The hybridisation of the central iodine atom | vedclass

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

Question

(A) The correct option is $A$.Explanation:$1$. For $IF_5$: Iodine has $7$ valence electrons. It forms $5$ sigma bonds with $F$ atoms and has $1$ lone

Vedclass · Accepted Answer

$sp^3d^2, sp^3d, sp^3$

Vedclass · Answer

(A) The correct option is $A$.Explanation:$1$. For $IF_5$: Iodine has $7$ valence electrons. It forms $5$ sigma bonds with $F$ atoms and has $1$ lone pair. Total electron pairs = $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridisation.$2$. For $I_3^-$: The central iodine atom has $7$ valence electrons,plus $1$ electron from the negative charge,totaling $8$. It forms $2$ sigma bonds with other $I$ atoms and has $3$ lone pairs. Total electron pairs = $2 + 3 = 5$,which corresponds to $sp^3d$ hybridisation.$3$. For $I_3^+$: The central iodine atom has $7$ valence electrons,minus $1$ electron from the positive charge,totaling $6$. It forms $2$ sigma bonds with other $I$ atoms and has $2$ lone pairs. Total electron pairs = $2 + 2 = 4$,which corresponds to $sp^3$ hybridisation.

The hybridisation of the central iodine atom in $IF_5$,$I_3^-$,and $I_3^+$ are respectively:

Similar Questions

The ammonium ion $(NH_4^+)$ is:

Discuss the hybridization and shape of $PCl_5$.

The hybrid orbitals used by central atoms in $BeCl_2$,$BCl_3$ and $CCl_4$ molecules are respectively

What are the hybridization states of carbon in diamond,graphite,and acetylene,respectively?

In which of the following molecules/ions $BF_3$,$NO_2^-$,$NH_2^-$ and $H_2O$,the central atom is $sp^2$ hybridised?

Vedclass Test Series

Exam Paper Generator

Online Exam Module