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CHAPTER 10
CHEMICAL BONDING I: BASIC CONCEPTS
PRACTICE EXAMPLES
1A (E) Mg is in group 2(2A), and thus has 2 valence electrons and 2 dots in its Lewis symbol. Ge is in group 14(4A), and thus has 4 valence electrons and 4 dots in its Lewis symbol. K is in group 1(1A), and thus has 1 valence electron and 1 dot in its Lewis symbol. Ne is in group 18(8A), and thus has 8 valence electrons and 8 dots in its Lewis symbol.
Mg Ge
K : Ne:
1B (E) Sn is in Family 4A, and thus has 4 electrons and 4 dots in its Lewis symbol. Br is in Family 7A with 7 valence electrons. Adding an electron produces an ion with 8 valence electrons. Tl is in Family 3A with 3 valence electrons. Removing an electron produces a cation with 2 valence electrons. S is in Family 6A with 6 valence electrons. Adding 2 electrons produces an anion with 8 valence electrons.
Sn
[ Br ]: :
[Tl ] +^ [ S ]: : 2
2A (E) The Lewis structures for the cation, the anion, and the compound follows the explanation. (a) Na loses one electron to form Na +^ , while S gains two to form S 2 ^.
Na 1 e Na
S 2e ^ [ S ]: :
Lewis Structure: [Na] [ S ]+^ : : 2 [Na]+
(b) Mg loses two electrons to form Mg 2+^ , while N gains three to form N 3 ^.
Mg 2 e Mg 2+ (^3)
N 3e ^ [ N ]: :
Lewis Structure: [Mg]^ 2+ [ N ]: ^ : 3 ^ [Mg]2+ [ N ]: :^3 [Mg]2+
2B (E) Below each explanation are the Lewis structures for the cation, the anion, and the compound. (a) In order to acquire a noble-gas electron configuration, Ca loses two electrons, and I gains one, forming the ions Ca^ 2+^ and I^. The formula of the compound is^ CaI^2.
Ca 2 e Ca 2+ : I.. e [: :]I
.. .. .. ^ ^ Lewis Structure: [: I :] [Ca]2+ [: I :]
(b) Ba loses two electrons and S gains two to acquire a noble-gas electron configuration, forming the ions Ba 2+^ and S 2 ^. The formula of the compound is BaS.
Ba 2 e Ba 2+ (^2) S 2e [: S :]
^ Lewis Structure: [Ba]2+ [: S :]..^2
(c) Each Li loses one electron and each O gains two to attain a noble-gas electron configuration, producing the ions Li +^ and O 2 ^. The formula of the compound is Li O 2.
Li -1 e Li .... (^2) O.. 2e [: O :]..
^ Lewis Structure: [Li] [: O :]+^ ..^2 [Li]+
3A (M) In the Br 2 molecule, the two Br atoms are joined by a single covalent bond. This bonding arrangement gives each Br atom a closed valence shell configuration that is equivalent to that for a Kr atom.
: Br Br :
In CH 4 , the carbon atom is covalently bonded to four hydrogen atoms. This arrangement gives the carbon atom a valence shell octet and each H atom a valence shell duet.
C H
H
H
H
In HOCl, the hydrogen and chlorine atoms are attached to the central oxygen atom through single covalent bonds. This bonding arrangement provides each atom in the molecule with a closed valence shell. H^ O^ Cl :
3B (M) The Lewis structure for NI 3 is similar to that of NH 3. The central nitrogen atom is attached to each iodine atom by a single covalent bond. All of the atoms in this structure get a closed valence shell.
: I N I :
: I :
The Lewis diagram for N 2 H 4 has each nitrogen with one lone pair of electrons, two covalent bonds to hydrogen atoms, and one covalent bond to the other nitrogen atom. With this arrangement, the nitrogen atoms complete their octets while the hydrogen atoms complete their duets.
N N H
H H
H
In the Lewis structure for C 2 H 6 , each carbon atom shares four pairs of electrons with three hydrogen atoms and the other carbon atom. With this arrangement, the carbon atoms complete their octets while the hydrogen atoms complete their duets.
H C C H
4A (E) The bond with the most ionic character is the one in which the two bonded atoms are the most different in their electronegativities. We find electronegativities in Figure 10- and calculate EN for each bond. Electronegativities: H = 2.1 Br = 2.8 N = 3.0 O= 3.5 P = 2.1 Cl = 3. Bonds : H—Br N—H N—O P—Cl EN values : 0.7 0.9 0.5 0. Therefore, the N—H and PCl bonds are the most polar of the four bonds cited.
