Find the number of unpaired electrons in each case

$$\begin{gathered} {\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}=x\mathit{ }{\left[{\mathrm{FeF}}_{\mathrm{6}}\right]}^{\mathit{3}\mathit{-}}=t \\ {\left[{\mathrm{MnCl}}_{\mathrm{6}}\right]}^{\mathit{3}\mathit{-}}=y\mathit{ }{\left[{\mathrm{CoF}}_{\mathrm{6}}\right]}^{\mathit{3}\mathit{-}}=u \\ {\left[\mathrm{Fe}\mathrm{(}\mathrm{CN}{\mathrm{)}}_{\mathrm{6}}\right]}^{\mathit{3}\mathit{-}}=z\mathit{ }\left[\mathrm{Co}{\left({\mathrm{H}}_{\mathrm{2}}\mathrm{O}\right)}_{\mathrm{3}}{\mathrm{F}}_{\mathrm{3}}\right]=v \end{gathered}$$

Hence, find the correct relationship between x, y, z, t, U, V

Number of unpaired electrons in the given complexes can be given as,

$\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]-=x\mathit{=}\mathit{2}$

${\left[{\mathrm{MnCl}}_{\mathrm{6}}\right]}^{\mathit{3}\mathit{-}}=y\mathit{=}\mathit{4}$

${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}=z\mathit{=}\mathit{1}$

${\left[{\mathrm{FeF}}_6\right]}^{3-}=t\mathit{=}\mathit{5}$

${\left[{\mathrm{CoF}}_6\right]}^{3-}=u\mathit{=}\mathit{4}$

$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{F}}_3\right]=v\mathit{=}\mathit{0}$

Here,

y - u = v

4 - 4 = 0

Here, the relation of A, B and C are correct.