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Topic 9 Chemical Bonding and Lewis Structures

Topic 9 Chemical Bonding and Lewis Structures. The Octet/Duet Rule.

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Topic 9 Chemical Bonding and Lewis Structures

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  1. Topic 9 Chemical Bonding and Lewis Structures

  2. The Octet/Duet Rule The chemical properties of atoms primarily depend on their valence configuration. They seem to have a tendency to be noble-gas-like meaning losing, gaining, or sharing electrons so that they end up being surrounded by the same number of valence electrons as noble gas atoms (8, except He which has 2). The tendency of atoms in a molecule to have eight electrons in their outer shell (two for hydrogen) is called the octet/duet rule.

  3. Lewis Symbols A Lewis symbol is a symbol in which the electrons in the valence shell of an atom or ion are represented by dots placed around the letter symbol of the element. Group VI Group I Group II Group III Group IV Group V Group VII Group VIII . . . . . . . . . . . . . . Ar S Cl Na . . Mg . . . . . . . . . Al Si P . . . . . . . . . . . • Electrons are place singly on each side before they are paired. • Note that the group number on the traditional periodic table indicates the number of valence electrons.

  4. Lewis Symbols for Ions For ions, dots are added or taken away depending on the charge, square brackets are drawn, and the charge is indicated in the upper right hand corner. The magnesium atom has two valence electrons; its Lewis symbol has two dots. Naturally-occurring magnesium ion has a charge of +2; this means two fewer electrons compared to a magnesium atom. Lewis symbol for magnesium ion has no dots and a +2 charge. [ Mg ] 2+ The fluorine atom has seven valence electrons; its Lewis symbol has seven dots. Naturally-occurring fluoride ion has a charge of -1; this means one additional electron compared to a fluorine atom. Lewis symbol for fluoride ion has 8 dots and a -1 charge. . . [ F ]- . . . . . . By following the octet/duet rule, we can say that the atoms end up as ions with a noble-gas-like configuration, or as being isoelectronic with a noble gas.

  5. Lewis Electron-Dot Formulas A Lewis electron-dot formula is an illustration used to represent the transfer of electrons during the formation of an ionic bond. • There is a transfer of electrons from the metal to the nonmetal giving each species a complete octet. An ionic bond involves the attraction of oppositely charged ions for each other. • As an example, let’s look at the transfer of electrons from magnesium to fluorine to form magnesium fluoride.

  6. : F . : : : F Mg . . . : : : : [ F ] [ F ] : : : : : : Lewis Electron-Dot Formulas • The magnesium has two electrons to give, whereas the fluorines have only one “vacancy” each. Consequently, magnesium can accommodate two fluorine atoms. - 2+ - Mg HW 64 code: dot

  7. Electron Configurations of Ions As metals lose electrons to form cations and establish a “noble gas” configuration (octet), the electrons are lost from the valence shell first. • For example, magnesium generally loses two electrons from its 3s subshell to look like neon. [Ne]3s2 [Ne]

  8. For example, zinc generally loses two electrons from its 4s subshell to adopt a “pseudo”-noble gas configuration. [Ar]4s23d10 [Ar]3d10 Electron Configurations of Ions Transition metals also lose electrons from the valence shell first, which is not the last subshell to fill according to the aufbau sequence.

  9. Covalent Bonds When two nonmetals bond, they often share electrons since they have similar attractions for them. This sharing of valence electrons is called a covalent bond. • These atoms will share sufficient numbers of electrons in order to achieve a noble gas electron configuration (octet/duet rule).

  10. . . : + H H H H Lewis Structures You can represent the formation of the covalent bond in H2 as follows: • This uses the Lewis symbols for the hydrogen atom and represents the covalent bond by a pair of dots. • To complete the duet for H, each H atom shares their electron between them so that they both have a duet.

  11. : H H Lewis Structures The shared electrons in H2 spend part of the time in the region around each atom. • In this sense, each atom in H2 has a helium configuration (complete duet). • H is found in nature as H2 not H because this arrangement of sharing electrons is more stable.

  12. : : . . : : : + H Cl H Cl : : Lewis Structures The formation of a bond between H and Cl to give an HCl molecule can be represented in a similar way. 8e- 2e- • Thus by sharing electrons, hydrogen has two valence electrons about it (as in He - duet) and Cl has eight valence electrons about it (as in Ar - octet).

