The hybridization states of the central atoms | vedclass

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(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central at

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

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(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $I_3^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.$2$. For $ICl_4^-$: Central atom $I$ has $V=7$,$M=4$,$A=1$. $H = \frac{1}{2}(7 + 4 + 1) = 6$. Hybridization is $sp^3d^2$.$3$. For $ICl_2^-$: Central atom $I$ has $V=7$,$M=2$,$A=1$. $H = \frac{1}{2}(7 + 2 + 1) = 5$. Hybridization is $sp^3d$.Thus,the hybridization states are $sp^3d, sp^3d^2, sp^3d$.

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 hybridization states of the central atoms of the ions $I_3^-$,$ICl_4^-$ and $ICl_2^-$ are respectively

Similar Questions

$sp^3$ hybridisation is found in

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:

In which of the following reactions is there no change in the hybridisation of the central atom?

In which hybridisation is an angular shape not possible?

Consider the following reaction: $MX_4 + X'_2 \to MX_4X'_2$. If the atomic number of $M$ is $52$ and $X$ and $X'$ are halogens and $X'$ is more electronegative than $X$. Then choose the correct statement regarding the given information.

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