Question

When two compounds $AC{l}_3$ and $DC{l}_3$ of two elements $A$ and $D$ are mixed together a compound $ADC{l}_6$ is formed. Structural analysis showed that $ADC{l}_6$ is an ionic compound. Given that $DC{l}_3$ is trigonal planar and $AC{l}_3$ 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)

$\Rightarrow AsDC{l}_3$ is trigonal planar, hence, ' $D$ ' belongs to boron family. $\Rightarrow AsAC{l}_3$ is trigonal pyramidal, hence, '$A$' belongs to nitrogen family. $\Rightarrow DC{l}_3+AC{l}_3\to {\left[DC{l}_2\right]}^{+}{\left[AC{l}_4\right]}^{-}$
or ${\left[AC{l}_2\right]}^{+}{\left[DC{l}_4\right]}^{-}$
If anionic part is ${\left[AC{l}_4\right]}^{-}$which has Sea - saw structure then cationic part will be ${\left[DC{l}_2\right]}^{+}$where hybridization of $D$ will be 'sp' and hence, shape will be linear.
$\Rightarrow$ If anionic part is $DC{l}_4^{-}$i.e, it is of tetrahedral shape the cation part will be $AC{l}_2^{+}$which will be bent with $s{p}^2$ hybridisation.