  13. Lewis Structures Formulas such as these are referred to as Lewis electron-dot formulas or Lewis structures. bonding pair : : : H Cl : 3 lone pairs • An electron pair is either a bonding pair(shared between two atoms) or a lone pair(an electron pair that is not shared).

  14. . . : + A B A B • It is, however, possible that both electrons are donated by one of the atoms. This is called a coordinate covalent bond. : : + A B A B Coordinate Covalent Bonds When bonds form between atoms that both donate an electron, you have covalent bond:

  15. Multiple Bonds In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared. • It is possible to share more than one pair of electrons between atoms. Due to the lack of available electrons, atoms must share more than one pair of electrons to obtain an octet. A double bond involves the sharing of two pairs of electrons between atoms. • In ethylene, C2H4, there are only 12 e- available. For each H to have a duet and each C to have an octetby only using 12 e-, the two C atoms must share 4 e- to make a double bond. 2e- 2e- H H : : 8e- : : C C or : : 2e- 8e- H H 2e-

  16. Multiple Bonds In acetylene, C2H2, there are only 10 e- available. For each H to have a duet and each C to have an octetby only using 10 e-, the two C atoms must share 6 e- to make a triple bond. Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms. 2 e- 2 e- 8e- 2e- 2e- ::: H H C C or : : 8e- 6e- Even though the duets and octets are met, only 10 e- were used. This is accomplished by the C atoms sharing 6 e-. Typically, a bonded pair is represented with a single line, two bonded pairs by a double bond (previous slide), and three bonded pairs by a triple bond as in this example.

  17. Bond Length and Bond Order The length and strength of the bond between atoms varies with amount of electron density present (number of bonded pairs). Single bonds are the longest and weakest bonds while triple bonds are the shortest and strongest bonds. The bond order, determined by the Lewis structure, is the number of pairs of electrons in a bond. • Bond length depends on bond order. • As the bond order increases, the bond gets shorter and stronger. Bond length Bond energy Bond Order 154 pm C C 346 kJ/mol 1 134 pm C C 602 kJ/mol 2 120 pm C C 835 kJ/mol 3

  18. Why does a multiple bond occur? There’s not enough electrons available in the structure for each atom to have an octet by themselves; therefore, they must share electrons. Let’s look at the molecule N2. There are only a total of 10 electrons (5 from each N) available to complete octets. . . . . . .   +  . . . . . . . . . . . . . . N N N N N N N N . . . . . . . . . . . . . . . . . . . . The nitrogens can each share an electron to form a bond between them. However, this arrangement will not give each nitrogen an octet (only 6e- around each N atom); therefore, a different arrangement is needed. The nitrogens can each share another pair of electrons to form a second bond between the atoms (making a double bond between atoms). However, this arrangement will still not give each nitrogen an octet (only 7e- around each N atom); therefore, a different arrangement is still needed. The nitrogens can once again share another pair of electrons to form a third bond between the atoms (making a triple bond between atoms). This arrangement gives each nitrogen an octet (8e- around each N atom) and only uses the available 10 electrons to do it.

  19. Polar Covalent Bonds A polar covalent bond is one in which the bonding electrons spend more time near one of the two atoms involved, basically the atoms do not share the electrons equally. • When the atoms are alike, as in the H-H bond of H2 , the bonding electrons are shared equally and a nonpolar covalent bond results. • When the two atoms are of different elements, their electronegativities may differ enough to cause the bonding electrons not to be shared equally, resulting in a “polar” bond (i.eHCl).

  20. Polar Covalent Bonds For example, the bond between carbon and oxygen in CO2 is considered polar because the shared electrons spend more time orbiting the oxygen atoms (the more electronegative atom). We say that the bond is polarized towards the more electronegative atom and that the bond is polar covalent. : : d+ d- d- : : • Since the electrons are not equally shared between the atoms, there is a partial negative charge on the oxygens (denotedd-)and a partial positive charge on the carbon (denotedd+).

  21. d+ d- d- Dipole Moment and Molecular Geometry The product of the magnitude of the charge on an atom and the bond length between the atoms is called the dipole moment (m) of the bond. By definition, molecules are electrically neutral.  However, many a molecule has a region, or pole, bearing a net partial positive charge, and the opposite region, or pole, bearing a net partial negative charge. As long as the charges do not cancel each other out, such a molecule has a dipole moment, which is equal to the vector sum of the dipole moments of all bonds in the molecule.  Such molecules are said to be polar because they possess a permanent dipole moment.