4B (E) The most polar bond is the one with the greatest electronegativity difference. Electronegativities: C = 2.5 S = 2.5 P = 2.1 O = 3.5 F = 4. Bonds: C — S C — P P — O O — F EN (^) values: 0.0 0.4 1.4 0. Therefore, the P — O bond is the most polar of the four bonds cited.
7A (E)
(a) A plausible Lewis structure for the nitrosonium cation, NO+, is drawn below: N O The nitrogen atom is triply bonded to the oxygen atom and both atoms in the structure possess a lone pair of electrons. This gives each atom an octet and a positive formal charge appears on the oxygen atom. (b) A plausible Lewis structure for N 2 H 5 +^ is given below:
The two nitrogen atoms have each achieved an octet. The right hand side N atom is surrounded by three bonding pairs and one lone pair of electrons, while the left hand side N atom is surrounded by four bonding pairs of electrons. Each hydrogen atom has completed its duet by sharing a pair of electrons with a nitrogen atom. A formal 1+ charge has been assigned to the left hand side nitrogen atom because it is bonded to four atoms (one more than its usual number) in this structure.
(c) In order to achieve a noble gas configuration, oxygen gains two electrons, forming the stable dianion. The Lewis structure for O2-^ is shown below.
O
2-
7B (M)
(a) The most likely Lewis structure for BF 4 -^ is drawn below:
B F
F
F
F
Four bonding pairs of electrons surround the central boron atom in this structure. This arrangement gives the boron atom a complete octet and a formal charge of -1. By virtue of being surrounded by three lone pairs and one bonding electron pair, each fluorine achieves a full octet.
(b) A plausible Lewis structural form for NH 3 OH+, the hydroxylammonium ion, has been provided below:
N O
H
H
H
H
By sharing bonding electron pairs with three hydrogen atoms and the oxygen atom, the nitrogen atom acquires a full octet and a formal charge of 1+. The oxygen atom shares one bonding electron pair with the nitrogen and a second bonding pair with a hydrogen atom.
H N
H
H
N
H
H
(c) Three plausible resonance structures can be drawn for the isocyanate ion, NCO -. The nitrogen contributes five electrons, the carbon four, oxygen six, and one more electron is added to account for the negative charge, giving a total of 16 electrons or eight pairs of electrons. In the first resonance contributor, structure 1 below, the carbon atom is joined to the nitrogen and oxygen atoms by two double bonds, thereby creating an octet for carbon. To complete the octet of nitrogen and oxygen, each atom is given a lone pair of electrons. Since nitrogen is sharing just two bonding pairs of electrons in this structure, it must be assigned a formal charge of 1-. In structure 2, the carbon atom is again surrounded by four bonding pairs of electrons, but this time, the carbon atom forms a triple bond with oxygen and just a single bond with nitrogen. The octet for the nitrogen atom is closed with three lone pairs of electrons, while that for oxygen is closed with one lone pair of electrons. This bonding arrangement necessitates giving nitrogen a formal charge of 2- and the oxygen atom a formal charge of 1+. In structure 3, which is the dominant contributor because it has a negative formal charge on oxygen (the most electronegative element in the anion), the carbon achieves a full octet by forming a triple bond with the nitrogen atom and a single bond with the oxygen atom. The octet for oxygen is closed with three lone pairs of electrons, while that for nitrogen is closed with one lone pair of electrons.
N C O N C O
Structure 1 Structure 2 Structure 3
N C O
8A (M) The total number of valence electrons in NOCl is 18 (5 from nitrogen, 6 from oxygen and 7 from chlorine). Four electrons are used to covalently link the central oxygen atom to the terminal chlorine and nitrogen atoms in the skeletal structure: NOCl. Next, we need to distribute the remaining electrons to achieve a noble gas electron configuration for each atom. Since four electrons were used to form the two covalent single bonds, fourteen electrons remain to be distributed. By convention, the valence shells for the terminal atoms are filled first. If we follow this convention, we can close the valence shells for both the nitrogen and the chlorine atoms with twelve electrons. N O Cl Oxygen is moved closer to a complete octet by placing the remaining pair of electrons on oxygen as a lone pair. N O Cl The valence shell for the oxygen atom can then be closed by forming a double bond between the nitrogen atom and the oxygen atom.
