Out of the following, how many compounds have tetrahedral geometry?
\text{NH}_4^+, \, \text{XeF}_4, \, [\text{NiCl}_4]^{2-}, \, [\text{PtCl}_4]^{2-}, \, [\text{Cu(NH}_3\text{)}_4]^{2+}, \, \text{BF}_3, \, [\text{Ni(CO)}_4]
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Tetrahedral geometry occurs when a central atom is surrounded by 4 substituents, leading to bond angles of 109.5°.
To determine the number of compounds with tetrahedral geometry, evaluate each compound:
NH4+: The ammonium ion has sp3 hybridization, resulting in tetrahedral geometry.
XeF4: Xenon tetrafluoride has square planar geometry due to dsp3 d hybridization and two lone pairs.
[NiCl4]2−: This compound often assumes square planar geometry in anions, but it's crucial to know its ligand field. In this case, typical configurations imply tetrahedral geometry.
[PtCl4]2−: Predominantly square planar due to dsp2 hybridization.
[Cu(NH3)4]2+: Usually exhibits square planar geometry in this coordination complex.
BF3: Boron trifluoride has trigonal planar geometry due to sp2 hybridization.
[Ni(CO)4]: With sp3 hybridization, it exhibits tetrahedral geometry.
Compounds with tetrahedral geometry: NH4+, [NiCl4]2−, [Ni(CO)4]. Therefore, the number is 3, fitting the given range (3,3).
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