{"id":36448,"date":"2025-06-25T07:23:39","date_gmt":"2025-06-25T07:23:39","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36448"},"modified":"2025-06-25T07:23:41","modified_gmt":"2025-06-25T07:23:41","slug":"what-is-the-hybridization-of-the-central-atom-in-no3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-hybridization-of-the-central-atom-in-no3\/","title":{"rendered":"What is the hybridization of the central atom in NO3"},"content":{"rendered":"\n<p>What is the hybridization of the central atom in NO3? Select one: a. sp b. sp2 c. sp3 d. spd e. sp3d2<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer: b. sp\u00b2<\/strong><\/p>\n\n\n\n<p>The hybridization of the central atom in the nitrate ion (NO\u2083\u207b) is <strong>sp\u00b2<\/strong>.<\/p>\n\n\n\n<p>To determine the hybridization of the central atom, which is nitrogen in this case, begin by drawing the Lewis structure of the nitrate ion. The nitrate ion consists of one nitrogen atom centrally bonded to three oxygen atoms. One of the oxygen atoms forms a double bond with nitrogen, while the other two form single bonds and carry a negative charge each. Due to resonance, the double bond is delocalized, and the three N-O bonds are equivalent.<\/p>\n\n\n\n<p>Now consider the electron regions around the nitrogen atom. Nitrogen forms three sigma bonds with the three oxygen atoms. There are no lone pairs on nitrogen in the nitrate ion. Therefore, the nitrogen atom has three regions of electron density.<\/p>\n\n\n\n<p>According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, an atom with three regions of electron density adopts a <strong>trigonal planar<\/strong> geometry. The hybridization that corresponds to three regions of electron density is <strong>sp\u00b2<\/strong>.<\/p>\n\n\n\n<p>Hybrid orbitals result from the mixing of atomic orbitals to form new orbitals that accommodate bonding. In sp\u00b2 hybridization, one s orbital and two p orbitals mix to form three equivalent sp\u00b2 hybrid orbitals. These orbitals lie in a plane at 120\u00b0 angles, which matches the geometry observed in NO\u2083\u207b.<\/p>\n\n\n\n<p>This sp\u00b2 hybridization helps explain the planar shape and equal bond lengths in the nitrate ion due to resonance. All three N\u2013O bonds in nitrate are identical because the electron density is evenly distributed among them, and the hybrid orbitals enable this delocalization of pi electrons over the structure.<\/p>\n\n\n\n<p>In conclusion, based on the number of regions of electron density and the resulting molecular geometry, the hybridization of the nitrogen atom in NO\u2083\u207b is <strong>sp\u00b2<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the hybridization of the central atom in NO3? Select one: a. sp b. sp2 c. sp3 d. spd e. sp3d2 The correct answer and explanation is: Correct Answer: b. sp\u00b2 The hybridization of the central atom in the nitrate ion (NO\u2083\u207b) is sp\u00b2. To determine the hybridization of the central atom, which is [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-36448","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36448","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=36448"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36448\/revisions"}],"predecessor-version":[{"id":36450,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36448\/revisions\/36450"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36448"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36448"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36448"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}