A $\pi$-bond is formed by the sidewise overlap of p-orbitals. The formation of a $\pi$-bond in addition to a $\sigma$-bond (forming a double bond) or two $\pi$-bonds (forming a triple bond) increases the electron density between the two bonded atoms. This increased electron density leads to a stronger attraction between the nuclei and the shared electrons, pulling the nuclei closer together. Consequently, the bond length decreases as the bond order increases.

For example, the C-C single bond length is approximately 1.54 Å, the C=C double bond length is approximately 1.34 Å, and the C≡C triple bond length is approximately 1.20 Å. This clearly shows that the formation of $\pi$-bonds leads to a decrease in bond length.

• A) increases bond length: This is incorrect. The formation of $\pi$-bonds increases bond order, which strengthens the bond and decreases bond length.
• B) decreases bond length: This is correct. The additional electron density from $\pi$-bonds pulls the nuclei closer, resulting in a shorter bond.
• C) distorts the geometry of molecule: This is incorrect. While $\pi$-bonds restrict rotation and influence geometry (e.g., planar for sp2, linear for sp), they don't necessarily 'distort' it in a negative sense; rather, they define a specific, stable geometry.
• D) makes homoatomic molecules more reactive: This is not universally true. While $\pi$-bonds are sites of reactivity (e.g., addition reactions), the overall reactivity depends on many factors, and increased bond strength can sometimes lead to less reactivity in certain contexts. For instance, N$\equiv$N is very stable due to the strong triple bond.

Correct Answer: (B)