Lecture 16_Dipole moment_Lewis Structures_Formal Charge_Intro Resonance

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Oct 30, 2023

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Announcements for October 6 th , 2023 Our objective is to determine how Lewis structures are constructed for complex molecular systems The plan: • Dipole moment and % ionic character • Complex Lewis Structures • Formal charge • Resonance intro Friday reading: Chapter 8, Sections 4 – 5 Next week reading : Chapter 8, Sections 6 – 8; Chapter 2, Section 8 (polyatomic ions); and Chapter 9, Sections 1, 2, 4, 5 • Homework 5 due Monday, October 9 th by 11:59 PM

Dipole moment ( റ 𝜇 ) = charge x distance: The dipole moment ( റ 𝜇 ) of a molecule is a direct measure of molecular polarity Bond dipole EN(H) = 2.1 EN(F) = 4.0 ΔEN = 1.9 (Polar covalent) • Molecules with a net imbalance of charge are polar • ΔEN and molecular shape (later) can be qualitatively used to predict polar vs non-polar • റ 𝜇 is an experimentally measured vector quantity (direction and size) that expresses charge separation (Q) across a distance ( റ 𝑟 ) Q = _____________________= δ റ 𝑟 = ______________________ ΔEN tells us whether a ____________________ exists (ΔEN > 0) റ 𝑟 = 0.92 Å റ 𝜇 = 6.1 x 10 -30 Coulomb-meter (Cm) δ = δ - δ + 𝑄 = റ 𝜇 റ 𝑟 = 6.1𝑥10 −30 ?? 9.2𝑥10 −11 ? An e - has a charge of -1.602x10 -19 C. What is the ‘fractional’ e - charge transfer in HF? റ𝜇 = 𝑄റ 𝑟

We can quantify how far from ideally covalent polar covalent bonds stray using % ionic character % 𝑖??𝑖? = r = 0.92 Å റ 𝜇 = 6.1 x 10 -30 C ∙ m റ 𝜇 = 𝑄റ 𝑟 r = 0.92 Å e - e + = 1.5 x 10 -29 C ∙ m Q = 1.602x10 -19 C Measured dipole moment of HF Ideal dipole moment of point charges Partial e - sharing, mostly ionic interaction Valence orbitals are ‘fluffy’ and distort to share e - density e - density asymmetrically shared Possess permanent dipole moment (= deviation from ideal covalent bond)

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The experimentally measured dipole moment of HCl is 3.6x10 -30 Cm. Calculate the partial charges (in units of the fundamental electronic charge, e - ) on each atom and the % ionic character of the bond. The bond length of HCl is 1.27 Å and the charge of an e - is 1.602x10 -19 C. (1m = 10 10 Å) 1. Rearrange the dipole moment equation and solve for the partial charges in units of C. 2. Convert the partial charge into units of ‘fractional’ electronic charge. δ = ______ δ = ______ 3. Calculate the dipole moment of ideal point charges. 4. Calculate the% ionic character.

The unit of Debye (D) is the experimentally preferred unit of reporting molecular dipole moments. Experiments have shown that the dipole moments of H 3 C — Cl and H 3 C — F are 1.87 D and 1.81 D, respectively. Propose a reason for this result. The Lewis structures for each molecule are given below. (1 D = 3.335x10 -30 Cm) 2. Consider relative EN differences between C and each halogen. Do you expect the parameter from (1) to be greater or smaller for C — F (in comparison to C — Cl)? 1. In the dipole moment equation, which parameter is related to EN difference? 3. Are the dipole moments consistent with your expectations? 4. What parameter remains? Use this to justify the experimental result.

Sharing of e - density always occurs: Bond polarity is a “sliding scale” between mostly ionic to non-polar covalent Δ EN = 2.1 Δ EN = 1.8 Δ EN = 1.5 Δ EN = 1.2 Δ EN = 0.9 Δ EN = 0.5 Δ EN = 0 Ionic Mostly ionic Mostly polar covalent Polar covalent Polar covalent Polar covalent Non-polar covalent

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Lewis dot structures are graphical representations of molecular connectivity What is the Lewis structure for OF 2 ? Step 1: Place the element capable of forming the most bonds in the center and the remaining as far apart geometrically but relative to central atom O: [He] 2s 2 2p 4 F: [He] 2s 2 2p 5 2 more electrons 1 more electron Step 2: Count total number of valence electrons Valence e - count O → F → Step 3: Connect each surrounding atom to the central atom by single bonds and subtract two electrons per bond Total e - = Step 4: Place electrons as lone pairs on surrounding atoms to achieve octet (duet for H only!) Step 5: IF NECESSARY, place remaining electrons as lone pairs on central atoms to achieve octet Bonding e - = LP e - for surrounding atoms = Step 6: IF NECESSARY, form multiple bonds to achieve octet or reduce formal charges 2 bonds 1 bond We can predict connectivity of molecular compounds containing more than one element by knowing: (i) chemical formula and (ii) valence electron count for each contributing atom _______ total valence e - Typically, least EN atom

