Lecture 21, Chapters 8.1 - 8.3 Vocab

25 July 2022
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Simple Polarity Example
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-If the positive end and negative end are apart and different, then it is polar -If the positive end and negative end are on top of each other and in the center, then it is nonpolar
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Amplitude of Polarity
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Described by dipole moment(µ = qr), For µ is the dipole moment, q is the overall charge of molecule and r is the distance between the charges, If dipole moment increases, then the polarity of the molecule increases as well
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When r(distance between charges) = 0
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µ is always equal to 0, So the object is nonpolar
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When q(overall charge) = 0
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µ is always equal to 0 too, So the object is nonpolar
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When r(distance between charges) ≠ 0
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If q equals 0, it is nonpolar If q does not equal 0, it is polar
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When q(overall charge) ≠ 0
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If r equals 0, it is nonpolar If r does not equal 0, it is polar
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Polarity of a Bond
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-A bond is the attraction between two atoms -r is never equal to 0, because there is a distance between the two atoms in a molecule -We need to look at q, need to look at EN or the ability for the atoms to attract electrons If ΔEN = 0(neutral), each atom has the same EN, then the atoms equally share the electrons, so the bond is nonpolar, These bonds are between two of the same atoms If ΔEN is not equal to 0, each atom has a different EN, Each atom has a partial charge, The electrons will be closer to the atom with the higher EN and partial negative charge(δ-), which would result in a polar bond
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Polarity of Molecules(ΔEN ≠ 0)
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This means q does not equal 0 for most molecules, except for molecules with the same atom such as S8, but we are disregarding them: -We need to look at the quantity of r(distance) -Some atoms will be partially positive charged(δ+) and some will be partially negative charged(δ-) We have too look at the arrangement of the electrons, We can identify: -The center of the positive charges -The center of the negative charges We have to look at the geometry of the molecules: -If the positive center is superimposed on the negative center, right on top, then r = 0, and it is a nonpolar molecule -If the positive center is separated from the negative center, then r is not equal to 0, and it is a polar molecule Remember Polarity is always determined by: µ = qr
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Polarity of XeF
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-We don't even need to do the Lewis Structure, if the molecule is made up of two different atoms, then it is polar, F has the higher EN so would attract the electrons more and they would be closer to its center than Xe -Remember, A molecule can contain multiple bonds polar bonds, but the molecule can still be nonpolar because of its arrangement
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Polarity of XeF4
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Lewis Structure: 1.) 36 total electrons 2.) F F-Xe-F F 3.) •• ••F•• •• ••F••-Xe-••F•• •• •• ••F•• 4.) •• ••F•• •• ••F••-••Xe••-••F•• •• •• ••F•• -Xe has 6 electron groups, It electron geometry would be Octahedral, but remember when it has lone pairs, lone pairs take up space(but less than bonds), So we want them as far as possible, So the 2 lone pairs will be on opposite sides(top and bottom) in the molecular geometry. -Xe has 6 electron groups, 4 bonding pairs, and 2 lone pairs, So its Molecule Geometry would be Square Planar -Because of the polar bonds Xe is δ+ and the Fs are δ-, but Since they are all in the same plane, the bonding pairs of F cancel out, and the lone pairs cancel out, Since the positive center in Xe and the negative center in F are on top of each other and cancel each other out, It is nonpolar, It has all polar bonds but because of the arrangement, the molecule is nonpolar
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Is XeF3Cl polar or nonpolar?
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Lewis Structure: 1.) 36 total electrons 2.) •• ••Cl•• •• ••F••-Xe-••F•• •• •• ••F•• 3.) •• ••Cl•• •• ••F••-••Xe••-••F•• •• •• ••F•• -Xe has 6 electron groups, 4 bonding pairs, and 2 lone pairs, So its Molecular Geometry is Square Planar too -It has all polar bonds in the same plane, but they do not cancel out because the electrons are more attracted to the F bonds than the Cl bonds(F has a higher EN than Cl), They all have partial negative charges and Xe has a partial positive charge, but the electrons are pulled closer to the F atoms then the Cl atom, the Cl atom's bond is shorter and not as strong as the Fs, which means the negative charge is not in the center and is more towards the F that is opposite of the Cl, The sum of charges, q, is not equal to 0, so it is a Polar molecule
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Is the following XeF2Cl2 polar or nonpolar?
