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(C) $1$. In $[Cr(NH_3)_6]Cl_3$,the oxidation state of $Cr$ is $+3$. The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.$2$. Since $NH_3$ is a st

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The complex is an outer orbital complex.

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(C) $1$. In $[Cr(NH_3)_6]Cl_3$,the oxidation state of $Cr$ is $+3$. The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.$2$. Since $NH_3$ is a strong field ligand,it causes pairing,but for $d^3$ configuration,there are already $3$ unpaired electrons in $3d$ orbitals. The hybridization is $d^2sp^3$,which corresponds to an inner orbital complex,not an outer orbital complex.$3$. The geometry is octahedral. The complex is paramagnetic due to the presence of $3$ unpaired electrons.$4$. It reacts with $AgNO_3$ to give $3$ moles of $AgCl$ (white precipitate) because there are $3$ chloride ions outside the coordination sphere.$5$. Therefore,the statement that it is an outer orbital complex is incorrect.

Which statement is incorrect regarding the complex $[Cr(NH_3)_6]Cl_3$?

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