201. A carbon atom in a Lewis structure has four single bonds and no lone pair. Its formal charge is
ⓐ. \(0\)
ⓑ. \(-1\)
ⓒ. \(+1\)
ⓓ. \(+4\)
Correct Answer: \(0\)
Explanation: \( \textbf{Formula used:} \)
\[
FC=V-L-\frac{B}{2}
\]
\( \textbf{For carbon:} \) \(V=4\).
\( \textbf{Non-bonding electrons:} \) No lone pair means \(L=0\).
\( \textbf{Bonding electrons:} \) Four single bonds contain \(8\) bonding electrons.
\( \textbf{Substitution:} \)
\[
FC=4-0-\frac{8}{2}
\]
\( \textbf{Simplification:} \)
\[
FC=4-4=0
\]
\( \textbf{Final answer:} \) The formal charge is \(0\). This is why carbon in \(\mathrm{CH_4}\) is formally neutral.
202. The central carbon atom in the Lewis structure \(\mathrm{O=C=O}\) has formal charge
ⓐ. \(-2\)
ⓑ. \(0\)
ⓒ. \(-1\)
ⓓ. \(+2\)
Correct Answer: \(0\)
Explanation: \( \textbf{Central atom:} \) Carbon is bonded to two oxygen atoms by two double bonds.
\( \textbf{Formal-charge relation:} \)
\[
FC=V-L-\frac{B}{2}
\]
\( \textbf{Valence electrons of carbon:} \) \(V=4\).
\( \textbf{Non-bonding electrons on carbon:} \) \(L=0\).
\( \textbf{Bonding electrons around carbon:} \) Two double bonds contain \(2\times4=8\) bonding electrons.
\( \textbf{Substitution:} \)
\[
FC=4-0-\frac{8}{2}
\]
\( \textbf{Calculation:} \)
\[
FC=4-4=0
\]
\( \textbf{Final answer:} \) Carbon has formal charge \(0\). The double-bond structure of \(\mathrm{CO_2}\) avoids unnecessary charge separation.
203. Two possible skeleton-based descriptions for \(\mathrm{CO_2}\) are compared.
Case 1: \(\mathrm{O=C=O}\)
Case 2: \(\mathrm{^-O-C\equiv O^+}\)
The generally preferred Lewis description is
ⓐ. Case 2, because every structure with charges is always more stable
ⓑ. Case 2, because oxygen should always carry positive formal charge
ⓒ. Case 1, because it minimizes formal charge separation
ⓓ. both cases are identical in formal charge distribution
Correct Answer: Case 1, because it minimizes formal charge separation
Explanation: A good Lewis structure usually has small formal charges and minimal charge separation, provided octets are satisfied. In \(\mathrm{O=C=O}\), carbon and both oxygen atoms have formal charge \(0\). The charged structure \(\mathrm{^-O-C\equiv O^+}\) has separated formal charges. Although it can be drawn as a contributor in a broader discussion, it is not the best simple Lewis structure for \(\mathrm{CO_2}\). When a neutral structure with complete octets is available, it is usually preferred over a charge-separated alternative.
204. The formal charges in one contributor of \(\mathrm{CO_3^{2-}}\) are most suitably described as
ⓐ. all three oxygen atoms have \(0\), and carbon has \(-2\)
ⓑ. single-bonded O atoms: \(-1\); double-bonded O: \(0\)
ⓒ. all three oxygen atoms have \(-2\), and carbon has \(+4\)
ⓓ. one oxygen has \(+2\), and the ion has no negative charge
Correct Answer: single-bonded O atoms: \(-1\); double-bonded O: \(0\)
Explanation: In a common contributor of \(\mathrm{CO_3^{2-}}\), carbon has one double bond to oxygen and two single bonds to oxygen atoms. A singly bonded oxygen with three lone pairs has formal charge \(-1\). A double-bonded oxygen with two lone pairs has formal charge \(0\). Carbon has formal charge \(0\) in this contributor. The two \(-1\) formal charges add to the overall \(2-\) charge of the carbonate ion.
205. A nitrate ion contributor has one \(\mathrm{N=O}\) bond and two \(\mathrm{N-O}\) single bonds. If each singly bonded oxygen carries \(-1\), the formal charge on nitrogen in that contributor must be
ⓐ. \(-1\)
ⓑ. \(+1\)
ⓒ. \(0\)
ⓓ. \(+2\)
Correct Answer: \(+1\)
Explanation: \( \textbf{Overall ion charge:} \) \(\mathrm{NO_3^-}\) has total charge \(-1\).
\( \textbf{Formal charges on two singly bonded oxygens:} \) Each is \(-1\).