N O Cl
This structure obeys the requirement that all of the atoms end up with a filled valence shell, but is much poorer than the one derived in Example 10-8 because it has a positive formal charge on oxygen, which is the most electronegative atom in the molecule. In other words, this structure can be rejected on the grounds that it does not conform to the third rule for determining plausibility of a Lewis structure based on formal charges, which states that "negative formal charges should appear on the most electronegative atom, while any positive formal charge should appear on the least electronegative atom."
formal charges of the same type (two 1+ formal charges) on adjacent atoms, as well as negative formal charges on carbon, which is not the most electronegative element in the molecule.
N N
H
H C
2-
9A (D) The skeletal structure for SO 2 has two terminal oxygen atoms bonded to a central sulfur atom. Sulfur has been selected as the central atom by virtue of its being the most electropositive atom in the molecule. It turns out that two different Lewis structures of identical energy can be derived from the skeletal structure described above. First we determine that SO 2 has 18 valence electrons (6 from each atom). Four of the valence electrons must be used to covalently bond the three atoms together. The remaining 14 electrons are used to close the valence shell of each atom. Twelve electrons are used to give the terminal oxygen atoms a closed shell. The remaining two electrons (14 -12 = 2) are placed on the sulfur atom, affording the structure depicted below:
O S^ O
At this stage, the valence shells for the two oxygen atoms are closed, but the sulfur atom is two electrons short of a complete octet. If we complete the octet for sulfur by converting a lone pair of electrons on the right hand side oxygen atom into a sulfur-to-oxygen -bond, we end up generating the resonance contributor (A) shown below:
O S O (A)
Notice that the structure has a positive formal charge on the sulfur atom (most electropositive element) and a negative formal charge on the left-hand oxygen atom. Remember that oxygen is more electronegative than sulfur, so these charges are plausible. The second completely equivalent contributor, (B), is produced by converting a lone pair on the left-most oxygen atom in the structure into a -bond, resulting in conversion of a sulfur-oxygen single bond into a sulfur-oxygen double bond:
O S O (B)
Neither structure is consistent with the observation that the two S-O bond lengths in SO (^2) are equal, and in fact, the true Lewis structure for SO 2 is neither (A) nor (B), but rather an equal blend of the two individual contributors called the resonance hybrid (see below).
O S^ O O S O
O S O
9B (D) The skeletal structure for the NO 3 -^ ion has three terminal oxygen atoms bonded to a central nitrogen atom. Nitrogen has been chosen as the central atom by virtue of being the most electropositive atom in the ion. It turns out that three contributing resonance structures of identical energy can be derived from the skeletal structure described here. We begin the process of generating these three structures by counting the total number of valence electrons in the NO 3 -^ anion. The nitrogen atom contributes five electrons, each oxygen contributes six electrons, and an additional electron must be added to account for the 1- charge on the ion. In total, we must account for 24 electrons. Six electrons are used to draw single covalent bonds between the nitrogen atom and three oxygen atoms. The remaining 18 electrons are used to complete the octet for the three terminal oxygen atoms:
O
O
2+
N O
O
O
At this stage the valence shells for the oxygen atoms are filled, but the nitrogen atom is two electrons short of a complete octet. If we complete the octet for nitrogen by converting a lone pair on O 1 into a nitrogen-to-oxygen -bond, we end up generating resonance contributor (A):
1 2 N^ O 3 (A)
O
O
Notice the structure has a 1+ formal charge on the nitrogen atom and a 1- on two of the oxygen atoms (O 2 and O 3 ). These formal charges are quite reasonable energetically. The second and third equivalent structures are generated similarly; by moving a lone pair from O 2 to form a nitrogen to oxygen (O 2 ) double bond, we end up generating resonance contributor (B), shown below. Likewise, by converting a lone pair from oxygen (O 3 ) into a -bond with the nitrogen atom, we end up generating resonance contributor (C), also shown below.
2 3 (B)
1
N O
O
O 2 N^ O (C)
O
O
1
3
None of these individual structures ((A), (B), or (C)) correctly represents the actual bonding in the nitrate anion. The actual structure, called the resonance hybrid, is the equally weighted average of all three structures (i.e. 1/3(A) + 1/3(B) + 1/3(C)):
1 (^2 )
N O
O
O 23
1
N O
O
O 2 N O
O
O
1
3
(^23)
1 N O
O
O Resonance Hybrid
of electrons and two lone pairs, resulting in a tetrahedral electron-group geometry and a bent molecular shape around the O atom, with a C — O — H bond angle of slightly less than 109.5^.
