Chlorine atom,in its third excited state,reac | vedclass

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

Question

(D) The ground state electronic configuration of $Cl$ $(Z=17)$ is $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^1$.In the first excited state,one electron from $3p$ m

Vedclass · Accepted Answer

$ClF_7$,pentagonal bipyramidal

Vedclass · Answer

(D) The ground state electronic configuration of $Cl$ $(Z=17)$ is $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^1$.In the first excited state,one electron from $3p$ moves to $3d$,giving $3$ unpaired electrons.In the second excited state,one electron from $3s$ moves to $3d$,giving $5$ unpaired electrons.In the third excited state,one more electron from $3p$ moves to $3d$,resulting in $7$ unpaired electrons $(3s^1, 3p_x^1, 3p_y^1, 3p_z^1, 3d^3)$.These $7$ unpaired electrons react with $7$ fluorine atoms to form $ClF_7$.The hybridization is $sp^3d^3$,which corresponds to a pentagonal bipyramidal geometry.

Chlorine atom,in its third excited state,reacts with fluorine to form a compound $X$. The formula and shape of $X$ are :

Similar Questions

Describe the change in hybridization (if any) of the $Al$ atom in the following reaction:
$AlCl_3 + Cl^{-} \to AlCl_4^-$

Is there any change in the hybridisation of $B$ and $N$ atoms as a result of the following reaction?
$BF_{3} + NH_{3} \rightarrow F_{3}B \cdot NH_{3}$

Choose the molecules in which hybridization occurs in the ground state?
$(i)$ $BCl_3$ $(ii)$ $NH_3$
$(iii)$ $PCl_3$ $(iv)$ $BeF_2$
The correct answer is

Hybridisation involves

The hybridization in sulphur dioxide is

Vedclass Test Series

Exam Paper Generator

Online Exam Module