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Chemistry Question on coordination compounds

Aluminium reacts with NaOH and forms compound 'X'. If the coordination number of aluminium in 'X' is 6, the correct formula of X is

A

[Al(H2O)4(OH)2]+[Al(H_2O)_4(OH)_2]^+

B

[Al(H2O)3(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⟶Sodium  meta  aluminate2NaAlO3+3H22Al + 2NaOH + 2H_2O \longrightarrow ^{2NaAlO_3}_{Sodium\ \\\ meta\ \\\ aluminate}+3H_2
Sodium meta alurninate, thus formed, is soluble in water and changes into the complex [Al(H2O)2(OH)4]−[Al(H_2O)_2(OH)_4]^- in which coordination number of Al is 6.