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 $3p$

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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 $3d$. In the second excited state,one electron from $3p$ moves to $3d$. In the third excited state,one electron from $3s$ moves to $3d$,resulting in $7$ unpaired electrons $(3s^1, 3p_x^1, 3p_y^1, 3p_z^1, 3d^1, 3d^1, 3d^1)$. Thus,$Cl$ can form $7$ bonds with $F$ 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 :

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