\( \textbf{Total from the two singly bonded oxygens:} \)
\[
(-1)+(-1)=-2
\]
\( \textbf{Double-bonded oxygen:} \) It has formal charge \(0\).
\( \textbf{Let nitrogen formal charge be \(x\):} \)
\[
x+(-2)+0=-1
\]
\( \textbf{Solving:} \)
\[
x=+1
\]
\( \textbf{Final answer:} \) Nitrogen has formal charge \(+1\). This keeps the sum of formal charges equal to the overall nitrate charge.
206. A formal-charge calculation for an atom uses \(V=5\), \(L=2\), and \(B=6\). The formal charge is
ⓐ. \(0\)
ⓑ. \(-1\)
ⓒ. \(+1\)
ⓓ. \(+2\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given values:} \) \(V=5\), \(L=2\), and \(B=6\).
\( \textbf{Formal-charge relation:} \)
\[
FC=V-L-\frac{B}{2}
\]
\( \textbf{Substitution:} \)
\[
FC=5-2-\frac{6}{2}
\]
\( \textbf{Half of bonding electrons:} \)
\[
\frac{6}{2}=3
\]
\( \textbf{Final calculation:} \)
\[
FC=5-2-3=0
\]
\( \textbf{Final answer:} \) The formal charge is \(0\). The term \(\frac{B}{2}\) is used because bonding electrons are shared equally for formal-charge bookkeeping.
207. The statement that best separates formal charge from oxidation state is
ⓐ. formal charge and oxidation state are always numerically identical
ⓑ. formal charge is used only for ionic solids, while oxidation state is used only for covalent gases
ⓒ. oxidation state is calculated using \(FC=V-L-\frac{B}{2}\)
ⓓ. formal charge uses a Lewis structure; oxidation state uses electronegativity rules
Correct Answer: formal charge uses a Lewis structure; oxidation state uses electronegativity rules
Explanation: Formal charge is a bookkeeping charge based on a particular Lewis structure. It assigns half of the bonding electrons to each bonded atom. Oxidation state follows a different convention, usually assigning bonding electrons to the more electronegative atom. Therefore, the two values need not be the same. Confusing them can lead to wrong structure selection and wrong charge interpretation in covalent species.
208. For choosing between two Lewis structures with similar octet satisfaction, the better structure often places negative formal charge on
ⓐ. the least electronegative atom in every case
ⓑ. the more electronegative atom
ⓒ. hydrogen whenever possible
ⓓ. the atom with the largest mass only
Correct Answer: the more electronegative atom
Explanation: When formal charges cannot be avoided, a negative formal charge is usually more acceptable on a more electronegative atom. More electronegative atoms can better accommodate electron density. Positive formal charge is usually less favourable on a highly electronegative atom, though the whole structure must still be checked. This rule is used along with octet completion and minimization of charge separation. It is a preference rule, not permission to ignore the total electron count.
209. A proposed Lewis structure has formal charges \(+2\), \(-1\), and \(-1\) on three atoms in a neutral molecule. Another possible structure has all formal charges \(0\). If both satisfy octets, the second structure is usually preferred because it
ⓐ. has less formal charge separation
ⓑ. uses more electrons than available
ⓒ. makes the molecule ionic by force
ⓓ. gives every atom the same electronegativity
Correct Answer: has less formal charge separation
Explanation: Lewis structures with smaller formal charges are generally preferred when other requirements are satisfied. A structure with all formal charges \(0\) avoids charge separation. The first structure has separated \(+2\), \(-1\), and \(-1\) formal charges even though the total is \(0\). Total charge alone is not enough; distribution of formal charge also matters. A neutral molecule is usually best represented by the structure with the least unnecessary formal charge.
210. In one contributor of \(\mathrm{O_3}\), the central oxygen has one single bond, one double bond, and one lone pair. The formal charge on the central oxygen is
ⓐ. \(-1\)
ⓑ. \(0\)
ⓒ. \(+1\)
ⓓ. \(+2\)
Correct Answer: \(+1\)
Explanation: \( \textbf{Atom considered:} \) The central atom is oxygen.
\( \textbf{Valence electrons of oxygen:} \) \(V=6\).
\( \textbf{Non-bonding electrons:} \) One lone pair gives \(L=2\).
\( \textbf{Bonding electrons:} \) One single bond has \(2\) electrons, and one double bond has \(4\), so \(B=6\).