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Given this Lewis Structure: •• ••F•• •• ••F••-••Xe••-••Cl•• •• ••Cl •• •• -Because of the 6 electron groups, 4 bonding pairs and 2 lone pairs, Its Molecular Geometry is Square Planar, It is Polar because the Cl bonds are shorter because the Fs are opposite of them and have a stronger pull and attraction to the electrons(higher EN), The overall negative charged center is not on top of Xe and is closer to the Fs
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Is the following XeF2Cl2 polar or nonpolar?, This time however both Fs are oppoiste each other and both Cls are opposite from each other in their arrangement of the molecule
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Given this Lewis Structure: •• ••F•• •• ••Cl••-••Xe••-••Cl•• •• •• ••F•• -Remember, Polarity is determined by the structure of molecules, Because of the 6 electron groups, 4 bonding pairs and 2 lone pairs, Its Molecular Geometry is Square Planar, But it is now Nonpolar because of the arrangement, the Cls and Fs are directly opposite of each other(same ENs, so cancel out charges), The overall negative charged center is on top of the positive charged center in Xe, so it is nonpolar
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Is H2O polar or nonpolar?
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Lewis Structure: 1.) 8 total valence electrons 2.) H-O-H 3.) H-••O••-H -O has 4 electron groups, 2 bonding pairs, and 2 lone pairs, Its Electron Geometry is Tetrahedral, and its Molecular Geometry is Bent, which means the H atoms are not directly across from each other so don't cancel out, O has a higher EN so it is partially negative charged and the Hs are partially positive charged, Since the Hs are not directly opposite of each other, the positive center is between them and not on top of the negative charged center on the O, so the molecule is Polar
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Which of the following molecules is polar? BeCl2 CCl4 CO2 SF2
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SF2 Lewis Structure: 1.) 20 total valence electrons 2.) F-S-F 3.) •• ••F••-S-••F•• •• 4.) •• ••F••-••S••-••F•• •• -S has 4 electron groups, 2 bonding pairs, and 2 lone pairs, So its Molecular Geometry is Bent, The Fs have a partial negative charge(higher EN) but are not directly across from each other so the negative center is not on top of the positive center of the S atom, it is between the F atoms -Therefore, SF2 is polar, similar to H2O -The rest of the molecules have their positive charged centers and negative charged centers on top of each other, so are Nonpolar
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Valence Orbital Bond Theory
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-Have to look at the shapes of orbitals -s has 1 orbital, ball-shaped -p has 3 orbitals, dumbbell-shaped, but the orientation is different(either point to X, Y, or Z), the angle between them is 90° -The 3 p orbitals look like a dumbbell going left and right on the X-axis(px), one going up and down on the Y-axis(py), and one going front and back on the Z-axis(pz) -d has 5 orbitals and f has 7 orbitals and are much more complex When two 1s orbitals(ex. H) come together: -When they are sufficiently far apart, they have no interaction -Atoms begin to interact as they move closer together -When they bond, there is a optimum distance to achieve the lowest overall energy of the system(this is always preferred), this is the H-H bond length -They are now bonded, and at the overlapping bond, there is now two electrons, one spinning up and one spinning down, they now have a complete shell together in the covalent bond -They come in together and are overlapping, more dense, so we have a higher potential of finding the electrons in the overlapping bond
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Two Types in Bond Overlapping
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Sigma(σ)-bond: end-to-end overlapping -such as in H2(s-s orbital), HCl(s-p orbital), and F2 (p-p orbital) bonding with one electron Pi(π)-bond: side-to-side overlapping -there is two p-p orbital bonds and the sides of one p-p bond connect to the sides of the other p-p bond to form this bond
