For octahedral $Mn(II)$ and tetrahedral $Ni(II)$ complexes,consider the following statements:
$(I)$ Both the complexes can be high spin.
$(II)$ $Ni(II)$ complex can very rarely be low spin.
$(III)$ With strong field ligands,$Mn(II)$ complexes can be low spin.
$(IV)$ Aqueous solution of $Mn(II)$ ions is yellow in color.

Using valence bond theory,explain the following in relation to the complexes given below: $[Mn(CN)_{6}]^{3-}, [Co(NH_{3})_{6}]^{3+}, [Cr(H_{2}O)_{6}]^{3+}, [FeCl_{6}]^{4-}$
$(i)$ Type of hybridisation.
$(ii)$ Inner or outer orbital complex.
$(iii)$ Magnetic behaviour.
$(iv)$ Spin only magnetic moment value.

Briefly explain the Valence Bond Theory $(VBT)$.

Which one of the following is square planar in structure and has diamagnetic property?

The octahedral diamagnetic low spin complex among the following is

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $MCl_2 \cdot 6H_2O (X)$ and $NH_4Cl$ gives an octahedral complex $Y$ in the presence of air. In aqueous solution,complex $Y$ behaves as a $1:3$ electrolyte. The reaction of $X$ with excess $HCl$ at room temperature results in the formation of a blue coloured complex $Z$. The calculated spin-only magnetic moment of $X$ and $Z$ is $3.87 \ B.M.$,whereas it is zero for complex $Y$.
Among the following options,which statement$(s)$ is(are) correct?
$[A]$ The hybridization of the central metal ion in $Y$ is $d^2sp^3$
$[B]$ $Z$ is a tetrahedral complex
$[C]$ Addition of silver nitrate to $Y$ gives only two equivalents of silver chloride
$[D]$ When $X$ and $Z$ are in equilibrium at $0^{\circ}C$,the colour of the solution is pink