12B (M) The Lewis structure is drawn below. With four electron groups surrounding each, the electron-group geometries of N, the central C, and the right-hand O are all tetrahedral. The H — N — H bond angle and the H — N — C bond angles are almost the tetrahedral angle of 109.5^ , made a bit smaller by the lone pair. The H — C — N angles, the H — C — H angle and the HCC angles all are very close to109.5^. The C — O — H bond angle is made somewhat smaller than 109.5^ by the presence of two lone pairs on O. Three electron groups surround the right-hand C, making its electron-group and molecular geometries trigonal planar. The O — C — O bond angle and the O — C — C bond angles all are very close to120^.
13A (M) Lewis structures of the three molecules are drawn below. Around the S in the SF 6
molecule are six bonding pairs of electrons, and no lone pairs. The molecule is octahedral; each of the S-F bond moments is cancelled by one on the other side of the molecule.
S
F
F F
F F
F
O O
H
H
C C
H
H H
H
SF 6 is nonpolar. In H 2 O 2 , the molecular geometry around each O atom is bent; the bond moments do not cancel. H O 2 2 is polar. Around each C in C H 2 4 are three bonding pairs of electrons; the molecule is planar around each C and planar overall. The polarity of each —CH 2 group is cancelled by the polarity of the other H C— 2 group. C H 2 2 is nonpolar.
13B (M) Lewis structures of the four molecules are drawn below and we can consider the three C — H bonds and the one C — C bond to be nonpolar. The three C — Cl bonds are tetrahedrally oriented.
Cl C
Cl C H
H
Cl H
P
Cl Cl Cl Cl
Cl C
Cl
Cl
H H N^ H
H
H
If there were a fourth C — Cl bond on the left-hand C, the bond dipoles would cancel out, producing a nonpolar molecule. Since it is not there, the molecule is polar. A similar argument is made for NH 3 , where three tetrahedrally-oriented N — H polar bonds are not balanced by a fourth, and for CH 2 Cl 2 , where two tetrahedrally oriented C — Cl bonds
H N
H
C
H
H
C
O
O H
are not balanced by two others. This leaves PCl 5 as the only nonpolar species; it is a highly symmetrical molecule in which individual bond dipoles cancel out.
14A (M) The Lewis structure of CH Br has all single bonds. From Table 10.2, the length of a 3
C — H bond is 110 pm. The length of a C — Br bond is not given in the table. A reasonable value is the average of the C — C and Br — Br bond lengths.
C — Br
C — C Br — Br pm pm = pm
C Br
H
H
H
14B (E) In Table 10.2, the C=N bond length is 128 pm, while the CN bond length is 116 pm. The observed CN bond length of 115 pm is much closer to a carbon-nitrogen triple bond. This can be explained by using the following Lewis structure: | S— C N |where there is a formal negative charge on the sulfur atom. This molecule is linear according to VSEPR theory.
15A (M) We first draw Lewis structures for all of the molecules involved in the reaction.
2 H H + O O H O H
Break 1 O=O + 2H—H = 498 kJ/mol + (2 × 436 kJ/mol) = 1370 kJ/mol absorbed Form 4H—O = (4 × 464 kJ/mol) = 1856 kJ/mol given off Enthalpy change = 1370 kJ / mol 1856 kJ / mol = 486 kJ / mol
15B (M) The chemical equation, with Lewis structures, is:
N N (^) + H H H N H H
Energy required to break bonds 12 N N + 23 H — H
= 0.5b 946 kJ / mol g b+ 1.5 436 kJ / mol g= 1.13 10 3 kJ / mol
Energy realized by forming bonds = 3 N — H = 3 389 kJ / mol = 1.17 10 3 kJ / mol H = 1.13 10 3 kJ / mol 1.17 10 3 kJ / mol = 4 101 kJ / mol of NH 3. Thus, H (^) f = 4 101 kJ / mol NH 3 (Appendix D value is H (^) f = 46.11 kJ / mol NH ) 3
16A (M) The reaction below,
CH 3 C=O CH (g) + H (g) 3 2 CH 3 2 CH OH (g)
Involves the following bond breakages and formations: Broken: 1 C=O bond (736 kJ/mol) Broken: 1 H–H bond (436 kJ/mol) Formed: 1 C–O bond (360 kJ/mol) Formed: 1 C–H bond (414 kJ/mol) Formed: 1 O–H bond (464 kJ/mol)
The Lewis structures of PCl 3 and PCl 5 are shown below.