  22. d+ d- d- Dipole Moment and Molecular Geometry The dipole moment of a molecule, being a vector, has a direction, which is shown using the following symbol: The arrowhead of the symbol is pointed toward the negative pole and the plus sign toward the positive pole. A molecule may not have a dipole moment despite containing polar bonds that do.  For example, each of the two carbon-oxygen bonds in CO2 has a dipole moment, but the CO2 molecule has no dipole moment because the dipole moments of the two carbon-oxygen bonds are identical in magnitude and opposite in direction, resulting in a vector sum of zero. Thus, molecules that are perfectly symmetric have a zero dipole moment. These molecules are considered nonpolar. polar bond net dipole moment polar molecule polar bonds no net dipole moment (opposite dipoles) nonpolar molecule

  23. : N H H H Dipole Moment and Molecular Geometry Molecules that exhibit any asymmetry in the arrangement of electron pairs would have a nonzero dipole moment. The NH3 molecule below is considered polar because the sum of its dipole moments do not cancel out; in fact, in this particular molecule all dipoles are in the same direction and additive. Therefore, NH3 is a very polar molecule. d- d+ d+ d+ Note: the shape of this molecule is in the form of a pyramid.

  24. Polar Covalent Bonds How do you determine the electronegativity of an atom to decide if and in what direction a dipole moment exists? Electronegativity (EN) is a measure of the ability of an atom in a molecule to draw bonding electrons to itself. • In general, electronegativity increases from the lower-left corner to the upper-right corner of the periodic table (Fluorine highest value). • The current electronegativity scale, developed by Linus Pauling, assigns a value of 4.0 to Fluorine and a value of 0.7 to Francium. • Metals tend to have values < 1.9 while nonmetals tend to have values > 2.1.

  25. Electronegativities Note value of EN for H Electronegativity differences between atoms that are greater than 1.8 are considered ionic bonds; generally occurs with metal-nonmetal bonds. Electronegativity differences between atoms that are greater than 0.5 but less than 1.8 are considered polar covalent bonds; generally occurs with nonmetal-nonmetal bonds. Electronegativity differences between atoms that are larger than 0 but less than 0.5 are considered covalent bonds but are essentially nonpolar in nature. Metallic bonding involves bonding between metal atoms in a solid and is a covalent bond.

  26. Polar Covalent Bonds • When this difference is small (less than 0.5), the bond is nonpolar (i.e. H-C DEN = 2.5 - 2.1 = 0.4). Basically, the electrons are shared similarly between the atoms. • When this difference is large (greater than 0.5 but less than 1.8), the bond is considered polar (i.e. H-ClDEN = 3.0 - 2.1 = 0.9). The electrons are not shared equally and spend more time with one atom over the other. • If the difference exceeds approximately 1.8, sharing of electrons is no longer possible and the bond becomes ionic. Ionic bond is ultimate in polarity, one species gets all electrons; no sharing of electrons (transferred) and formation of charged species occurs. • Generally, N, O, F, Cl, Br and another nonmetal (excluding these) are polar bonds (DEN 0.5 to 1.8); other nonmetal-nonmetal combinations are nonpolar (DEN <0.5). The absolute value of the difference in electronegativity of two bonded atoms gives a rough measure of the polarity of the bond. HW 65 code: en

  27. Writing Lewis Dot Formulas The Lewis structure is a visual representation of how the atoms in a molecule or polyatomic ion are sharing valence electrons. In a Lewis structure, • Each atom is represented by its symbol • All valence electrons must be represented by dots or lines • Each dot represents one valence electron • A line is equivalent to two dots • Lines or dots drawn between symbols of two atoms represent valence electrons shared by the two atoms (bonding pair) • For polyatomic ions, the Lewis structure must be enclosed in square brackets with the charge of the ion as a superscript.

  28. By following the rules below, the Lewis Structure can be written for covalent compounds and polyatomic ions: 1.  Count the total number of valence electrons; adjust for charge if an ion: add one electron for each negative charge, subtract one electron for each positive charge. 2.  Draw skeleton of structure - typically least EN atom is the central atom (most obvious one most of the time, except H is never it – H always on outside). 3.  Complete the octet around each atom surrounding the central atom (except H only needs 2 electrons, Be only needs 4 electrons, B and Al typically have 6 electrons for full “octet”).

