$[Cu(NH_3)_4]^{2+}$ shows the following hybridization:

Match each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is

$LIST-I$	$LIST-II$
$P. dsp^2$	$1. [FeF_6]^{4-}$
$Q. sp^3$	$2. [Ti(H_2O)_3Cl_3]$
$R. sp^3d^2$	$3. [Cr(NH_3)_6]^{3+}$
$S. d^2sp^3$	$4. [FeCl_4]^{2-}$
	$5. Ni(CO)_4$
	$6. [Ni(CN)_4]^{2-}$

Which of the following complexes is diamagnetic and square planar?

Which of the following statements are true about $[CoF_6]^{3-}$ ion?
$I$. The complex has octahedral geometry.
$II$. Coordination number of $Co$ is $3$ and oxidation state is $+6$.
$III$. The complex is $sp^3d^2$ hybridised.
$IV$. It is a high spin complex.

Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2-}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Observe the following complexes: $[Mn(CN)_6]^{3-}, [Fe(CN)_6]^{3-}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}, [Fe(CN)_6]^{4-}, [CoF_6]^{3-}$. From the above,the number of inner orbital complexes with paramagnetic nature is