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NEET · Chemistry · STD 12 - 5. Co-ordination chemistry
Aluminlum chloride in acidified aqueous solution forms a complex \(\mathrm{A}\), in which hybridisate of \(Al\) is \(B\) What are '\(A\)' and '\(B\)', respectively?
- A \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}, \mathrm{sp}^{3} \mathrm{d}^{2}\)
- B \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{3+}, \mathrm{sp}^{3}\)
- C \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{3+}, \mathrm{dsp}^{2}\)
- D \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3++}, \mathrm{d}^{2} \mathrm{sp}^{3}\)
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}, \mathrm{sp}^{3} \mathrm{d}^{2}\)
Step-by-step Solution
Detailed explanation
\(AlCl_{3}\) in acidified aqueous solution form a \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) \(\mathrm{Al}^{+3}=[\mathrm{Ne}]\)
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