The type of hybrid orbitals used by the chlor | vedclass

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

Question

(A) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electro

Vedclass · Accepted Answer

$sp^3$

Vedclass · Answer

(A) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $ClO_2^-$,$V = 7$ (for $Cl$),$M = 0$ (oxygen is divalent),$C = 0$,and $A = 1$.$H = \frac{1}{2}(7 + 0 - 0 + 1) = \frac{8}{2} = 4$.$A$ value of $4$ corresponds to $sp^3$ hybridization.Therefore,the chlorine atom in $ClO_2^-$ is $sp^3$ hybridized,resulting in an angular (bent) shape due to the presence of two lone pairs.

The type of hybrid orbitals used by the chlorine atom in $ClO_2^-$ is

Similar Questions

Compare the carbon-carbon $\sigma$ bond length in ethane, ethylene and acetylene.

The shape of $CH_3^+$ species is

What is the hybridization of $Xe$ in $XeF_2$?

Hybridisation states of $C$ in $CH_3^+$ and $CH_4$ are

Match the following:
| List-$I$ (Molecule/ion) | List-$II$ (Hybridisation) |
| :--- | :--- |
| $(A)$ $ICl_4^-$ | $(P)$ $sp^3$ |
| $(B)$ $NO_3^-$ | $(Q)$ $sp^3d$ |
| $(C)$ $PCl_4^+$ | $(R)$ $sp^3d^2$ |
| $(D)$ $SiF_6^{2-}$ | $(S)$ $sp^2$ |
| | $(T)$ $sp$ |

Vedclass Test Series

Exam Paper Generator

Online Exam Module