Question

Statement I : Aluminium reacts with excess of \( \mathrm{NaOH} \) to form \( [\mathrm{Al(OH)_6}]^{3-} \)
Statement II : For the complex \( [\mathrm{Fe(H_2O)_6}]^{3+} \) : \( \left[ \left( d_{xy} = d_{yz} = d_{zx} \right) < \left( d_{x^2-y^2} = d_{z^2} \right) \right] \)
and for the complex \( [\mathrm{FeCl_4}]^{2-} \) : \( \left[ \left( d_{xy} = d_{yz} = d_{zx} \right) > \left( d_{x^2-y^2} = d_{z^2} \right) \right] \)

• A. Both Statement I and Statement II are correct
• B. Statement I is correct but Statement II is incorrect.
• C. Statement I is incorrect but Statement II is correct.
• D. Both Statement I and Statement II are incorrect

Hint:

Remember the two key facts: in octahedral complexes, \( t_{2g} \) is lower than \( e_g \), while in tetrahedral complexes the order is reversed. Also, aluminium in excess \( \mathrm{NaOH} \) commonly forms aluminate, \( [\mathrm{Al(OH)_4}]^- \).

Answer 1

The Correct Option is C

Step 1: Check Statement I about aluminium in excess \( \mathrm{NaOH} \).
Aluminium is an amphoteric metal, so it reacts with both acids and bases. When aluminium reacts with excess aqueous \( \mathrm{NaOH} \), it forms sodium aluminate along with evolution of hydrogen gas. In aqueous medium, the complex species generally written is tetrahydroxoaluminate: \[ [\mathrm{Al(OH)_4}]^- \] and not \( [\mathrm{Al(OH)_6}]^{3-} \).
A representative reaction is: \[ 2\mathrm{Al} + 2\mathrm{NaOH} + 6\mathrm{H_2O} \rightarrow 2\mathrm{Na[Al(OH)_4]} + 3\mathrm{H_2} \] Therefore, Statement I is incorrect.
Step 2: Check the splitting pattern for \( [\mathrm{Fe(H_2O)_6}]^{3+} \).
The complex \( [\mathrm{Fe(H_2O)_6}]^{3+} \) is an octahedral complex. In an octahedral crystal field, the five \( d \)-orbitals split into two sets: \[ t_{2g} : d_{xy}, d_{yz}, d_{zx} \] \[ e_g : d_{x^2-y^2}, d_{z^2} \] In octahedral geometry, the \( t_{2g} \) set lies at lower energy, while the \( e_g \) set lies at higher energy. Hence: \[ (d_{xy} = d_{yz} = d_{zx}) < (d_{x^2-y^2} = d_{z^2}) \] So, the first part of Statement II is correct.
Step 3: Check the splitting pattern for \( [\mathrm{FeCl_4}]^{2-} \).
The complex \( [\mathrm{FeCl_4}]^{2-} \) is a tetrahedral complex. In tetrahedral crystal field splitting, the order is reversed compared to octahedral splitting. Here: \[ e : d_{x^2-y^2}, d_{z^2} \] are lower in energy, and \[ t_2 : d_{xy}, d_{yz}, d_{zx} \] are higher in energy.
Thus, for a tetrahedral complex: \[ (d_{xy} = d_{yz} = d_{zx}) > (d_{x^2-y^2} = d_{z^2}) \] So, the second part of Statement II is also correct. Hence, Statement II is correct.
Step 4: Conclusion.
Statement I is incorrect because aluminium in excess \( \mathrm{NaOH} \) forms \( [\mathrm{Al(OH)_4}]^- \), not \( [\mathrm{Al(OH)_6}]^{3-} \).
Statement II is correct because the given orbital energy ordering matches the standard crystal field splitting for octahedral and tetrahedral complexes.
Therefore, the correct option is: \[ (C)\ \text{Statement I is incorrect but Statement II is correct.} \] Final Answer: Statement I is incorrect but Statement II is correct.