The pair in which both the species have the same magnetic moment (spin only value) is:
Answer & explanation
Correct answer: option 2
The correct answer is option (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.