KI + MnO_2 + H_2SO_4 to Vapours (B) + other p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: $2KI + MnO_2 + 2H_2SO_4 	o K_2SO_4 + MnSO_4 + 2H_2O + I_2(g)$. Thus,$(B) = I_2$.Step $2$: $I_2 + K_2S_2O_8 	o K_2SO_4 + I_2(SO_4)$ (or

Vedclass · Accepted Answer

$sp^3d$

Vedclass · Answer

(A) Step $1$: $2KI + MnO_2 + 2H_2SO_4 	o K_2SO_4 + MnSO_4 + 2H_2O + I_2(g)$. Thus,$(B) = I_2$.Step $2$: $I_2 + K_2S_2O_8 	o K_2SO_4 + I_2(SO_4)$ (or similar oxidation reaction). Here $(D)$ is $I_2$.Step $3$: $I_2 + KI 	o KI_3$. The brown complex $(E)$ is $KI_3$,which contains the triiodide ion $I_3^-$.Step $4$: In $I_3^-$,the central iodine atom has $3$ lone pairs and $2$ bond pairs,resulting in $5$ electron pairs. The hybridisation is $sp^3d$.

$KI + MnO_2 + H_2SO_4 \to$ Vapours $(B) +$ other products
Vapours $(B) + K_2S_2O_8 \to (C) + (D)$
$(D) \xrightarrow{(B)}$ Brown $(E)$
Hybridisation of central atom in $(E)$ is

Similar Questions

Among $[I_3]^{+}$,$[SiO_4]^{4-}$,$SO_2Cl_2$,$XeF_2$,$SF_4$,$ClF_3$,$Ni(CO)_4$,$XeO_2F_2$,$[PtCl_4]^{2-}$,$XeF_4$,and $SOCl_2$,the total number of species having $sp^3$ hybridised central atom is. . . . .

Which of the following pairs have the same type of hybridization but different shapes?

In which of the following compounds does the $C$ atom use $sp^3$ hybrid orbitals for bonding?

The geometry of the molecule with $sp^3d^2$ hybridised central atom is

Total number of species from the following with central atom utilising $sp^2$ hybrid orbitals for bonding is.................... $NH_3, SO_2, SiO_2, BeCl_2, C_2H_2, C_2H_4, BCl_3, HCHO, C_6H_6, BF_3, C_2H_4Cl_2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$KI + MnO_2 + H_2SO_4 \to$ Vapours $(B) +$ other productsVapours $(B) + K_2S_2O_8 \to (C) + (D)$$(D) \xrightarrow{(B)}$ Brown $(E)$Hybridisation of central atom in $(E)$ is