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

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(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

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$ClF_7$,pentagonal bipyramidal

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(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.

List-$I$ (molecule)	List-$II$ (hybridization,shape)
$A. XeO_3$	$I. sp^3$; pyramidal
$B. XeF_2$	$II. sp^3d$; linear
$C. XeOF_4$	$III. sp^3d^2$; square pyramidal
$D. XeF_6$	$IV. sp^3d^3$; distorted octahedral

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

Similar Questions

In $Cl_2O_{6(l)} + HF \to P + Q$. If $H$ of acid $HF$ attaches with $Q$,then identify the correct option for hybridization of $Cl$ atom and $\angle OClO$ in the $P$ and $Q$ ions.

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Match List-$I$ with List-$II$.

List-$I$ (molecule) List-$II$ (hybridization,shape)

$A. XeO_3$ $I. sp^3$; pyramidal

$B. XeF_2$ $II. sp^3d$; linear

$C. XeOF_4$ $III. sp^3d^2$; square pyramidal

$D. XeF_6$ $IV. sp^3d^3$; distorted octahedral

Choose the correct answer from the options given below.

In ethene,the bond angle$(s)$ is/are

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