Question

Given below are two statements:
Statement I: The number of paramagnetic species among $[\mathrm{CoF_6}]^{3-}$, $[\mathrm{TiF_6}]^{3-}$, $\mathrm{V_2O_5}$ and $[\mathrm{Fe(CN)_6}]^{3-}$ is 3.
Statement II: $ \mathrm{K_4[Fe(CN)_6]}<\mathrm{K_3[Fe(CN)_6]}<\mathrm{[Fe(H_2O)_6]SO_4 \cdot H_2O}<\mathrm{[Fe(H_2O)_6]Cl_3} $ is the correct order in terms of number of unpaired electrons.
Choose the correct answer from the options given below:

• A. Both Statement I and Statement II are false
• B. Statement I is false but Statement II is true
• C. Statement I is true but Statement II is false
• D. Both Statement I and Statement II are true

Hint:

Strong field ligands like CN$^-$ cause electron pairing, while weak field ligands like H$_2$O increase paramagnetism.

Answer 1

The Correct Option is D

To solve the given question, we need to analyze the paramagnetic properties and electron configurations of the species involved in Statement I, and the order of unpaired electrons in Statement II.

Analyzing Statement I:

The number of paramagnetic species among \([\mathrm{CoF_6}]^{3-}\), \([\mathrm{TiF_6}]^{3-}\), \(\mathrm{V_2O_5}\), and \([\mathrm{Fe(CN)_6}]^{3-}\) needs to be determined:

• \([\mathrm{CoF_6}]^{3-}\): Cobalt in this complex is in the +3 oxidation state. In this oxidation state, the electron configuration is \(3d^6\). With fluoride as a weak field ligand, it is high spin with 4 unpaired electrons, making it paramagnetic.
• \([\mathrm{TiF_6}]^{3-}\): Titanium here is in the +3 oxidation state, giving it a \(3d^1\) configuration, which includes 1 unpaired electron, making it paramagnetic.
• \(\mathrm{V_2O_5}\): Vanadium in vanadium pentoxide is in the +5 oxidation state without d electrons (since the contributing vanadium was \(3d^0\ 4s^0\)), leaving it diamagnetic (no unpaired electrons).
• \([\mathrm{Fe(CN)_6}]^{3-}\): Iron in this complex is in the +3 oxidation state. The electronic configuration is \(3d^5\). With CN^- as a strong field ligand, it forms a low spin complex resulting in 1 unpaired electron, which makes it paramagnetic.

Hence, three out of the four species are paramagnetic, validating Statement I as true.

Analyzing Statement II:

The statement compares the number of unpaired electrons in various iron complexes:

• \(\mathrm{K_4[Fe(CN)_6]}\): Fe²⁺ in low spin state due to CN^-, electronic configuration \(3d^6\), 0 unpaired electrons.
• \(\mathrm{K_3[Fe(CN)_6]}\): Fe³⁺ in low spin state, electronic configuration \(3d^5\), 1 unpaired electron.
• \(\mathrm{[Fe(H_2O)_6]SO_4 \cdot H_2O}\): Fe²⁺ in high spin state due to weak field ligand H₂O, electronic configuration \(3d^6\), 4 unpaired electrons.
• \(\mathrm{[Fe(H_2O)_6]Cl_3}\): Fe³⁺ in high spin state, electronic configuration \(3d^5\), 5 unpaired electrons.

The order of complexes based on increasing number of unpaired electrons is indeed: \(\mathrm{K_4[Fe(CN)_6]} (0) < \mathrm{K_3[Fe(CN)_6]} (1) < \mathrm{[Fe(H_2O)_6]SO_4 \cdot H_2O} (4) < \mathrm{[Fe(H_2O)_6]Cl_3} (5)\). Thus, Statement II is true.

Based on the analysis above, since both statements are true, the correct option is: Both Statement I and Statement II are true.