In which of the following sets, all the given species are isostructural?

To determine which set contains species that are all isostructural, we need to recall that isostructural means having the same molecular geometry. We will analyze each option step by step.

Starting with Option A: $$CO_2$$, $$NO_2$$, $$ClO_2$$, $$SiO_2$$.

Carbon dioxide ($$CO_2$$) has a linear structure with bond angle 180° due to sp hybridization. Nitrogen dioxide ($$NO_2$$) has an odd number of valence electrons (17 total), resulting in a bent structure with a bond angle of approximately 134°. Chlorine dioxide ($$ClO_2$$) also has an odd number of valence electrons (19 total), leading to a bent structure with a bond angle of about 118°. Silicon dioxide ($$SiO_2$$) is not a discrete molecule; it forms a network solid with tetrahedral coordination around silicon atoms. Thus, the species have different geometries: linear, bent, bent, and tetrahedral network. They are not isostructural.

Next, Option B: $$PCl_3$$, $$AlCl_3$$, $$BCl_3$$, $$SbCl_3$$.

Phosphorus trichloride ($$PCl_3$$) has a central phosphorus atom with three bonds and one lone pair, resulting in a trigonal pyramidal geometry. Aluminum trichloride ($$AlCl_3$$) in its monomeric form (considered for molecular geometry) has aluminum with three bonds and no lone pairs, giving trigonal planar geometry. Boron trichloride ($$BCl_3$$) is trigonal planar due to three bonds and no lone pairs on boron. Antimony trichloride ($$SbCl_3$$) is similar to $$PCl_3$$, with a central antimony atom having three bonds and one lone pair, so trigonal pyramidal. Thus, we have trigonal pyramidal ($$PCl_3$$ and $$SbCl_3$$) and trigonal planar ($$AlCl_3$$ and $$BCl_3$$). They are not all isostructural.

Moving to Option C: $$BF_3$$, $$NF_3$$, $$PF_3$$, $$AlF_3$$.

Boron trifluoride ($$BF_3$$) has three bonds and no lone pairs on boron, so trigonal planar. Nitrogen trifluoride ($$NF_3$$) has nitrogen with three bonds and one lone pair, leading to trigonal pyramidal geometry. Phosphorus trifluoride ($$PF_3$$) is similar to $$NF_3$$, with phosphorus having three bonds and one lone pair, so trigonal pyramidal. Aluminum trifluoride ($$AlF_3$$) in its monomeric form has aluminum with three bonds and no lone pairs, resulting in trigonal planar geometry. Thus, we have trigonal planar ($$BF_3$$ and $$AlF_3$$) and trigonal pyramidal ($$NF_3$$ and $$PF_3$$). They are not all isostructural.

Finally, Option D: $$BF_4^-$$, $$CCl_4$$, $$NH_4^+$$, $$PCl_4^+$$.

Tetrafluoroborate ion ($$BF_4^-$$) has boron with four bonds and no lone pairs (due to the negative charge), so tetrahedral geometry. Carbon tetrachloride ($$CCl_4$$) has carbon with four bonds and no lone pairs, tetrahedral. Ammonium ion ($$NH_4^+$$) has nitrogen with four bonds and no lone pairs (due to the positive charge), tetrahedral. Tetrachlorophosphonium ion ($$PCl_4^+$$) has phosphorus with four bonds and no lone pairs (positive charge removes the lone pair), tetrahedral. All species have tetrahedral geometry.

Hence, only Option D has all species with the same molecular geometry (tetrahedral). So, the answer is Option D.