Question

Given below are two statements:
Statement I: The number of species among \( SF_4 \), \( NH_4^+ \), \( [NiCl_4]^{2-} \), \( XeF_4 \), \( [PtCl_4]^{2-} \), \( SeF_4 \) and \( [Ni(CN)_4]^{2-} \), that 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 the central atom.
In the light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is false but Statement II is true
• D. Statement I is true but Statement II is false

Hint:

Remember that 4d and 5d transition metals (like Pt) typically form square planar complexes with coordination number 4, regardless of the ligand strength.

Answer 1

The Correct Option is C

   Step 1: Evaluate Statement I (Geometry) 
Check the geometry of each species:

• \(SF_4\): \(sp^3d\), See-saw (Not Tetrahedral)
• \(NH_4^+\): \(sp^3\), Tetrahedral.
• \(Ni(CO)_4\): \(sp^3\), Tetrahedral.
• \(XeF_4\): \(sp^3d^2\), Square Planar.
• \([PtCl_4]^{2-}\): \(dsp^2\), Square Planar.
• \(SeF_4\): \(sp^3d\), See-saw.
• \([Ni(CN)_4]^{2-}\): \(dsp^2\), Square Planar.

Total tetrahedral species = 2 (\(NH_4^+, Ni(CO)_4\)). Statement I claims "3". 
Thus, Statement I is False. Step 2: Evaluate Statement II (Octet Rule) 

• \(NO\): Nitrogen has 7 valence \(e^-\), Oxygen 6. Total 11. Odd electron species. Incomplete octet on N (7 electrons).
• \(BeH_2\): Be has 2 valence \(e^-\). Forms 2 bonds. Total 4 valence \(e^-\). Incomplete octet.
• \(BF_3\): B has 3 valence \(e^-\). Forms 3 bonds. Total 6 valence \(e^-\). Incomplete octet.
• \(AlCl_3\): Al has 3 valence \(e^-\). Forms 3 bonds. Total 6 valence \(e^-\). Incomplete octet. All species have an incomplete octet.

Thus, Statement II is True. Final Answer: Option 3.