\( \textbf{Formal-charge formula:} \)
\[
FC=V-L-\frac{B}{2}
\]
\( \textbf{Substitution:} \)
\[
FC=6-2-\frac{6}{2}
\]
\( \textbf{Calculation:} \)
\[
FC=6-2-3=+1
\]
\( \textbf{Final answer:} \) The central oxygen has formal charge \(+1\). In the same contributor, the singly bonded terminal oxygen carries \(-1\), so the neutral \(\mathrm{O_3}\) molecule has total formal charge \(0\).
211. Resonance is needed when
ⓐ. atoms in a molecule physically exchange positions every instant
ⓑ. a compound changes from ionic to metallic bonding
ⓒ. all formal charges in a structure are impossible to calculate
ⓓ. one Lewis structure cannot show the electron distribution
Correct Answer: one Lewis structure cannot show the electron distribution
Explanation: Some molecules and ions cannot be represented well by only one Lewis structure. In such cases, two or more resonance contributors are drawn. The contributors have the same arrangement of atoms but different placement of electrons. The actual species is represented as a resonance hybrid, not as one contributor changing rapidly into another. Resonance is especially useful for species such as \(\mathrm{O_3}\), \(\mathrm{CO_3^{2-}}\), and \(\mathrm{NO_3^-}\).
212. In resonance contributors, the feature that must remain unchanged is
ⓐ. the positions of all lone pairs
ⓑ. the positions of all multiple bonds
ⓒ. the formal charge on every atom
ⓓ. the positions of atoms
Correct Answer: the positions of atoms
Explanation: Resonance contributors differ only in electron placement. The atomic skeleton must remain the same in all contributors. Lone pairs, \(\pi\)-bonds, and formal charges may appear in different positions depending on the contributor. If atoms change positions, the structures are not resonance contributors; they are different arrangements or species. This is why resonance is described as electron delocalization over a fixed framework.
213. A learner says that carbonate ion rapidly flips among three separate structures. The better statement is that carbonate ion is
ⓐ. a resonance hybrid with delocalized electron density
ⓑ. a mixture of three different compounds
ⓒ. a structure in which atoms continuously exchange positions
ⓓ. a molecule with one permanent double bond and two permanent single bonds only
Correct Answer: a resonance hybrid with delocalized electron density
Explanation: The three resonance contributors of \(\mathrm{CO_3^{2-}}\) are ways to represent electron distribution. The real carbonate ion is not flipping among separate structures. Instead, its electron density is delocalized over the three \(\mathrm{C-O}\) bonds. This makes the three \(\mathrm{C-O}\) bonds equivalent in the actual ion. The resonance hybrid is more accurate than any single contributor.
214. Resonance in \(\mathrm{O_3}\) explains why the two \(\mathrm{O-O}\) bond lengths are
ⓐ. equal and intermediate between single and double bonds
ⓑ. one pure single bond and one pure double bond permanently
ⓒ. both longer than ordinary single bonds
ⓓ. unrelated to electron distribution
Correct Answer: equal and intermediate between single and double bonds
Explanation: A single Lewis contributor of \(\mathrm{O_3}\) shows one \(\mathrm{O-O}\) single bond and one \(\mathrm{O=O}\) double bond. Another equivalent contributor places the double bond on the other side. The real ozone molecule is a resonance hybrid, so the two bonds are equivalent. Each bond has partial double-bond character. This gives bond lengths intermediate between ordinary single and double \(\mathrm{O-O}\) bonds.
215. The three major resonance contributors of \(\mathrm{CO_3^{2-}}\) are called equivalent because they
ⓐ. have different positions of carbon and oxygen atoms
ⓑ. contain different total numbers of valence electrons
ⓒ. same atomic arrangement and equal relative stability
ⓓ. represent three separate ions present in equal amounts
Correct Answer: same atomic arrangement and equal relative stability
Explanation: Equivalent resonance contributors have the same arrangement of atoms and the same energy or relative stability. In \(\mathrm{CO_3^{2-}}\), the double bond can be drawn to any one of the three oxygen atoms while the atom positions remain unchanged. Each contributor has the same pattern of formal charges, only placed on different oxygen atoms. The actual ion is a resonance hybrid with all three \(\mathrm{C-O}\) bonds equivalent. The contributors are drawing tools for delocalized electrons, not separate carbonate ions.
216. A resonance contributor is usually more important when it
ⓐ. changes the positions of atoms to reduce bond length
ⓑ. uses more electrons than the species actually has
ⓒ. satisfies octets and has smaller formal charge separation
ⓓ. places negative formal charge on the least electronegative atom in every case
Correct Answer: satisfies octets and has smaller formal charge separation
Explanation: Important resonance contributors usually obey the correct total electron count and satisfy octets where possible. Structures with smaller formal charges and less charge separation usually contribute more strongly. When negative formal charge is present, it is generally better placed on a more electronegative atom. A structure that changes atom positions is not a resonance contributor. Resonance comparison is therefore based on electron placement quality, not on changing the molecular skeleton.
