Chemical Bonding and Molecular Structure Part 3

Concept:Molecular geometry depends on electron pairs and ligand field strength.The octet rule may be incomplete for elements with fewer than 8 valence electrons.Explanation:Statement I:Check each species for tetrahedral geometry.$SF_4$: seesaw (4 bonds + 1 lone pair) – not tetrahedral.$NH_4^+$: tetrahedral (4 bonds, no lone pair) – yes.$[NiCl_4]^{2-}$: $Ni^{2+}$ ($d^8$) with weak field $Cl^-$ gives tetrahedral – yes.$XeF_4$: square planar (4 bonds + 2 lone pairs) – no.$[PtCl_4]^{2-}$: $Pt^{2+}$ ($d^8$) with weak field $Cl^-$ gives square planar – no.$SeF_4$: seesaw (4 bonds + 1 lone pair) – no.$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ ($d^8$) with strong field $CN^-$ gives square planar – no.Thus only 2 species are tetrahedral, not 3. So Statement I is false.Statement II:Check octet around central atom in each molecule.$NO_2$: odd‑electron molecule; N has 5 + 2 = 7 electrons, incomplete octet.$BeH_2$: Be has only 4 electrons (2 bonds).$BF_3$: B has only 6 electrons (3 bonds).$AlCl_3$: monomeric Al has 6 electrons; though it dimerizes, the statement is considered true.All have incomplete octet. So Statement II is true.Answer:Statement I is false but Statement II is true.Correct option: C.