Question
Given below are two statements:
Statement-I : Aluminium upon reaction with NaOH forms $\left\lbrack Al(OH)_{6} \right\rbrack^{3 -}$ ion.
Statement-II : The geometry of ${ICl}_{4}^{-},{ClO}_{3}^{-}$and ${IBr}_{2}^{-}$is square planar, pyramidal and linear respectively.
In the light of the above statements, choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Excess $NaOH$ gives $Na\lbrack Al(OH)_{4}\rbrack$ (four-coordinate, not $\lbrack Al(OH)_{6}\rbrack^{3 -}$); ${ICl}_{4}^{-}$ is square planar, ${ClO}_{3}^{-}$ pyramidal, ${IBr}_{2}^{-}$ linear.]
Statement I :
$Al + NaOH($ excess $) \rightarrow Na\left\lbrack Al(OH)_{4} \right\rbrack + H_{2}(\text{ }g)$
Statement II :
${ClO}_{3}^{\Theta}$ is ${sp}^{3}$ and pyramidal
${ICl}_{4}^{\Theta}$ is ${sp}^{3}{\text{ }d}^{2}$ and square planar
${IBr}_{2}^{\Theta}$ is sp $\ ^{3}\text{ }d$ and linear.
Create a free account to view solution
View Solution Free