Question

Out of the following, how many compounds have tetrahedral geometry? \[ {NH4^+,\ XeF4,\ [NiCl4]^{2-},\ [PtCl4]^{2-},\ [Cu(NH3)4]^{2+},\ BF3,\ [Ni(CO)4} \]

Hint:

For geometry determination:
Check hybridisation of the central atom
\(sp^3 \Rightarrow\) tetrahedral (if no lone-pair distortion)
Many \(d^8\) metal complexes are square planar

Answer 1

Correct Answer: 3

To determine how many compounds have a tetrahedral geometry, we need to analyze the molecular geometry of each compound using VSEPR theory or the corresponding coordination configuration: 

• \(NH_4^+\): Ammonium ion has 4 bonding pairs and no lone pairs around nitrogen. According to VSEPR, this results in a tetrahedral geometry.
• \(XeF_4\): Xenon tetrafluoride has 4 bonding pairs and 2 lone pairs. VSEPR predicts a square planar geometry.
• \([NiCl_4]^{2-}\): Nickel in the +2 oxidation state with 4 Cl⁻ ions forms a tetrahedral compound because it involves sp³ hybridization with no lone pairs affecting the geometry.
• \([PtCl_4]^{2-}\): Platinum(II) forms a square planar complex typically due to dsp² hybridization.
• \([Cu(NH_3)_4]^{2+}\): Copper forms a square planar complex in its +2 oxidation state.
• \(BF_3\): Boron trifluoride has trigonal planar geometry due to 3 bonding pairs around boron.
• \([Ni(CO)_4]\): Nickel tetracarbonyl is tetrahedral, with sp³ hybridization as it uses zero oxidation state nickel.

Analyzing these, \(NH_4^+\), \([NiCl_4]^{2-}\), and \([Ni(CO)_4]\) have tetrahedral geometries. Therefore, the number of compounds with tetrahedral geometry is 3, which falls within the given range of 3 to 3.