Statement-I: Bond angle of BF₃ and NF₃ are different.

Statement-II: BF₃ and NF₃ are having different shape.

To evaluate the two statements we compare the molecular geometry and bond angles of BF₃ and NF₃.

Step 1: Determine the shape of BF₃
BF₃ has three fluorine atoms bonded to a central boron atom. Boron provides three bonding pairs and no lone pairs. According to VSEPR theory, a molecule with three regions of electron density adopts a trigonal planar geometry. In a trigonal planar molecule the bond angles are all $120°$.

Step 2: Determine the shape of NF₃
NF₃ has three fluorine atoms and one lone pair on the central nitrogen atom. This gives four regions of electron density (three bonding pairs + one lone pair). VSEPR predicts a trigonal pyramidal shape, similar to NH₃. The presence of the lone pair compresses the bond angle to about $102°$ (slightly less than the ideal tetrahedral angle of $109.5°$).

Step 3: Compare the bond angles (Statement I)
The bond angle in BF₃ is $120°$ while the bond angle in NF₃ is roughly $102°$. Because these values are not equal, the statement “Bond angle of BF₃ and NF₃ are different” is true.

Step 4: Compare the shapes (Statement II)
BF₃ is trigonal planar; NF₃ is trigonal pyramidal. Since the two molecules adopt different geometrical arrangements, the statement “BF₃ and NF₃ are having different shape” is also true.

Thus both statements are correct, which matches the provided answer choice.