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2.6 Resonance and Formal Charge 2.7 VSEPR 2 Oct 2019 Agenda Per 2

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2.6 Resonance and Formal Charge 2.7 VSEPR 2 Oct 2019 Agenda Per 2 2.6 Resonance and Formal Charge 2.7 VSEPR 2 Oct 2019 Agenda Per 2 ● Topic 2.5 Lewis Diagrams exceptions ● Topic 2.6 Formal Charge and Resonance structures ● Topic 2.7 VSEPR and molecular shape - make a start 2 Oct 2019 Agenda Per 4 ● Topic 2.6 Resonance and Formal Charge - finish up - why resonance important ● VSEPR and molecular shape Writing Lewis Structures 1) Add the valence electrons 3- from all the atoms. PO4 Ve = 5 + 4(6) +3 = 32 2) Use a pair of electrons to form a bond between each O pair of bound atoms. 3) Arrange the remaining O - P - O electrons to satisfy the duet rule for hydrogen O and the octet rule for the Re = 32 - 8 = 24 second row elements. Writing Lewis Structures 1) Add the valence electrons PO 3- Ve = 5 + 4(6) +3 = from all the atoms. 4 2) UseHang a pair on of a electrons to minute - 32 form a bond between each doesn’t P O pairhave of bound 10 atoms. 3) Arrangeshared the remaining O - P - O electronselectrons??? to satisfy the duet?!! rule for hydrogen O and the octet rule for the second row elements. Re = 32 - 8 = 24 Writing Lewis Structures 1) Add the valence electrons PO 3- Ve = 5 + 4(6) +3 = from all the atoms. 4 2) UseHang a pair on of a electrons to minute - 32 form a bond between each doesn’t P O pairhave of bound 10 atoms. 3) Arrangeshared the remaining O - P - O electronselectrons??? to satisfy the duet?!! rule for hydrogen O and the octet rule for the second row elements. Re = 32 - 8 = 24 Exceptions to the Octet Rule p. 358 Boron - tends to form compounds in which the boron has fewer than eight electrons around it BF3(g) electron deficient Reacts violently (energetically) with electron rich molecules such as water and ammonia - which have lone pairs. Exceptions to the Octet Rule p. 358 Elements in period 3 and beyond exceed the octet rule. SF6 6 + 6(7) = 48 valence electrons 6 single bonds = 12 electrons 48 - 12 = 36, 3 lone pairs on each of the 6 F atoms. Exceeding the Octet rule Structure uses all 48 valence electrons F has 8 - obeys octet rule BUT S has 12 electrons around it... Exceeding the Octet rule What valence orbitals are available for second and third row elements? 2s2p 3s3p3d LE model assumes empty 3d orbitals can be used to accommodate extra electrons. Lewis Structures: Comments about Octet Rule p. 360 - Second row elements C, N, O, and F should always be assumed to obey the octet rule - B and Be often have fewer than eight electrons around them in their compounds. Electron deficient compounds - very reactive - 3rd row and heavier elements often satisfy the octet rule, but can exceed the octet rule whether actually use d orbitals is a matter of controversy beyond scope of this course Lewis Structures: Comments about Octet Rule p. 360 - so what do we do? - Satisfy the octet rule for the atoms first. If electrons remain, place them on the elements that have available d orbitals (elements in Period 3 or beyond). Odd-Electron Molecules LE model is based on pairs of electrons - does not cover odd electron cases seamlessly. NO 5 + 6 = 11 electrons More detailed Rules for drawing Lewis structures 1. Count the total valence electrons (Ve): Determine the total number of valence electrons in the molecule or ion according to the chemical formula. If the molecule is charged, add an electron for each negative charge and subtract an electron for each positive charge. 2. Draw a symmetrical skeleton structure: Draw a symmetrical skeletal structure of the molecule or ion by forming a single bond between the central atom and each outside atom. NOTE: each bond uses two valence electrons. Central atoms are usually the atoms with highest number of valence electrons, the largest atoms, or the least electronegative atom. H and the halogens are usually outside atoms. Do not put more than four atoms around a central atom unless the central atom is third period or lower. 3. Count the total number bonding of electrons (Be): Each bond or line (bonding electrons) represents two electrons. 4. Subtract the bonding electrons (Be) from valence electron (Ve): Ve – Be = Re (Remaining electrons). 5. Assign the remaining electrons (Re) to the atoms to satisfy the octet rule (for H, duet rule): Start the assignment with the terminal atoms (outside atoms) then the central atom(s). 