogen-valence-3-the-resulting-product-lead-iv-nitride-is-most-accurately-written-as-pb3n4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/when-lead-iv-combines-with-nitrogen-valence-3-the-resulting-product-lead-iv-nitride-is-most-accurately-written-as-pb3n4\/","title":{"rendered":"When lead (IV) combines with nitrogen (valence = -3), the resulting product, lead (IV) nitride, is most accurately written as Pb3N4"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">When lead (IV) combines with nitrogen (valence = -3), the resulting product, lead (IV) nitride, is most accurately written as Pb3N4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">The correct formula for <strong>lead (IV) nitride<\/strong> is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Pb\u2083N\u2084<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the correct chemical formula for a compound formed between lead (IV) and nitrogen, we need to consider the charges (valencies) of both ions and balance them so the compound is electrically neutral.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Determine the charges<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lead (IV)<\/strong> means lead is in the +4 oxidation state. Its ion is <strong>Pb\u2074\u207a<\/strong>.<\/li>\n\n\n\n<li>Nitrogen typically forms a nitride ion, <strong>N\u00b3\u207b<\/strong>, because it gains 3 electrons to complete its octet.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Step 2: Use the crisscross method<\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">To balance the charges, we use the <strong>crisscross method<\/strong>, where the magnitude of each ion\u2019s charge becomes the subscript of the other ion.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pb\u2074\u207a \u2192 subscript becomes 3<\/li>\n\n\n\n<li>N\u00b3\u207b \u2192 subscript becomes 4<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">This gives us: <strong>Pb\u2083N\u2084<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This means:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 lead (IV) ions each with a +4 charge contribute a total of <strong>+12<\/strong><\/li>\n\n\n\n<li>4 nitride ions each with a -3 charge contribute a total of <strong>-12<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So, total charge: <strong>+12 + (-12) = 0<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The compound is electrically neutral, which is required for all stable ionic compounds.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Naming the Compound<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Since lead is a transition metal and can have multiple oxidation states, we use Roman numerals to indicate its oxidation state. In this case, lead is in the <strong>+4<\/strong> state, so the correct name is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Lead (IV) nitride<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Lead (IV) ion = Pb\u2074\u207a<\/li>\n\n\n\n<li>Nitride ion = N\u00b3\u207b<\/li>\n\n\n\n<li>Balanced formula = <strong>Pb\u2083N\u2084<\/strong><\/li>\n\n\n\n<li>Name = <strong>Lead (IV) nitride<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">This correctly reflects the chemical combination of lead in a +4 state with nitrogen in a -3 state, forming a neutral ionic compound.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-442.jpeg\" alt=\"\" class=\"wp-image-226495\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When lead (IV) combines with nitrogen (valence = -3), the resulting product, lead (IV) nitride, is most accurately written as Pb3N4 The Correct Answer and Explanation is: The correct formula for lead (IV) nitride is: Pb\u2083N\u2084 Explanation To determine the correct chemical formula for a compound formed between lead (IV) and nitrogen, we need to [&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 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