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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
p Block Elements (Group 13 & 14)

NTA Abhyas 2020 · Chemistry · p Block Elements (Group 13 & 14)

The dissolution of Al (OH) _ 3 by a solution of NaOH results in the...

The dissolution of Al(OH)3Al{\left(OH\right)}_{3} by a solution of NaOHNaOH results in the formation of:

Options

  1. A[ Al (H2O)4(OH)2]+{\left[\text{ Al }{\left({H}_{2}O\right)}_{4}{\left(OH\right)}_{2}\right]}^{+}
  2. B[Al(H2O)3(OH)3]\left[Al{ \left({H}_{2}O\right)}_{3}{ \left(OH\right)}_{3}\right]
  3. C[ Al (H2O)2(OH)4]–{\left[\text{ Al }{\left({H}_{2}O\right)}_{2}{\left(OH\right)}_{4}\right]}^{–}
  4. D[Al(H2O)6(OH)3]\left[Al{ \left({H}_{2}O\right)}_{6}{ \left(OH\right)}_{3}\right]

Correct answer

C. [ Al (H2O)2(OH)4]–{\left[\text{ Al }{\left({H}_{2}O\right)}_{2}{\left(OH\right)}_{4}\right]}^{–}

Step-by-step solution

Al(OH)3Al{\left(OH\right)}_{3} dissolve in NaOHNaOH solution to give Al(OH)4–Al{\left(OH\right)}_{4}^{–} ion which is supposed to have the octahedral complex species [Al(OH)4(H2O)2]–{\left[Al{\left(OH\right)}_{4}{\left({H}_{2}O\right)}_{2}\right]}^{–} in aqueous solution

\text{Al}{\left(\text{OH}\right)_{{3}} + \text{NaOH}\left(\text{aq}\right)\to {\left[\text{Al}{\left(\text{OH}\right)_{{4}} {\left({H}_{2} \text{O}\right)_{{2}}\right]}^{–} \left(\text{aq}\right)+ {\text{Na}^{+}\left(\text{aq}\right)
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