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Chemistry Question on Group 13 Elements

Aluminium reacts with NaOHNaOH and forms compound 'XX'. If the coordination number of aluminium in 'XX' is 66, the correct formula of XX is

A

[Al(H2O)4(OH)2]+[Al(H_2O)4(OH)_2]^+

B

[Al(H2O3)(OH)3][Al(H_2O_3) (OH)_3]

C

[Al(H2O)2(OH)4]−[Al(H_2O)_2 (OH)4]^-

D

[Al(H2O)6](OH)3[Al(H_2O)_6](OH)_3

Answer

[Al(H2O)2(OH)4]−[Al(H_2O)_2 (OH)4]^-

Explanation

Solution

2Al+2NaOH+2H2O⟶2NaAlO2+3H2sodium meta aluminate2 Al +2 NaOH +2 H _{2} O \longrightarrow \underset{\text{sodium meta aluminate}}{2 NaAlO _{2}+3 H _{2}}
Sodium meta aluminate, thus formed, is soluble in water and changes into the complex [Al(H2O)2(OH)4]−\left[ Al \left( H _{2} O \right)_{2}( OH )_{4}\right]^{-}, in which coordination number of AlAl is 66 .