Chemistry - Complete P-block Question with Solution | TestHub

Detailed analysis:

A. Hydrolysis of TeF₆:

TeF₆ + 6H₂O -> H₆TeO₆ + 6HF

H₆TeO₆ is telluric acid, a weak acid. HF is a weak acid.

For neutralization, 6 moles of NaOH are needed for 6HF and 2 moles of NaOH for H₆TeO₆ (as it's a diprotic acid in this context, forming TeO₄²⁻). So, 8 moles of NaOH are required, not 2. Thus, statement A is incorrect.

B. Bond lengths in BF₃ and BF₂NH₂:

In BF₃, the B-F bond has partial double bond character due to pπ-pπ backbonding from F to B.

In BF₂NH₂, the nitrogen atom also has a lone pair that can participate in backbonding with boron. This competition for backbonding reduces the double bond character of the B-F bond in BF₂NH₂ compared to BF₃. Therefore, the B-F bond length in BF₃ is shorter than in BF₂NH₂. Thus, statement B is incorrect.

C. Non-bonding orbitals in CO:

CO is a heteronuclear diatomic molecule. Due to the difference in electronegativity between carbon and oxygen, the atomic orbitals of carbon and oxygen have different energy levels. Consequently, the non-bonding molecular orbitals (which are primarily localized on one atom) will also have different energy levels. Thus, statement C is correct.

D. Hydrolysis of SbCl₃:

SbCl₃ + H₂O <=> SbOCl (s) + 2HCl

This is an equilibrium reaction. According to Le Chatelier's principle, adding excess HCl (a product) will shift the equilibrium to the left (backward direction), thereby suppressing the hydrolysis of SbCl₃. Thus, statement D is correct.

The final answer is C, D