${\mathrm{Mn}}^{2+}$ forms a complex with ${\mathrm{Br}}^{-}$ ion. The magnetic moment (spin only) of the complex is $5.92 \mathrm{B}.\mathrm{M}.$ The probable formula and geometry of the complex are, respectively.

Consider the following complex ions, P, Q and R.

$\mathrm{P}={\left[{\mathrm{FeF}}_6\right]}^{3-},\mathrm{Q}={\left[\mathrm{V}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}, \mathrm{R}={\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ 

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is

For octahedral $\mathrm{Mn}\left(\mathrm{II}\right)$ and tetrahedral $\mathrm{Ni}\left(\mathrm{II}\right)$ complexes, consider the following statements :

$\left(\mathrm{I}\right)$ both the complexes can be high spin.

$\left(\mathrm{II}\right)$ $\mathrm{Ni}\left(\mathrm{II}\right)$ complex can very rarely be of low spin.

$\left(\mathrm{III}\right)$ with strong field ligands, $\mathrm{Mn}\left(\mathrm{II}\right)$ complexes can be low spin.

$\left(\mathrm{IV}\right)$ aqueous solution of $\mathrm{Mn}\left(\mathrm{II}\right) \mathrm{ions}$ is yellow in color.

The correct statements are :

Choose the complex, which is paramagnetic

(i) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$

(ii) ${\mathrm{K}}_3\left[\mathrm{Cr}{\left(\mathrm{CN}\right)}_6\right]$

(iii) ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$

(iv) ${\mathrm{K}}_2\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]$