Ni(CO)4 is high spin complex or low spin complex ? It is tetrahedral molecule so it should be high spin complex ( as tetrahedral complex are always high spin )

Q. Tetrahedral complexes are generally high spin, this is because:

Q. $\left[Fe\right(CN{)}_6{]}^{-3}$ is low spin complex but $\left[Fe\right(H_{2}O{)}_6{]}^{+3}$ is high spin complex. Explain.

Q. The first row transition metal complexes having tetrahedral geometry are high spin due to:

Q. Column-1 represents the formula of the complex.
Column-2 represents the hybridization state of the complex and
Column-3 represents the nature of the complex.
$\begin{array}{cccccc}& ColumnI& & Column2& & Column3\\ \left(I\right)& \left[Ni\right(CO{)}_4]& \left(i\right)& s{p}^3& \left(P\right)& Diamagnetic\left(Highspin\right)\\ \left(II\right)& [PtC{l}_4{]}^{2-}& \left(ii\right)& s{p}^3{d}^2& \left(Q\right)& Paramagnetic\left(Highspin\right)\\ \left(III\right)& \left[Ni\right(CO{)}_4]& \left(iii\right)& ds{p}^2& \left(R\right)& Diamagnetic\left(Lowspin\right)\\ \left(IV\right)& K_2\left[Co\right(CN{)}_4]& \left(iv\right)& d^{2}s{p}^3& \left(S\right)& Paramagnetic\left(Lowspin\right)\end{array}$
Which of the following set is correct for a diamagnetic tetrahedral complex?