Magnetic moment of ${ }_{26}{\mathrm{Fe}}^{2+}$ ion is same as:

For ${ }_{26}{\mathrm{Fe}}^{2+}$, n=4.

So magnetic moment,

$\mu =\sqrt{4(4+2)}=4.9BM$

For ${ }_{28}{\mathrm{Ni}}^{4+}$, n=4 as,

${ }_{28}{\mathrm{Ni}}^{4+}=3{\mathrm{d}}^4$

Now for ${ }_{24}{\mathrm{Cr}}^{2+}$, n=4 as,

${ }_{24}{\mathrm{Cr}}^{2+}=3{\mathrm{d}}^4$

Thus, all options are correct.