The pair in which both species have same magnetic moment (spin only...

The pair in which both species have same magnetic moment (spin only value) is

Options

A $\left[\mathrm{Cr}$ \left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, $\left[\mathrm{CoCl}_4\right]^{2-}$
B $\left[\mathrm{Cr}$ \left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
C $\left[\mathrm{Mn}$ \left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
D $\left[\mathrm{CoCl}_4\right]^{2-},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ .

B. $\left[\mathrm{Cr}$ \left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Spin only magnetic moment,

\mu_s=\sqrt{n(n+2)}

where

n=

number of unpaired electrons. Number of unpaired electrons in

\mathrm{Cr}^{2+}\left([\mathrm{Ar}] 3 d^4\right)

is 4 , in

\mathrm{Co}^{2+}\left([\mathrm{Ar}] 3 d^7\right)

is 3 , in

\mathrm{Fe}^{2+}\left([\mathrm{Ar}] 3 d^6\right)

is 4 , in

\mathrm{Mn}^{2+}\left([\mathrm{Ar}] 3 d^5\right)

is 5 . As the number of unpaired electrons in

\mathrm{Cr}^{2+}

and

\mathrm{Fe}^{2+}

are same, hence \left[\mathrm{Cr}

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

and \left[\mathrm{Fe}

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

will have same magnetic moment.