Question

Given below are two statements:
Statement I: The number of species among
\(SF_4, NH_4^+, Ni(CO)_4, XeF_4, [PtCl_{4}]^{2-, SeF_4, [Ni(CN)_4]^{2-}\) that have tetrahedral geometry is 3.
Statement II: In the set \(NO, BeH_2, BF_3, AlCl_3\), all the molecules have incomplete octet around 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: \begin{itemize} \item \(SF_4\): \(sp^3d\), See-saw (Not Tetrahedral). \item \(NH_4^+\): \(sp^3\), Tetrahedral. \item \(Ni(CO)_4\): \(sp^3\), Tetrahedral. \item \(XeF_4\): \(sp^3d^2\), Square Planar. \item \([PtCl_4]^{2-}\): \(dsp^2\), Square Planar. \item \(SeF_4\): \(sp^3d\), See-saw. \item \([Ni(CN)_4]^{2-}\): \(dsp^2\), Square Planar. \end{itemize} 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)
\begin{itemize} \item \(NO\): Nitrogen has 7 valence \(e^-\), Oxygen 6. Total 11. Odd electron species. Incomplete octet on N (7 electrons). \item \(BeH_2\): Be has 2 valence \(e^-\). Forms 2 bonds. Total 4 valence \(e^-\). Incomplete octet. \item \(BF_3\): B has 3 valence \(e^-\). Forms 3 bonds. Total 6 valence \(e^-\). Incomplete octet. \item \(AlCl_3\): Al has 3 valence \(e^-\). Forms 3 bonds. Total 6 valence \(e^-\). Incomplete octet. \end{itemize} All species have an incomplete octet. Thus, Statement II is True. Final Answer: Option 3.