Which is more basic $M g (O H)_{2}$ or $A l (O H)_{3}$ ?

Open in App

Solution

$A l (O H)_{3}$ is more basic than $B e (O H)_{2}$ since the size of Al atom is more than that of Be, the force of attraction between $A l^{3 +}$ ion and $O H^{-}$ will be less than that between $B e^{2 +}$ and $O H^{-}$ ion. Hence Al can loose the $O H^{-}$ ion easily than by Be.

Suggest Corrections

Similar questions

B e (O H)_{2}, M g (O H)_{2}, A l (O H)_{3}

B (O H)_{3}

(1)

Z n (O H)_{2}

(2)

C r (O H)_{3}

(3)

M g (O H)_{2}

(4)

A l (O H)_{3}

N H_{4} O H

N H_{4} C l

M g (O H)_{2}

A l (O H)_{3}

K_{s p}

M g (O H)_{2} = 4 \times 10^{- 12}, A l (O H)_{3} = 1 \times 10^{- 33}

10^{- 17}

B e (O H)_{2}, M g (O H)_{2}, A l (O H)_{3}, L i O H, B (O H)_{3},

A l (O H)_{3}, P b (O H)_{2}, M g (O H)_{2}, Z n (O H)_{2}

Join BYJU'S Learning Program