Step 1: Understanding the Question:
The reaction involves chlorobenzene undergoing an electrophilic aromatic substitution (EAS) reaction, specifically nitration, using a standard nitrating mixture (concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$). We must predict the products based on the directing effects of the chlorine atom.
Step 2: Detailed Explanation:
The chlorine atom ($-\text{Cl}$) attached to the benzene ring exhibits two competing electronic effects:
1. A strong $-I$ effect (Inductive effect): It pulls electron density away from the ring through the sigma bond because it is highly electronegative, making the ring less reactive than pure benzene overall (mildly deactivating).
2. A significant $+R$ effect (Resonance effect): The lone pairs of electrons on the chlorine atom can delocalize into the benzene pi-system. When evaluating resonance structures, this delocalization explicitly increases electron density at the ortho and para positions relative to the chlorine atom.
While the $-I$ effect dominates overall reactivity (making it deactivating), the $+R$ effect dictates the specific orientation of the incoming electrophile. The incoming nitronium ion ($\text{NO}_2^+$) will therefore seek the relatively electron-rich ortho and para positions.
As a result, nitration of chlorobenzene yields a mixture of the ortho and para substituted isomers.
The products are 1-chloro-2-nitrobenzene (ortho isomer, minor due to steric hindrance) and 1-chloro-4-nitrobenzene (para isomer, major product due to symmetry and lack of steric hindrance).
Step 3: Final Answer:
A mixture of 1-chloro-2-nitrobenzene and 1-chloro-4-nitrobenzene is formed, matching option (c).