Correct Answer:<\/strong><\/p>\n\n\n\n Hydrogen has one valence electron, fluorine has seven. They share one pair (single bond), giving hydrogen two electrons (full shell) and fluorine eight (full octet), with three lone pairs left on fluorine.<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n The Lewis structure is the simplest way to represent bonding by showing atoms sharing electrons to achieve stable configurations. HF\u2019s single bond satisfies the duet rule for hydrogen and the octet for fluorine. However, it does not explain energy levels or orbital interactions. The molecular orbital approach shows that bonding in HF involves the mixing of H\u2019s 1s with F\u2019s 2p_z orbital, creating bonding and antibonding orbitals, while fluorine\u2019s 2s and the two 2p orbitals perpendicular to the bond remain nonbonding because they do not have symmetry or suitable energy match to overlap with hydrogen\u2019s 1s orbital. The polarity of HF, revealed both in Lewis and MO descriptions, arises from fluorine\u2019s high electronegativity, which pulls electron density toward itself. This polarity makes HF highly reactive with both electrophiles and nucleophiles: nucleophiles attack the proton (hydrogen), while electrophiles can accept electrons from the fluorine\u2019s lone pairs. This combination of Lewis and MO perspectives gives a comprehensive understanding of HF\u2019s structure, bonding, and reactivity. Draw the Lewis structure for HF. Identify the bond order and discuss the polarity of the molecule based on electronegativity: Draw the molecular orbital (MO) diagram for HF (see Figure 2.14). Compare and contrast the Lewis and MO descriptions of HF. For any orbitals which are nonbonding; describe why they are nonbonding: Explain how an […]<\/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-40483","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40483","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=40483"}],"version-history":[{"count":3,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40483\/revisions"}],"predecessor-version":[{"id":41121,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40483\/revisions\/41121"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40483"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40483"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40483"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
rCopyEdit
H : F\n<\/code><\/pre>\n\n\n\n\n
Since there is one single bond, the bond order is 1.<\/li>\n\n\n\n
Fluorine is much more electronegative than hydrogen (F \u2248 3.98, H \u2248 2.20), causing the shared electrons to be pulled toward fluorine. This creates a partial negative charge on fluorine and a partial positive charge on hydrogen. Therefore, HF is a highly polar molecule with a dipole pointing toward fluorine.<\/li>\n\n\n\n\n
\n
\n
The 2s of fluorine and the 2p_x and 2p_y orbitals remain nonbonding because they have no suitable counterpart in hydrogen\u2019s 1s to interact with (either they are too low in energy like 2s or have no directional overlap like 2p_x and 2p_y).<\/li>\n\n\n\n
The Lewis structure shows a single shared pair of electrons completing octets, but does not account for the nonbonding orbitals explicitly. The MO diagram shows both the bonding interaction and the presence of nonbonding fluorine orbitals, offering a more detailed picture of electron distribution.<\/li>\n\n\n\n
Because fluorine carries a partial negative charge, an electrophile (electron-pair acceptor) would attack the fluorine atom. Conversely, a nucleophile (electron-pair donor) would be attracted to the partially positive hydrogen atom, attacking it because it is electron-deficient.<\/li>\n<\/ol>\n\n\n\n
<\/p>\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1267.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"