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

Compounds with Tetrahedral Geometry 

Given Compounds:

• \( \text{NH}_4^+ \) (Ammonium ion)
• \( \text{XeF}_4 \) (Xenon tetrafluoride)
• \( [\text{NiCl}_4]^{2-} \) (Nickel(II) chloride complex)
• \( [\text{PtCl}_4]^{2-} \) (Platinum(II) chloride complex)
• \( [\text{Cu(NH}_3)_4]^{2+} \) (Copper(II) complex)
• \( \text{BF}_3 \) (Boron trifluoride)
• \( [\text{Ni(CO)_4}] \) (Nickel(0) complex)

Step-by-Step Analysis:

1. \( \text{NH}_4^+ \) (Ammonium ion): It has 4 bonding pairs around nitrogen and no lone pairs, which gives it a tetrahedral geometry.

2. \( \text{XeF}_4 \) (Xenon tetrafluoride): It has 4 fluorine atoms and 2 lone pairs on Xenon, giving it a square planar geometry.

3. \( [\text{NiCl}_4]^{2-} \) (Nickel(II) chloride complex): Nickel has 4 chloride atoms and adopts a square planar geometry.

4. \( [\text{PtCl}_4]^{2-} \) (Platinum(II) chloride complex): Platinum in this complex also has 4 chloride ions and adopts a square planar geometry.

5. \( [\text{Cu(NH}_3)_4]^{2+} \) (Copper(II) complex): Copper has 4 ammonia molecules and adopts a square planar geometry.

6. \( \text{BF}_3 \) (Boron trifluoride): It has 3 fluorine atoms and adopts a trigonal planar geometry.

7. \( [{Ni(CO)_4}] \) (Nickel(0) complex): Nickel has 4 carbon monoxide molecules and adopts a tetrahedral geometry.

Summary of Tetrahedral Geometry Compounds:

• \( \text{NH}_4^+ \) (Ammonium ion)
• \([{Ni(CO)_4}]\) (Nickel(0) complex)

Answer: Out of the listed compounds, 2 compounds have tetrahedral geometry.