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

\([Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}\)

The correct answer is (2) \([Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}\).

The magnetic moment (\( \mu \)) of a species is related to the number of unpaired electrons. The formula for the magnetic moment in terms of the number of unpaired electrons (\( n \)) is given by:

\[ \mu = \sqrt{n(n+2)} \]

where \( n \) is the number of unpaired electrons.

Now, let's analyze each species in the given pairs:

1. \([Cr(H_2O)_6]^{2+}, [CoCl_4]^{2-}\):
   - \(Cr^{2+}\) has \(4\) unpaired electrons (\( n = 4 \)).
   - \(Co^{2-}\) has \(4\) unpaired electrons (\( n = 4 \)).

2. \([Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}\):
   - \(Cr^{2+}\) has \(4\) unpaired electrons (\( n = 4 \)).
   - \(Fe^{2+}\) has \(4\) unpaired electrons (\( n = 4 \)).

3. \([Mn(H_2O)_6]^{2+}, [Cr(H_2O)_6]^{2+}\):
   - \(Mn^{2+}\) has \(3\) unpaired electrons (\( n = 3 \)).
   - \(Cr^{2+}\) has \(4\) unpaired electrons (\( n = 4 \)).

4. \([CoCl_4]^{2-}, [Fe(H_2O)_6]^{2+}\):
   - \(Co^{2-}\) has \(7\) unpaired electrons (\( n = 7 \)).
   - \(Fe^{2+}\) has \(4\) unpaired electrons (\( n = 4 \)).

Now, let's compare the pairs to see which one has the same magnetic moment:

The pair in which both species have the same magnetic moment is: 2. \([Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}\)

Both \(Cr^{2+}\) and \(Fe^{2+}\) have \(4\) unpaired electrons (\( n = 4 \)), so they will have the same magnetic moment according to the given formula.

Therefore, the correct answer is option 2.