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Hybrid Atomic Orbitals
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O2 -The Lewis Structure is •• ••O=O•• •• -The electron configuration of O is [He]2s2 2p4 -The 2p orbital for the first O would have three electrons spinning up in each orbital(pX, pY, and pZ) and would have one electron spinning down in the pX orbital -Remember between each orbital the angle is 90°, pX, pY, and pZ are perpendicular to each other -The 2p orbital for the second O would be the opposite, It would have three electrons spinning up in each orbital(pX, pY, and pZ) but would have the last electron spinning down in the pZ orbital -When both Os bond, their pY orbitals form an end-to-end bond(σ-bond), and their pZ orbitals form a side-to-side bond(π-bond), the pX orbitals do not bond -So our new theory, The Valence Orbital Bond Theory says that these orbitals form angles of 90°, but the Molecular Geometry says the shape would be Trigonal Planar(two invisible lone pairs would be Linear but same still has same angles as Trigonal Planar) and would say that the angles between the orbitals are 120° -Which angle is preferred?, The Molecular Geometry Angle is preferred, Why, Because It is closer to the actual angle, Need to Hybridize orbitals to get these angles in the molecular structure
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sp3 Hybridization
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-To Hybridized Orbitals means to mix two orbitals together -Say We Have CH4 -The electron configuration of C would be [He] 2s2 2p2, It would have One 2s orbital and One 2p orbital, The 2s orbital has one electron spinning up and one electron spinning down, The 2p orbital has two electrons spinning up, one in 2pX and one in 2pY, and zero in the 2pZ, So it has a total of Four valence electrons -When the orbitals mix together, they generate a new, hybridized orbital, sp3 hybrid -The energy of the sp3 hybrid is between the 2s and 2p orbital but is closer to that of 2p but slighlty less, Remember atoms want the lowest energy to be most stable, and achieve this optimum distance to do so -After mixing the orbitals, the Molecular Geometry is Tetrahedral(remember, orbitals become hybridized to resemble molecular geometry and have the same angles, the ones left over are still at their original 90° p orbitals) but there are none left unhybridized in this type, four sp3 hybridized orbitals, that are all equal in energy -There are still four valence electrons but they are rearranged among the four sp3 orbitals, one spinning up in each -After hybridization, the four electrons are split up into each orbital, One in the s orbital, One in the pX orbital, One in the pY orbital, and One in the pZ orbital, which make up the hybridized sp3 orbital, Now C can overlap with the four Hs(they each have one electron in the s orbital and each can fill the four sp3 orbitals of C to completely fill)
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sp Hybridization
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-Lewis Structure for BeCl2 is: •• ••Cl••-Be-••Cl•• •• -The atomic orbital of Be looks like this, It is [He] 2s2, It has both electrons in the 2s orbital, One spinning up and one spinning down, Its 2p orbital is empty but when hybridizing it wants it to fill the second electron in, so each orbital can overlap with that of the bonding atom -After hybridization, It steals the 2pX orbital from the 2p orbital to form a sp hybrid, there are two 2p orbitals(pY and pZ) left over that are unhybridized, the new hybridized orbital still has 2 electrons, But one is spinning up in the s and one is spinning up in the pX to form the two sp hybridized orbitals -The Molecular Geometry of the hybridized orbitals is linear(2 groups), so the sp hybridized orbitals form an angle of 180° because they are linear to each other, but the two 2p orbitals that are left are still perpindicular(90°) -The two new sp orbitals can now overlap with an electron from each Cl orbital to fill it
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sp2 Hybridization