  29. 4.  Count the number of electrons placed in step 3. 5. Compare the number of electrons placed to that available (subtract placed in #4 from available in #1). [step 1 available – step 4 placed] a.) equal – verify that the central atom has 8 electron around it(watch exceptions: B, Al, Be); i.) If central atom has 8e-, then you’re finished; except could check formal charge (learn later) ii.) If central atom does not have 8e-, it indicates that the electrons must be shared differently (deficient in electrons needed) and a multiple bond is present (2e- fewer than an octet suggests a double bond; four fewer suggests a triple bond or two double bonds). To obtain a multiple bond, move one or two electron pairs (depending on whether the bond is to be double or triple) from a surrounding atom to the bond connecting the central atom. Typically multiple bonds are done with C, N, O, and S but NOT with H or halogens.

  30. b.) positive number – verify that the central atom has 8 electron around it (watch exceptions : B, Al, Be); i.) If central atom does not have 8e-, then add electrons to the central atom to obtain an octet and account for these in your comparison. ii.) If you have an octet around the central atom and there are still electrons left over, place the extra electrons on central atom. Typically this occurs when d orbitals are available (n=3 or higher) which allows for more than an octet (expanded octet). Typically, C – has 4 bonds in Lewis structure H – 1 bond in Lewis structure Halogen – 1 bond, 3 lone pairs in Lewis structure

  31. Writing Lewis Structures • Count valence electrons • H – 1e- • Cl– 7e- • Total – 8e- available • Draw skeleton with least EN atom as central atom (we will call Cl the central atom since H can’t be the central atom) • H Cl • Complete octets around each atom surrounding central atom (H has 2e-) • H :Cl • Count number of electrons placed: • 2e- placed • Subtract e- placed from available: • 8 available – 2 placed = 6e- still available to place • 5b. Positive number of e- leftover with no octet around the central atom; therefore, add e- to complete octet around central atom and account for electrons added • 8 available – 8 placed = 0 HCl . . . . . . . . H Cl - Final Lewis structure except typically we change bonded pairs to lines instead of leaving as dots

  32. Writing Lewis Structures • Count valence electrons • H – 2 atoms x 1e- = 2e- • O – 6e- • Total – 8e- available • Draw skeleton with least EN atom as central atom (we will call O the central atom since H can’t be the central atom) • H O H • Complete octets around each atom surrounding central atom (H has 2e-) • H : O : H • Count number of electrons placed: • 4e- placed • Subtract e- placed from available: • 8 available – 4 placed = 4e- still available to place • 5b. Positive number of e- leftover with no octet around the central atom therefore, add e- to complete octet around central atom and account for electrons added • 8 available – 8 placed = 0 H2O H O H . . . . . . . . - - Final Lewis structure except typically we change bonded pairs to lines instead of leaving as dots

  33. Count valence electrons • H – 3 atoms x 1e- = 3e- • N – 5e- • Total – 8e- available • Draw skeleton with least EN atom as central atom (we will call N the central atom since H can’t be the central atom) • H N H • H • Complete octets around each atom surrounding central atom (H has 2e-) • H : N : H • H • Count number of electrons placed: • 6e-placed • Subtract e- placed from available: • 8 available – 6 placed = 2e-still available to place • 5b. Positive number of e- leftover and there is no octet around the central atom; therefore, add e- to complete octet around central atom and account for electrons added • 8 available – 8 placed = 0 NH3 : . . . . H N H . . . . • Finally change bonded pairs to lines and the structure is complete. H

  34. Writing Lewis Structures Caution: A Lewis structure is a two-dimensional representation of a molecule or ion; its intent is to show how the valence electrons of atoms are distributed in the molecule. A Lewis structure does not necessarily depict the actual shape of the molecule or ion; however, we can use it to deduce the actual shape which will be covered in another section.

  35. : : N O H H : H H H Writing Lewis Structures For example, the previous Lewis structures we have drawn (H2O and NH3) do not reflect the actual shape of the molecules. The 2D drawing for water gives the appearance that water is linear (180o between atoms) while in reality the shape is angular or bent (<109.5o between atoms). In ammonia, the actual 3D shape is triangular pyramidal. 3D molecular Note: dotted line indicates species is going into the plane of the paper while a wedge indicates species is coming out of the paper. 2D Lewis 3D molecular - . . . H H N H H O H . . . - - - - 2D Lewis

