$A$ complex with the composition $[MA_3B]^{n \pm}$ is found to have no geometrical isomers. The possible structure$(s)$ of the complex is/are (where $A$ and $B$ are monodentate ligands):

Match the following:

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

The correct match is:

Draw the structure of the $[Cr(CO)_6]$ complex.

Identify the $CORRECT$ set of details from the following:
$A$. $[Co(NH_{3})_{6}]^{3+}$: Inner orbital complex; $d^{2}sp^{3}$ hybridized
$B$. $[MnCl_{6}]^{3-}$: Outer orbital complex; $sp^{3}d^{2}$ hybridized
$C$. $[CoF_{6}]^{3-}$: Outer orbital complex; $d^{2}sp^{3}$ hybridized
$D$. $[FeF_{6}]^{3-}$: Outer orbital complex; $sp^{3}d^{2}$ hybridized
$E$. $[Ni(CN)_{4}]^{2-}$: Inner orbital complex; $sp^{3}$ hybridized
Choose the correct answer from the options given below:

The hybridisation and magnetic behavior of the complex $[Ni(CO)_4]$ are

According to Valence Bond Theory,the number of unpaired electrons present in $[MnCl_6]^{3-}$,$[Fe(CN)_6]^{3-}$ and $[Co(C_2O_4)_3]^{3-}$,respectively,are