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

Discuss the hybridization and shape of $PCl_5$.

What is the state of hybridisation of carbon in $(a)$ $CO_3^{2-}$,$(b)$ diamond,and $(c)$ graphite?

The carbon atoms in an acetylene molecule are in which state of hybridisation?

Which of the following hybridisation results in non-planar orbitals?

Explain the tetrahedral structure of $CH_4$ (methane) using its bonding structure.

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