Chemistry - Complete Pblock- Question with Solution | TestHub

The reaction is a Lewis acid-base reaction. XeF₆ acts as a Lewis base, donating a fluoride ion to MF₃, which acts as a Lewis acid.

Reaction:

$${\mathrm{XeF}}_6+{\mathrm{MF}}_3\to {\left[{\mathrm{XeF}}_5\right]}^{+}{\left[{\mathrm{MF}}_4\right]}^{-}$$

Option A: Hybridization of Xe in XeF₆ is sp³d³, and in ${\left[{\mathrm{XeF}}_5\right]}^{+}$ it is sp³d². This is correct.

Option B: Oxidation states:

In XeF₆: Xe is +6, F is -1.

In MF₃: M is +3, F is -1.

In ${\left[{\mathrm{XeF}}_5\right]}^{+}$: Xe is +6, F is -1.

In ${\left[{\mathrm{MF}}_4\right]}^{-}$: M is +3, F is -1.

Since no oxidation states change, it is not a redox reaction. This is correct.

Option C: Characteristics of M:

(i) MX₃ are Lewis acids. Boron halides (BF₃, BCl₃) are known Lewis acids due to an incomplete octet.

(ii) M₂H₆ forms bridge bonding and undergoes unsymmetrical cleavage with NH₃. Diborane (B₂H₆) is a classic example of a compound with bridge bonding that reacts with NH₃ via unsymmetrical cleavage.

(iii) MF₃ shows 2p$\pi$-2p$\pi$ back bonding. BF₃ exhibits 2p$\pi$-2p$\pi$ back bonding from F to B.

All characteristics match Boron (B). This is correct.

Option D: Order of Lewis acidic strength of halides of M (Boron):

The order is BCl₃ > BF₃ > BBr₃ > BI₃. This is due to the extent of back bonding. In BF₃, the strong 2p$\pi$-2p$\pi$ back bonding from F to B reduces the electron deficiency of B, making it a weaker Lewis acid compared to BCl₃ where back bonding is less effective. Therefore, the given order ${\mathrm{MF}}_3>{\mathrm{MCl}}_3>{\mathrm{MBr}}_3>{\mathrm{MI}}_3$ is incorrect.

Therefore, options A, B, and C are correct.