Question

When two compounds $ACl _{3}$ and $DCl _{3}$ of two elements $A$ and $D$ are mixed together a compound $ADCl _{6}$ is formed. Structural analysis showed that $ADCl _{6}$ is an ionic compound. Given that $DCl _{3}$ is trigonal planar and $ACl _{3}$ is trigonal pyramidal. If anion has see-saw shape then shape of cation formed is

• A. Linear
• B. Bent
• C. Pentagonal bipyramidal
• D. Trigonal planar

Answer 1

Answer: (A)

$\Rightarrow As DCl _{3}$ is trigonal planar, hence, ' $D$ ' belongs to boron family. $\Rightarrow As ACl _{3}$ is trigonal pyramidal, hence, '$A$' belongs to nitrogen family. $\Rightarrow DCl _{3}+ ACl _{3} \rightarrow\left[ DCl _{2}\right]^{+}\left[ ACl _{4}\right]^{-}$
or $\left[ ACl _{2}\right]^{+} \left[ DCl _{4}\right]^{-}$
If anionic part is $\left[ ACl _{4}\right]^{-}$which has Sea - saw structure then cationic part will be $\left[ DCl _{2}\right]^{+}$where hybridization of $D$ will be 'sp' and hence, shape will be linear.
$\Rightarrow$ If anionic part is $DCl _{4}^{-}$i.e, it is of tetrahedral shape the cation part will be $ACl _{2}^{+}$which will be bent with $sp ^{2}$ hybridisation.