The hybridisation of the central iodine atom | vedclass

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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.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

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

Similar Questions

The state of hybridisation of the central atom in the dimer of $BH_3$ and $BeH_2$ is

Which of the $C - C$ bond is strongest?

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

What is the bond angle around an atom that undergoes $sp^2$ hybridization?

Which of the following has a linear structure?

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