  36. Count valence electrons • F – 3 atoms x 7e- = 21e- • B – 3e- • Total – 24e- available • Draw skeleton with least EN atom as central atom (we will call B the central atom) • F B F • F • Complete octets around each atom surrounding central atom • : F : B : F : • F • Count number of electrons placed: • 24e- placed • Subtract e- placed from available: • 24 available – 24 placed = 0e- still available to place • 5a. No e-remaining, but there is no octet around the central atom; however, B is an exception where 6 electrons is a complete octet; finallychange bonded pairs to lines and the structure is complete. BF3 : : : : : . . . . : : . . . . : . . . . - . . F . . . . F B F - -

  37. Count valence electrons • F – 5 atoms x 7e- = 35e- • As – 5e- • Total – 40e- available • Draw skeleton with least EN atom as central atom (we will call As the central atom) • F F • F As F • F • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 40e- placed • Subtract e- placed from available: • 40 available – 40 placed = 0e- still available to place • 5a. No e- remaining but notice there are 10e- around the As central atom; Typically this occurs when d orbitals are available (n=3 or higher) which allows for more than an octet around central atom (expanded octet). • Finally change bonded pairs to lines and the structure is complete. AsF5 : : : F : : F : : : : : • : F : As : F : • F : : : : : :

  38. Count valence electrons Cl– 2 atoms x 7e- = 14e- • O – 6e- • C – 4e- • Total – 24e- available • Draw skeleton with least EN atom as central atom (we will call C the central atom) • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 24e- placed • Subtract e- placed from available: • 24 available – 24 placed = 0e- still available to place • 5a. No e- remaining but notice there are only 6e- around the C central atom; not an octet and not an exception. This indicates a deficiency in electrons and a multiple bond is present (not enough e- to complete octets for all atoms; therefore, must share some additional e- between atoms). • To obtain a multiple bond, we must move one electron pair from the surrounding O atom to the bond connecting the C and O atoms. Note this is not typically done with halogens. • Finally change bonded pairs to lines with a double bond • between C and O. All have octets and only used 24e-. : : COCl2 : : : : ClCl : : • C • O : : : :

  39. Count valence electrons H – 4 atoms x 1e- = 4e- • O – 6e- • C – 4e- • Total – 14e-available • Draw skeleton with least EN atom as central atom (we will call C the central atom; one H must be on outside of O because H can only have one bond) • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 14e- placed • Subtract e- placed from available: • 14 available – 14 placed = 0e- still available to place • 5a. No e- remaining and there are 8e- around the C central atom; therefore, this is the complete structure. • Finally change bonded pairs to lines. CH4O H H C O H H : : : : : : :

  40. Count valence electrons • O – 3 atoms x 6e- = 18e- • N – 5e- • Total – 23e- available • We need to account for the charge on the ion. The -1 charge means there is an additional electron available that we must add to our total. • Draw skeleton with least EN atom as central atom (we will call N the central atom). Note: since it is an ion must have brackets and charge. • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 24e- placed • Subtract e- placed from available: • 24 available – 24 placed = 0e- still available to place • 5a. No e- remaining but notice there are only 6e- around the N central atom; not an octet and not an exception. This indicates a deficiency in electrons and a multiple bond is present (not enough e- to complete octets for all atoms; therefore, must share some additional e- between atoms). • To obtain a multiple bond, we must move one electron pair from a surrounding O atom to the bond connecting the N and O atoms. • Finally change bonded pairs to lines with a double bond • between N and O. All have octets and only used 24e-. + 1e-from charge = 24e- available NO3- [ ]- : : O N O O : : : : : : : : : :

  41. Delocalized Bonding: Resonance • Which O gets the double bond in the nitrate ion? • It is possible to have more than one way of distributing valence electrons for a given molecule or ion. Each of these unique ways is called a resonance structure. The connectivity in resonance structures is the same; resonance structures represent the same molecule. The existence of two or more resonance structures generally indicates that none of the structures adequately represents reality. • Reality is probably a hybrid of the resonance structures. In the nitrate ion experiments show bond lengths to be similar and are a combination of the single and a double bond for all bonds (bonds shorter than a single bond but longer than a double bond). • Basically, there is delocalized bonding in which a bonding pair of electrons is spread over a number of atoms and not localized just between two atoms.

