In the following complex ions order of paramagnetism is :
P: [FeF6]3-
Q: [CoF6]3-
R: [V(H2O)6]3+
S: [Ti(H2O)6]3+
Answer & explanation
Correct answer: option 1
The correct answer is option 1. P > Q > R > S.
Here is the explanation for the order of Paramagnetism:
\([FeF_6]^{3-}\): The oxidation state of iron is +3, so it has 5 electrons in the 3d orbital. The fluoride ligand is a weak field ligand, so it does not cause the electrons to pair up. Therefore, \([FeF_6]^{3-}\) has 5 unpaired electrons and is the most paramagnetic complex ion.
\([CoF_6]^{3-}\): The oxidation state of cobalt is +3, so it has 6 electrons in the 3d orbital. The fluoride ligand is a weak field ligand, so it does not cause the electrons to pair up. Therefore, \([CoF_6]^{3-}\) has 4 unpaired electrons.
\([V(H_2O)_6]^{3+}\): The oxidation state of vanadium is +3, so it has 2 electrons in the 3d orbital. The water ligand is a weak field ligand, so it does not cause the electrons to pair up. Therefore, \([V(H_2O)_6]^{3+}\) has 2 unpaired electrons.
\([Ti(H_2O)_6]^{3+}\): The oxidation state of titanium is +3, so it has 1 electron in the 3d orbital. The water ligand is a weak field ligand, so it does not cause the electrons to pair up. Therefore, \([Ti(H_2O)_6]^{3+}\) has 1 unpaired electron.