The magnetic behavior of complex ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ and ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ is
 

It is given to find out the magnetic behavior of complex ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ and ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$. 
The oxidation state of Fe in the first and second compounds is +2 and +3 respectively. 

The electronic configuration of Fe is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^2$. 
In ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$, the oxidation state of Fe is +2. 
The electronic configuration becomes $3{\mathrm{d}}^6$. 
Cyanide ion is a strong field ligand, pairs up the electrons. Six electrons will pair up. 
In ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$, the oxidation state of Fe is +3. 
The electronic configuration becomes $3{\mathrm{d}}^5$ . 
Cyanide ion pairs up the four electrons and one electron remains unpaired. 

Hence the correct answer is  Diamagnetic , Paramagnetic