question 5: remember, on the actual exam, you cant search google images for this information! you must rely…

Explanation:

Step1: Draw resonance structures for ozone ($O_3$)

Ozone has a central oxygen atom double - bonded to one oxygen and single - bonded to another. The resonance structures are: $O = O - O$ and $O - O = O$. The formal charges are calculated as follows: For the central O in $O = O - O$, valence electrons of O is 6, non - bonding electrons is 2, and bonding electrons is 6. Formal charge = $6-(2 + \frac{6}{2})=1$. For the double - bonded O, formal charge = $6-(4+\frac{4}{2}) = 0$. For the single - bonded O, formal charge = $6-(6+\frac{2}{2})=-1$. The two resonance structures distribute the formal charges.

Step2: Draw resonance structures for $COOH^-$

The $COOH^-$ has a carbon atom double - bonded to one oxygen and single - bonded to another oxygen with a negative charge. The resonance structures are: $O = C - O^-$ and $O^- - C = O$. For the carbonyl O in $O = C - O^-$, formal charge = $6-(4+\frac{4}{2}) = 0$. For the single - bonded O with negative charge, formal charge = $6-(6+\frac{2}{2})=-1$.

Step3: Draw resonance structures for $NCS^-$

The possible resonance structures for $NCS^-$ are: $N≡C - S^-$, $^-N = C = S$, and $N - C≡S^-$. Calculate formal charges: In $N≡C - S^-$, for N, formal charge = $5-(2+\frac{6}{2}) = 0$, for C, formal charge = $4-(0+\frac{8}{2}) = 0$, for S with negative charge, formal charge = $6-(6+\frac{2}{2})=-1$. In $^-N = C = S$, for N with negative charge, formal charge = $5-(6+\frac{2}{2})=-1$, for C, formal charge = $4-(0+\frac{8}{2}) = 0$, for S, formal charge = $6-(4+\frac{4}{2}) = 0$. In $N - C≡S^-$, for N, formal charge = $5-(4+\frac{4}{2})=-1$, for C, formal charge = $4-(0+\frac{8}{2}) = 0$, for S with negative charge, formal charge = $6-(6+\frac{2}{2})=-1$. The most valid structure is usually the one with the least separation of formal charges and negative charge on the more electronegative atom. For $NCS^-$, $^-N = C = S$ is a more valid structure as S is more electronegative than N and can better accommodate the negative charge in the resonance hybrid.

Step4: Match diagrams for crystal lattice and alloys

• Ionic compound: Diagram (d) represents an ionic compound with alternating positive and negative ions.
• Interstitial alloy: Diagram (b) shows an interstitial alloy where smaller atoms fit into the holes of the lattice of larger atoms.
• Substitutional alloy: Diagram (e) shows a substitutional alloy where atoms of similar size replace each other in the lattice.
• Crystal lattice (metallic - electron sea model): Diagram (a) represents a metallic crystal lattice with a sea of delocalized electrons surrounding positive metal ions.
• Malleability is associated with metallic lattices (a) as the delocalized electrons allow the metal ions to slide past each other.
• Brittleness is associated with ionic compounds (d) as the strong electrostatic forces between ions can be disrupted by dislocation of ions.
• Conductivity is associated with metallic lattices (a) due to the presence of delocalized electrons.

Answer:

a) Resonance structures for ozone ($O_3$): $O = O - O$ and $O - O = O$ b) Resonance structures for $COOH^-$: $O = C - O^-$ and $O^- - C = O$ c) Resonance structures for $NCS^-$: $N≡C - S^-$, $^-N = C = S$, $N - C≡S^-$; most valid is $^-N = C = S$ d) Ionic compound: Diagram (d) e) Interstitial alloy: Diagram (b) f) Substitutional alloy: Diagram (e) g) Crystal lattice (metallic - electron sea model): Diagram (a) h) Malleability: Diagram (a) i) Brittleness: Diagram (d) j) Conductivity: Diagram (a)