Molecules with multiple bonds between 2 atoms What is the Lewis structure for H 2 CO ? Step 1: Identify central atom Step 2: Valence electron count Step 3: Connect central atom to bonding atoms Step 4: Place electrons as lone pairs on surrounding atoms to achieve octet (duet for H only!) Step 5: IF NECESSARY, place remaining electrons as lone pairs on central atoms to achieve octet Step 6: IF NECESSARY, form multiple bonds to achieve octet or reduce formal charges

Molecules with more than one central atom What is the Lewis structure for CH 3 OH (H 3 C — OH)? Step 1: Identify central atoms Step 2: Valence electron count Step 3: Connect central atoms to bonding atoms Step 4: Place electrons as lone pairs on surrounding atoms to achieve octet (duet for H only!) Step 5: IF NECESSARY, place remaining electrons as lone pairs on central atoms to achieve octet Step 6: IF NECESSARY, form multiple bonds to achieve octet or reduce formal charges CH 3 OH = Condensed Formula (Atomic symbols listed as they appear structurally; “collapsed”) H 3 C — OH = Condensed Formula showing explicit connection between central atoms

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What is the Lewis structure for CO 2 ? Step 1: Which atom can form the most bonds? Step 2: Valence electron count Step 3: Draw single bonds between central and bonding. Step 4: Place electrons as lone pairs on surrounding atoms to achieve octet (duet for H only!) Step 5: Place remaining electrons as lone pairs on central atoms to achieve octet Step 6: Form multiple bonds to achieve octet or reduce formal charges, if necessary C O O Is this octet abiding structure okay?

Formal charges are fictitious “ionic” charges an atom would have if bonding e - ’s were equally shared Formal charges are used to help make our decision • Cut bonds in ½, split shared e - ’s evenly, and count total valence e - (‘Covalent count’) How do we decide? • Minimize formal charge → Closer or equal to 0 is better! • Negative formal charges are favorable for more EN atoms (e.g., O) if needed Atomic 6 4 6 Covalent Atomic 6 4 6 Covalent 𝐹?𝑟?𝑎? ?ℎ𝑎𝑟𝑔? = ________________ ? − ????? − __________________ ? − ????? Greater ‘electrostatic energy’ Lower ‘electrostatic energy’ (neutral) Formal charge Formal charge

Two plausible Lewis structures for N 2 O are given below. Calculate the formal charges of each atom and identify which corresponds to the ‘best’ representation of the molecular structure. Atomic Covalent FC Atomic Covalent FC Does either structure have all FC’s = 0? Which structure puts the most negative FC on the most EN atom?

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F F EN considerations influence the assignment of formal charge • EN(F) = 4.0 and EN(B) = 2.0 • ΔEN is too big – B is not EN enough to tug from F • No octet for B! 1. Fluorine atoms ALWAYS have 0 or -1 formal charge values BF 3 : B F ? 2. Odd-electron molecules cannot abide by octet rule (1) Reduce formal charges to 0 • Two structures possible (2)Closest octet-abiding structure is ‘best’ NO (11 valence e - ):

Formal charge is a great tool for predicting reactivity Homodiatomic Heterodiatomic polar bond non-polar bond

Formal charges can suggest more than one Lewis structure contributes to the true molecular structure What is the Lewis structure for the cyanate anion, OCN - , a polyatomic ion? Which of these structures is ‘best’? *Sum of formal charges must equal net charge!* Octet abiding structures – use arrows to “push” electrons:

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Resonance structures are a valid set of two or more Lewis structure that describe electronic distribution Ozone (O 3 ) • Involve redistribution of lone pair and “double or triple bond” electrons ONLY • Formal charge rules are guidelines – they help to identify strongest contributing structure! O O O = symbolic transfer of e - ’s throughout molecule O O O −1 +2 −1 ✓ Formal charges reduced ✓ Negative formal charge on more EN atom ✓ Formal charges reduced ✓ Negative formal charge on more EN atom Both forms of O 3 are identical! True structure is an average of equivalent resonance structures ! Experiment shows structures are NOT interconverting! Both O — O bonds identical

True structure is average of all Lewis structures – octet abiding and lowest formal charge structures contribute most 𝑂 − 𝑂 ?𝑂 = Resonance structures indicate _______________ : • e - density spread over multiple covalent bond regions/bonding domains (i.e., lone pairs) Electrons described by orbitals (wavefunctions) – we can average functions! Resonance hybrid Neither O — O bond is a true double or single bond – we must define BOND ORDER • ‘Number’ of chemical bonds between two atoms (1 = single bond, 2 = double bond, etc) • Average of all identical bonds involved in resonance structures Same applies to formal charge (take the average!) ????𝑖?𝑔 𝑂 𝐹? = O — O bond properties between true single and double bond (Ex: H 2 O 2 and O 2 )

When can resonance structures be drawn? 1. When a lone pair is adjacent to a double (or triple) bond: 2. When two double (or triple) bonds are separated by a single bond:

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