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-The Lewis Structure for BF3 is: •• ••F•• •• ••F••-B-••F•• •• -The Molecular Geometry is Trigonal Planar(3 bonding groups) -The atomic orbitals in B before mixing looks like [He] 2s2 2p1, Two electrons spinning up and down in 2s orbital and One electron spinning up in the 2pX orbital, for a total of 3 valence electrons -The hybridized orbitals wants to grab the 2pX and 2pY orbitals so that the 3 valence electrons can be split up between each, forming the three sp2 orbitals, with one unused pZ orbital left over -The angle between the three hybridized sp2 orbitals is 120° because of the geometry, and the remaining 2pZ orbital is still at 90° and perpindicular, The hybridized orbitals can now overlap with the three F's orbitals
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sp3d Hybridization
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-The Lewis Structure for PCl5 is: •• ••Cl•• •• ••Cl••-P-••Cl•• •• •• ••Cl•• ••Cl•• •• -The atomic orbitals of P is [Ne] 3s2 3p3, It has a total of Five valence electrons with two spinning up and down in the 3s orbital, and three in the 3p orbital with one spinning up in the pX, pY, and pZ orbitals -When hybridizing, it wants to split up the Five valence electrons between the s, pX, pY, pZ, and now one d orbital to form five sp3d hybridized(each containing one electron spinning up orbitals(each containing one electron spinning up) with four d orbitals left unhybridized(still at 90°) -The geometry for the molecule is Trigonal Bipyramidal and uses the sp3d hybridization(5 electron groups) to allow overlap with the orbitals of the five Cl atoms, and for bonding to occur
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sp3d2 Hybridization
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-SF6 has a Lewis Structure of: •• ••F•• ••F•• •• •• ••F••-S-••F•• •• •• ••F•• ••F•• •• -It would form a shape of Octahedral(six electron groups) -Before Hybridization, S would have an electron configuration of [Ne] 2s2 2p4, With an electron spinning up and down in the s orbital, and an electron spinning up and down in the pX orbital, and one electron spinning up in the pY and pZ orbitals, for a total of 6 valence electrons -When hybridizing it takes two d orbitals to form six hybridized sp3d2 orbitals so it can split up its 6 electrons, with three d orbitals left over and unused(still at 90°) -It now is set up to overlap with the orbitals of the 6 F atoms
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Hybridization Table
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Two Electron Groups -Ex. BeCl2 -Electron Geometry: Linear -Atomic Orbitals Mixed: One s, One p -Hybridized Orbitals Formed: Two sp -Unhybridized Orbitals Left Over: Two p Three Electron Groups -Ex. BF3 -Electron Geometry: Trigonal Planar -Atomic Orbitals Mixed: One s, Two p -Hybridized Orbitals Formed: Three sp2 -Unhybridized Orbitals Left Over: One p Four Electron Groups -Ex. CH4 -Electron Geometry: Tetrahedral -Atomic Orbitals Mixed: One s, Three p -Hybridized Orbitals Formed: Four sp3 -Unhybridized Orbitals Left Over: None Five Electron Groups -Ex. PCl5 -Electron Geometry: Trigonal Bipyramidal -Atomic Orbitals Mixed: One s, Three p, One d -Hybridized Orbitals Formed: Five sp3d -Unhybridized Orbitals Left Over: Four d Six Electron Groups -Ex. SF6 -Electron Geometry: Octahedral -Atomic Orbitals Mixed: One s, Three p, Two d -Hybridized Orbitals Formed: Six sp3d2 -Unhybridized Orbitals Left Over: Three d
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The Lewis Structure of an Atom is Very Important
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It lets us know how many electron groups are in the central atom: -Knowing the number of electron groups helps us determine the shape through molecular geometry, which can be used to predict the polarity of a molecule, and can help us predict some other properties as well -Knowing the number of electron groups also helps us determine the type of hybridization that is used by the central atom(it only happens when needed)
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What is the Expected Hybridization of O in H2O?