P
Cl
Cl
Cl P
Cl
Cl Cl
Cl
Cl
Since the geometries of the two molecules differ, the orbital overlap between P and the surrounding Cl atoms will be different and therefore the P–Cl bonds in these two compounds will also be slightly different.
B. (M) (a) The structures are shown below, with appropriate geometries:
H 2 N
C
O
H
Formamide
H 2 C
N
OH
Formaldoxime
BE H NCOH 2 2 N H 1 N C 1 C O 1 C H
2(389) 305 736 414 2233 kJ/mol
BE H C 2 N OH 2 C H 1 C N 1 N O 1 O H
2(414) 615 222 464 2129 kJ / mol
Since BE of formamide is greater than that of formaldoxime, it is more stable, and its conversion endothermic.
(b) The experiment shows that the geometry around C is trigonal planar, and around N is trigonal pyramidal.
N
C
O
H
H
H
124 o
109 o
EXERCISES
Lewis Theory
1. (E)
(a) Kr^ (b) (^) Ge (c) (^) N (d) (^) Ga (e) (^) As (f) Rb
2. (E)
(a) (^) H - (b) Sn^
2+ (c) K^
(d) Br
(e) Se
(f) (^) Sc 3+
3. (E)
(a)
F Cl
(b)
I I
(c)
F
S
F
(d)
F N
F
F
(e)
H
Te H
4. (E) For simplicity, the 3 lone pairs on halogens are not shown in the structures below. Add 6 electrons per halogen atom. (a)
O C S
(b)
C C
O
H
H
H
H
(c)
C O
F
F
(d)
S O
Cl
Cl
(e)
H C C H
8. (M)
(a) In order to construct an H 3 molecule, one H would have to bridge the other two. This would place 3 electrons around the central H atom, which is more than the stable pair found around H in most Lewis structures. As well, one bond would be the normal 2 e -^ bond and the other would be a one electron bond, which is beyond Lewis theory (i.e., HH.^ H). (b) In HHe there would be three electrons between the two atoms, or three electrons around the He atom, one of which would be a nonbonding electron. Neither of these is a particularly stable situation. (c) He 2 would have either a double bond between two He atoms and thus four electrons around each He atom, or three electrons around each He atom (2 e-^ in a bond and an unpaired electron on each atom). Neither situation achieves the electron configuration of a noble gas. (d) H 3 O has an expanded octet (9 electrons) on oxygen; expanded octets are not possible for elements in the second period. Other structures place a multiple bond between O and H. Both situations are unstable.
9. (E)
(a) H^ H N^ O^ H has two bonds to (four electrons around) the
second hydrogen, and only six electrons around the nitrogen. A better Lewis structure is shown below.
(b) Here, Ca^ O^ is improperly written as a covalent Lewis structure. CaO is actually
an ionic compound. [Ca]
2+ O 2-
is a more plausible Lewis structure for CaO.
10. (E)
(a) O^ Cl O^ has 20 valence electrons, whereas the molecule ClO 2 has 19 valence
electrons. This is a proper Lewis structure for the chlorite ion, although the brackets and the minus charge are missing. A plausible Lewis structure for the molecule
ClO 2 is O^ Cl O
(b) C^ N^
has only six electrons around the C atom and two too few overall.
C N - is a more plausible Lewis structure for the cyanide ion.
H N
H O H
H O
H
H
11. (M) The answer is (c), hypochlorite ion. The flaws with the other answers are as follows:
(a) O^ C N^ ^ does not have an octet of electrons around C.
(b) [C^ C^ ]
does not have an octet around either C. Moreover, it has only 6 valence electrons in total while it should have 10, and finally, the sum of the formal charges on the two carbons doesn’t equal the charge on the ion. (d) The total number of valence electrons in NO is incorrect. No, being an odd-electron species should have 11 valence electrons, not 12.
12. (M)
(a) Mg^ O^ is incorrectly written as a covalent structure. One expects an ionic Lewis
structure, namely [Mg]
2+[ O ]2-
(b) (^) [ O N O]+^ has too many valence electrons—8.5 electron pairs or 17 valence
electrons—it should have b 2 6 + 5g 1 = 16valence electrons or 8 electron pairs. A
a plausible Lewis structure is [O^ N O] 2+ , which has 1+ formal charge on N and 0
formal charge of zero on each oxygen.
