Given below are two statements:
Statement I :
The number of species among $$SF_{4},NH_4^+,[NiCl_{4}]^{2-},XeF_{4},[PtCl_{4}]^{2-},SeF_{4}\text{ and }[Ni(CN)_{4}]^{2-}$$, tha t have tetrahedral geometry is 3.
Statement II :
In the set $$[NO_{2},BeH_{2},BF_{3},AlCl_{3}]$$ all the molecules have incomplete octet around central atom.
In the light of the above statements, choose the correct answer from the options given below :

We need to evaluate two statements about molecular geometry.

Statement I: Claims that 3 species from the list are tetrahedral: $$NH_4^+, [NiCl_4]^{2-}, SF_4, XeF_4, [PtCl_4]^{2-}, SeF_4, [Ni(CN)_4]^{2-}$$.

Let us check each:

- $$NH_4^+$$: 4 bond pairs, 0 lone pairs on N. Tetrahedral.

- $$[NiCl_4]^{2-}$$: Ni²⁺ is d⁸, Cl⁻ is a weak field ligand. Tetrahedral ($$sp^3$$).

- $$SF_4$$: S has 4 bond pairs + 1 lone pair. Shape is see-saw (not tetrahedral).

- $$XeF_4$$: Xe has 4 bond pairs + 2 lone pairs. Shape is square planar.

- $$[PtCl_4]^{2-}$$: Pt²⁺ is d⁸, strong crystal field. Square planar.

- $$SeF_4$$: Se has 4 bond pairs + 1 lone pair. Shape is see-saw.

- $$[Ni(CN)_4]^{2-}$$: Ni²⁺ is d⁸, CN⁻ is strong field. Square planar ($$dsp^2$$).

Only 2 species are tetrahedral, but Statement I claims 3. Statement I is FALSE.

Statement II: "$$NO_2, BeH_2, BF_3, AlCl_3$$ have incomplete octet around the central atom."

- $$NO_2$$: N has 17 valence electrons total; one N has 7 electrons around it (incomplete). Yes.

- $$BeH_2$$: Be has only 4 electrons (2 bonds). Yes, incomplete octet.

- $$BF_3$$: B has only 6 electrons (3 bonds). Yes, incomplete octet.

- $$AlCl_3$$: Al has only 6 electrons (3 bonds). Yes, incomplete octet.

Statement II is TRUE.

The correct answer is Option (4): Statement I is false but Statement II is true.