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-The Lewis Structure of H2O is H-••O••-H -It has four electron groups, two bonding pairs and two lone pairs, Its electron geometry would be Tetrahedral, but its molecular geometry would be Bent(so it is polar) -Since it has four electron groups, it would have a hybridization of sp3 -It would have four sp3 hybridized orbitals with no unhybridized orbitals left over
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Multiple Bonds
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There are three different types of bonds: -A single bond which has one σ-bond -A double bond which has one σ-bond and one π-bond -A triple bond which has one σ-bond and two π-bonds
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Single Bond
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-For example, H3C-CH3 -Its Lewis Structure is: H H H-C-C-H H H -Both Cs have four hybridized sp3 orbitals(four electron groups) and are both completely full with no unhybridized orbitals left over -Because they have four electron groups, they are Tetrahedral and form overlapping angles of 109.5° -There is no side-to-side(π) bonds because there are no unhybridized orbitals so this molecule only contains a single bond(one σ-bond) -Each C forms three σ-bonds with three of their sp3 orbitals with the three Hs(s1) and their last sp3 hybridized orbitals form a σ-bond bond together
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Double Bond
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-For example, H H C=C H H -Each C has three electron groups so each has three hybridized orbitals of sp2 with one p orbital left unhybridized -Its geometry is Trigonal Planar(three electron groups), Each sp2 has an angle of 120° and the p orbital left over has an angle of 90° -Each C forms a σ-bond with two Hs with two of their sp2 hybridized orbitals -When the two Cs interact their last sp2 hybridized orbitals form an σ-bond together, and their left over p orbitals form a π-bond together and overlap at right angles
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Triple Bond
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-For example, H-C≡C-H -Each C has two electron groups so each has two hybridized orbitals of sp with two p orbitals left unhybridized -Its geometry is Linear(two electron groups), Each sp angle forms an angle of 180° and the two p orbitals left over form angles of 90° -Each C uses one sp orbital to form a σ-bond with the H atom, and their last sp orbital to form a σ-bond together, they each have two p unhybridized orbitals left which form two π-bond side-by-side together at right angles
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Which of the following Contains Both Sigma and Pi Bonds? Nitrate ion, NO3- Carbon tetrachloride, CCl4 Water, H2O None of the above
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-NO 3- would have a Lewis Structure of: [•• ••O••-N=O••]- •• ••O•• -N and an O form a double bond which means they would have a σ-bond and a π-bond overlapping with each other -The other two molecules only contain single bonds so would only overlap with a σ-bond together
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Valence Bond Theory Definition
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Describes a covalent bond as the overlap of half-filled orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms, A covalent bond results when two conditions are met: 1.) An orbital of one atom overlaps an orbital on a second atom 2.) The single electrons in each orbital combine to form an electron pair -The mutual attraction between this negatively charged electron pair and the two atoms' positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond -The strength of a covalent bond depends on the extent of overlap of the orbitals involved, Orbitals that overlap extensively form bonds that are stronger than those that have less overlap -Two atoms keep coming closer until a bond is formed, giving it the least energy possible and most stable form
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Overlap
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Orbits on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space
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Sigma(σ) Bonds
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A covalent bond in which the electron density is concentrated in the region along the internucleuar axis(end-to-end), that is, a line between the nuclei would pass through the center of the overlap region -Single bonds in Lewis Structures are described as these in valence bond theory -These are stronger and have more overlap than π-bonds, but only one of these can be in a location(first one formed) compared to one or two π-bonds that are formed afterwards in double and triple bonds
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Pi(Ï€) Bonds
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A type of covalent bond that results from the side-by-side overlap of two p orbitals, The regions of orbital overlap lie on opposite sides of the internucular axis, Along the axis itself, there is a node, that is, a plane with no probability of finding an electron -Multiple bonds contain these and in addition to their σ-bond
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Hybridization Definition
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The process of combining wave functions for atomic orbitals, It is mathematically accomplished by the linear combination of atomic orbitals, The new orbitals that result are called hybrid orbitals -Valence Bond Theory must include a hybridization component to give accurate predictions -The following ideas are important in understanding hybridization: 1.) Hybrid orbitals do not exist in isolated atoms, They are formed only in covalently bonded atoms 2.) Hybrid orbitals have shapes and orientations that are very different from those of atomic orbitals in isolated atoms 3.) A set of hybrid orbitals is generated by combining atomic orbitals, The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set 4.) All orbitals in a set of hybrid orbitals are equivalent in shape and energy 5.) The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory 6.) Hybrid orbitals overlap to form σ-bonds, Unhybridized orbitals overlap to form π-bonds
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sp Hybridization Definition