6. Keep collapsing pair of electrons to form multiple bonds until octet rule is satisfied: You could form an additional bond between two atoms by collapsing one pair of electrons from the adjacent atom to satisfy the octet rule. Formal Charge Molecules or polyatomic ions containing atoms that can exceed the octet rule often have many nonequivalent Lewis structures that all comply with the rules for writing Lewis structures. How do we decide which structure best describes the actual bonding? Formal Charge Estimate the “charge” on each atom in the various possible Lewis structures and use these to select most appropriate structure(s). Formal Charge of an atom in a molecules is the difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule. To determine Formal Charge We need to know 1) The number of valence electrons on the free neutral atoms (which has zero net charge because the number of electrons equals the number of protons). 2) The number of valence electrons “belonging” to the atom in a molecule (lone pair electrons belong entirely to the atom). 3) Shared electrons are divided equally between 2 sharing atoms. SO 2- 4 Valence electrons O 6 + 4(6) + 2 = 32 O S O 32 - 8 = 24 Each O gets 3 lone O pairs SO 2- 4 To Find formal charge on each O 6 lone pair electrons + ½ (2) electrons shared in the bond 6 + ½(2) = 7 Formal charge = 6 - 7 = -1 SO 2- 4 For S No lone pairs 1- ½(8) electrons shared 1- 2+ in the 4 single bonds 1- = 4 1- Formal charge = 6 - 4 = +2 SO 2- 4 O with single bonds 2- 6 lone pair electrons + ½(2) electrons shared 1- 1- in the bond 6 + ½(2) = 7 Formal charge = 6 - 7 = -1 SO 2- 4 O with double bonds 2- 4 lone pair electrons + ½(4) electrons shared 1- 1- in the bond 4 + ½(4) = 6 Formal charge = 6 - 6 = 0 SO 2- 4 S 2- No lone pairs + ½(12) electrons shared in the 1- 1- bonds = 6 Formal charge = 6 - 6 = O Cautions Although formal charges are closer to actual atomic charges in molecules than are oxidation states (see oxidation-reduction reactions notes later), formal charges still provide only estimates of charge and should not be taken as actual atomic charges. Also, the evaluation of Lewis structures using formal charge ideas can lead to erroneous predictions. Tests based on experiments must be used to make the final decision on the correct description of the bonding in a molecule or polyatomic ion. To use formal charges to evaluate Lewis structures we assume that: 1) Atoms in molecules tend to achieve formal charges as close to zero as possible. 2) Any negative formal charges are expected to reside on the most electronegative atoms. SO 2- 4 2- 1- 1- 1- 1- 1- 2+ 1- S and 2 O Formal O Formal charge = -1 charge = 0 S Formal charge = +2 2 O = -1 SO 2- 4 2- Preferred structure Has lower formal charges and -1 formal charge is located on S and 2 O Formal electronegative Oxygen charge = 0 atoms 2 O = -1 2- SO4 and then look - rotate and 2- 2- ↔ 2 ways of drawing represent 2 different 3D arrangements - called Resonance structures - when more than one valid Lewis structure can be written for a particular molecule. Resonance (8.12) When more than one valid and equivalent Lewis structure is possible for a given molecule. The resulting electron structure of the molecule is given by the average of these resonance structures. The situation is usually represented by double-headed arrows: Notice in resonance structures that ● In all the resonance structures the arrangement of the nuclei is the same ● Only the placement of the electrons differs ● The arrows do not mean that the molecule “flips” from one resonance structure to another ● The arrows simply show that the actual structure is an average of the three resonance structures. Concept of resonance is necessary because ● Localized electron model postulates that electrons are localized (fixed) between a given pair of atoms BUT ● Nature does not really operate this way. ○ Electrons are really delocalized, they can move around the entire molecule. ● The valence electrons in the nitrate ion distribute themselves to provide equivalent N-O bonds. Concept of resonance is necessary to ● Compensate for the defective assumption of the localized electron model. BUT ● The model is so useful that we retain the concept of localized electrons and add resonance to allow the model to treat species such as nitrate ions. If bond lengths are shorter in molecular entities than predicted by LE theory atone, then resonance must be making the PE lower (a more stable state). Describe the electron arrangement in the nitrite - (NO2 ) anion using the localized electron model. - NO2 Ve = 5 + 2(6) + 1 = 18 Describe the electron arrangement in the nitrite - (NO2 ) anion using the localized electron model. - NO2 Ve = 5 + 2(6) + 1 = 18 O - N - O Re = 18 - 4 = 14 Describe the electron arrangement in the nitrite - (NO2 ) anion using the localized electron model.
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