  42. Delocalized Bonding: Resonance • When we draw resonance structures, we draw all the structures with double-headed arrows between them to indicate resonance. NO3- [ ]- [ ]- [ ]- : : : : : : O N O O O N O O O N O O : : : : : : : : : : : : : : : : : : Below is a shorter notation sometimes used to represent resonance: [ ]- O N O O

  43. Count valence electrons • H – 4 atoms x 1e- = 4e- • N – 5e- • Total – 9e- available • We need to account for the charge on the ion. The +1 charge means there is one less electron available that we must subtract from our total. • Draw skeleton with least EN atom as central atom (we will call N the central atom). Note: since it is an ion must have brackets and charge. • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 8e- placed • Subtract e- placed from available: • 8 available – 8 placed = 0e- still available to place • 5a. No e- remaining and there are 8e- around the C central atom; therefore, this is the complete structure. • Finally change bonded pairs to lines. - 1e-from charge = 8e- available NH4+ H H N H H [ ]+ : : : :

  44. Count valence electrons • F – 4 atoms x 7e- = 28e- • Xe– 8e- • Total – 36e- available • Draw skeleton with least EN atom as central atom (we will call Xe the central atom) • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 32e- placed • Subtract e- placed from available: • 36 available – 32 placed = 4e- still available to place • 5b. Positive number of e-remaining but there are already 8e- around the Xe central atom. Since Xe has d orbitals available (n=3 or higher), it can accommodate more than an octet around it (expanded octet); therefore, we will add the extra e- around the central atom. • Finally change bonded pairs to lines and the structure is complete. : XeF4 : : F : : : : : : : : • F Xe F • F : : : : : : :

  45. Count valence electrons H –1e- • N – 5e- • C – 4e- • Total – 10e-available • Draw skeleton with least EN atom as central atom (we will call C the central atom) • Complete octets around each atom • surrounding central atom • Count number of electrons placed: • 10e- placed • Subtract e- placed from available: • 10 available – 10 placed = 0e- still available to place • 5a. No e- remaining but notice there are only 4e- around the C central atom; not an octet and not an exception. This indicates a deficiency in electrons and a multiple bond is present (not enough e- to complete octets for all atoms; therefore, must share some additional e- between atoms). • To obtain a multiple bond, we must move one electron pair from the surrounding N atom to the bond connecting the C and N atoms. Note this cannot be done with H. • This only increases the number of e- around the C atom to 6; therefore we must do this again. • Finally change bonded pairs to lines with a triple bond • between C and N. All have octets (H duet) and only used 10e-. HCN : : : • H C N : : HW 66 code: lewis

  46. Formal Charge and Lewis Structures In certain instances, more than one feasible Lewis structure can be illustrated for a molecule. For example, : : H C N H N C or • The concept of “formal charge” can help discern which structure is the most likely.

  47. Formal Charge and Lewis Structures The formal charge of an atom is determined by using the following formula: (note: formal charge is not the oxidation number of the atom) Formal Charge (F.C.) = F–B – 2L where F is the number of valence electrons in the free atom (neutral, not bonded to another atom). B is the number of bonding pairs attached to the atom (single bond counts as 1, double bond counts as 2, and triple bond counts as 3). L is the number of lone pairs around the atom.

  48. Formal Charge and Lewis Structures The most likely structure is the one with the least number of atoms carrying formal charge. Structures with small formal charges (such as 0, +1, -1) are more likely to be stable or significant than those with larger formal charges (such as +2, -2). Structures where neighboring atoms have formal charges of the same sign are less likely to be stable or significant. If neighboring atoms have opposite formal charges, we would expect the more electronegative atom to have the negative formal charge and the less electronegative atom to have the positive formal charge.

  49. Formal Charge and Lewis Structures formal charge : : H C N H N C or 0 0 0 0 +1 -1 H: F.C. = 1 valence e-– 1 bond– 2(0 lone pairs) = 0 C: F.C. = 4 valence e-– 4 bonds– 2(0 lone pairs) = 0 N: F.C. = 5 valence e-– 3 bonds – 2(1 lone pairs) = 0 H: F.C. = 1 valence e-– 1 bond– 2(0 lone pairs) = 0 N: F.C. = 5 valence e-– 4 bonds – 2(0 lone pairs) = +1 C: F.C. = 4 valence e-– 3 bonds – 2(1 lone pairs) = -1 • In this case, the structure on the left (C central atom) is the most stable or significant one based on the least number of atoms carrying a formal charge. The structure on the right is also not selected because the least electronegative atom, C, has a negative formal charge while the most electronegative atom, N, has a positive one.

  50. Bond Length and Bond Order Bond length (or bond distance) is the distance between the nuclei in a bond. • Knowing the bond length in a molecule can sometimes give clues as to the type of bonding present. • Covalent radii are values assigned to atoms such that the sum of the radii of atoms “A” and “B” approximate the A-B bond length.

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