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-The Be atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms, There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be-Cl bonds, To accommodate these two electron domains, two of the Be atom's four valence orbitals will mix to yield two hybrid orbitals, This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry, In this figure, the set of sp orbitals appear similar in shape to the original p orbital, but there is an important difference, The number of atomic orbitals combined always equals the number of hybrid orbitals formed, The p orbital is one orbital that can hold up to two electrons, The sp set is two equivalent orbitals that point 180° from each other, The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled, In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ-bonds(there are two unhybridized 2p orbitals left over at a right angle), When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals, The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons, Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be-Cl bonds
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sp2 Hybridization Definition
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The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of sp2 hybrid orbitals and one unhybridized p orbital, This arrangement results from sp2 hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry, The observed structure of the borane molecule, BH3, suggests sp2 hybridization for boron in this compound, The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms, We can illustrate this comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH3, as shown in the orbital energy level diagram, We redistribute the three valence electrons of the boron atom in the three sp2 hybrid orbitals, and each boron electron pairs with a hydrogen electron when B-H bonds form
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sp3 Hybridization Definition
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The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four sp3 hybrid orbitals, The hybrid results from the mixing of one s orbital and all three p orbitals that produces four identical sp3 hybrid orbitals, Each of these hybrid orbitals point toward a different corner of a tetrahedron, A molecule of methane, CH4, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tertrahedron, The carbon atom in methane exhibits sp3 hybridization, We illustrate the orbitals and electron distribution in an isolated carbon atom in the bonded atom CH4, The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C-H bonds form, In a methane molecule, the 1s orbital of each of the four hydrogen atoms overlaps with one of the four sp3 orbitals of the carbon atom to form a σ-bond, This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH4
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sp3d and sp3d2 Hybridization Definition
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To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals(the s orbital, the three p orbitals, and one of the d orbitals), which gives us five sp3d hybrid orbitals, With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals(the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives us six sp3d2 hybrid orbitals, These hybridizations are only possible for atoms that have d orbitals in their valence subshells(that is, not those in the first or second period) -In a molecule of phosphorus pentachloride, PCl5, there are five P-Cl bonds(thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid, We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp3d2 hybrid orbitals that are involved in the P-Cl bonds -The sulfur atom in sulfur hexafluoride, SF6, exhibits sp3d2 hybridization, A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom, There are no lone pairs of electrons on the central atom, To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and the two of the 3d orbitals form six equivalent sp3d2 hybrid orbitals, each directed toward a different corner of an octahedron
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Assignment of Hybrid Orbitals to Central Atoms
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The hybridization of an atom is determined based on the number of regions of electron density that surround it or electron groups, The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown, These arrangements are identical to those of the electron-pair geometries predicted by the VSEPR theory, VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed, To find the hybridization of a central atom, we can use the following guidelines: 1.) Determine the Lewis Structure of the molecule 2.) Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region(electron groups) 3.) Assign the set of hybridized orbitals that corresponds to this geometry(sp for two groups, sp2 for three groups, sp3 for four groups, sp3d for five groups, and sp3d2 for six groups of electrons on the central atom) -Remember: We only invoke hybridization where it is necessary to explain the observed structures and there angles between their electron orbitals, Sometimes they will display angles that are predicted so we won't need to use hybridized orbitals
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Multiple Bonds Definition
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-In Ethene, it shows that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms, The three bonding regions form a trigonal planar electron-pair geometry, Thus we expect the σ-bonds from each carbon atom are formed using a set of sp2 hybrid orbitals that result from hybridization of two of the 2p orbitals and the 2s orbital, These orbitals form the C-H single bonds and the σ-bond in the C=C double bond, The π-bond in the C=C double bond results from the overlap of the third(remaining) 2p orbital on each carbon atom that is not involved in hybridization, This unhybridized p orbital is perpindicular to the plane of the sp2 hybrid orbitals, Thus the unhybridized 2p orbitals overlap in a side-by-side fashion, above and below the internucleur axis and form a π-bond -In Acetylene, It has molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom, which in a linear molecule, The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ-bond between the carbon atoms, The remaining sp orbitals form σ-bonds with hydrogen atoms, The two unhybridized p orbitals per carbon are positioned such that they overlap side-by-side, hence, form two π-bonds, The two carbon atoms of acetylene are thus bound together by one σ-bond and two π-bonds, giving a triple bond