(c) (^) [Cl ]+[ O ]2-[ (^) Cl] +is written as an ionic structure, even though we expect a covalent
structure between nonmetallic atoms. A more plausible structure is Cl^ O Cl
(d) In the structure [^ S C N]
- neither S nor C possesses an octet of electrons. In
addition, there are only 7 pairs of valence electrons in this structure or 14 valence electrons. There should be 6 + 4 + 5 +1 = 16 valence electrons, or 8 electron pairs. Two structures are possible. S^ C N^ has a formal charge of 1- on N and is
preferred over (^) S C N , with its formal charge of 1- on S, which is less
electronegative than N.
Ionic Bonding
13. (E)
(a) [: Cl :] [Ca]2+ [: Cl :]
(b) [Ba]2+ [: S :].. 2
(c) [Li] [: O :]+ .. 2 [Li]+
(d) [Na] [: F :]+ ..
Formal Charge
17. (M)
(a) computations for: no. valence e–
- no. lone-pair e– 12 no. bond-pair e– formal charge
H—
—C
C
(b) computations for: no. valence e–
- no. lone-pair e– 12 no. bond-pair e– formal charge
=O
—O( 2)
C
(c) computations for: no. valence e–
- no. lone-pair e– 12 no. bond-pair e– formal charge
—H(7)
side C(2) 4
central C 4
18. (M)
(a) The formal charge on each I is 0, computed as follows: no. valence e–^ = 7 –no. lone-pair e–^ = – 12 no. bond-pair e–^ = – formal charge = 0
(b) computations for: no. valence e–
- no. lone-pair e– 12 no. bond-pair e– formal charge
=O
—O
=S—
(c) computations for: no. valence e–
- no. lone-pair e– 12 no. bond-pair e– formal charge
=O
—O
=N—
19. (M) There are three features common to formal charge and oxidation state. First, both indicate how the bonding electrons are distributed in the molecule. Second, negative formal charge (in the most plausible Lewis structure) and negative oxidation state are generally assigned to the more electronegative atoms. And third, both numbers are determined by a set of rules, rather than being
determined experimentally. Bear in mind, however, that there are also significant differences. For instance, there are cases where atoms of the same type with the same oxidation state have different formal charges, such as oxygen in ozone, O 3. Another is that formal charges are used to decide between alternative Lewis structures, while oxidation state is used in balancing equations and naming compounds. Also, the oxidation state in a compound is invariant, while the formal charge can change. The most significant difference, though, is that whereas the oxidation state of an element in its compounds is usually not zero, its formal charge usually is.
20. (M) The most common instance in which formal charge is not kept to a minimum occurs in the
case of ionic compounds. For example, in (^) Mg O the formal charge on Mg is 1+ and on O it
is 1-, while in the ionic version [Mg]2+^ [ (^) O]^2 -^ , formal charges are 2+ and 2-, respectively.
Additionally, in some resonance hybrids, formal charge is not minimized. In order to have bond lengths agree with experimental results, it may not be acceptable to create multiple bonds. Yet a third instance is when double bonds are created to lower formal charge, particularly when this results in the octet rule being violated. For instance, all the Cl — O bonds in ClO (^4) are best represented using single bonds. Although including some double bonds would minimize formal charges, the resulting structure is less desirable because the octet for Cl has been exceeded and this requires the input of additional energy.
21. (E) FC = # valence e–^ in free atom – number lone-pair e –^ – ½ # bond pair e – (a) Central O in O 3 : 6 – 2 – 3 = + (b) Al in AlH 4 –^ : 3 – 0 – 4 = – (c) Cl in ClO 3 –^ : 7 – 2 – 5 = 0 (d) Si in SiF 6 2-^ : 4 – 0 – 6 = – (e) Cl in ClF 3 : 7 – 4 – 3 = 0 22. (M) (a) H 2 NOH is favored because same charges don’t reside adjacent to each other.
N O H
H
H
O N H
H
H
FC(N) = 5-2-3 = 0
FC(O) = 6-2-2 = +
FC(N) = 5-2-2 = +
FC(O) = 6-2-3 = +
(b) S=C=S is favored, because FC on everything is zero.
S C S C S S
FC(C) = 4-0-4 = 0 FC(S) = 6-4-2 = 0
FC(C) = 4-4-2 = -
FC(=S=) = 6-0-4 = +2; FC(=S) = 6-4-2 = 0