217. In ozone, \(\mathrm{O_3}\), two equivalent resonance contributors show one \(\mathrm{O-O}\) single bond and one \(\mathrm{O=O}\) double bond. The average bond order of each \(\mathrm{O-O}\) bond in the resonance hybrid is
ⓐ. \(1.0\)
ⓑ. \(1.5\)
ⓒ. \(2.0\)
ⓓ. \(3.0\)
Correct Answer: \(1.5\)
Explanation: \( \textbf{Resonance contributors:} \) Each contributor has one single \(\mathrm{O-O}\) bond and one double \(\mathrm{O=O}\) bond.
\( \textbf{Bond order of single bond:} \) A single bond has bond order \(1\).
\( \textbf{Bond order of double bond:} \) A double bond has bond order \(2\).
\( \textbf{Equivalent sharing:} \) In the resonance hybrid, the double-bond character is spread equally over the two \(\mathrm{O-O}\) bonds.
\( \textbf{Average bond order:} \)
\[
\frac{1+2}{2}=1.5
\]
\( \textbf{Final answer:} \) Each \(\mathrm{O-O}\) bond has average bond order \(1.5\). This explains why the two bonds have equal length intermediate between ordinary single and double bonds.
218. Each \(\mathrm{CO_3^{2-}}\) resonance contributor has one \(\mathrm{C=O}\) bond and two \(\mathrm{C-O}\) single bonds. The average \(\mathrm{C-O}\) bond order in the resonance hybrid is
ⓐ. \(1\)
ⓑ. \(\frac{3}{2}\)
ⓒ. \(2\)
ⓓ. \(\frac{4}{3}\)
Correct Answer: \(\frac{4}{3}\)
Explanation: \( \textbf{Bonding in one contributor:} \) One \(\mathrm{C=O}\) double bond has bond order \(2\).
\( \textbf{Other two bonds:} \) Two \(\mathrm{C-O}\) single bonds have total bond order \(1+1=2\).
\( \textbf{Total bond order over three \(\mathrm{C-O}\) links:} \)
\[
2+1+1=4
\]
\( \textbf{Equivalent bonds in hybrid:} \) Resonance makes the three \(\mathrm{C-O}\) bonds equivalent in actual \(\mathrm{CO_3^{2-}}\).
\( \textbf{Average bond order per \(\mathrm{C-O}\) bond:} \)
\[
\frac{4}{3}
\]
\( \textbf{Final answer:} \) The average \(\mathrm{C-O}\) bond order is \(\frac{4}{3}\). The hybrid does not contain one fixed double bond and two fixed single bonds.
219. The nitrate ion, \(\mathrm{NO_3^-}\), has three equivalent resonance contributors with one \(\mathrm{N=O}\) bond in each contributor. The average \(\mathrm{N-O}\) bond order is
ⓐ. \(\frac{3}{4}\)
ⓑ. \(\frac{5}{3}\)
ⓒ. \(2\)
ⓓ. \(\frac{4}{3}\)
Correct Answer: \(\frac{4}{3}\)
Explanation: \( \textbf{One contributor:} \) It contains one \(\mathrm{N=O}\) double bond and two \(\mathrm{N-O}\) single bonds.
\( \textbf{Bond-order sum in one contributor:} \)
\[
2+1+1=4
\]
\( \textbf{Number of equivalent \(\mathrm{N-O}\) bonds in the hybrid:} \) There are three.
\( \textbf{Average bond order:} \)
\[
\frac{4}{3}
\]
\( \textbf{Resonance interpretation:} \) The double-bond character is delocalized over the three \(\mathrm{N-O}\) bonds.
\( \textbf{Final answer:} \) Each \(\mathrm{N-O}\) bond has average bond order \(\frac{4}{3}\). This fractional value belongs to the resonance hybrid, not to any one drawn contributor.
220. Resonance energy is best described as
ⓐ. energy barrier for rotation about a single bond
ⓑ. lattice energy released when an ionic crystal forms
ⓒ. energy lowering due to electron delocalization
ⓓ. energy assigned to one contributor without delocalization
Correct Answer: energy lowering due to electron delocalization
Explanation: Resonance energy expresses the stabilization gained through electron delocalization. The resonance hybrid is lower in energy than any single contributing Lewis structure. This does not mean the molecule is physically changing from one contributor to another. The contributors are only limiting drawings, while the actual species has delocalized electron density. Larger resonance energy generally means greater resonance stabilization.
