Chemistry - CHEMICAL BONDING Question with Solution | TestHub

⇒ As DCl₃is trigonal planar, hence, 'D' belongs to boron family.

⇒ As ACl₃is trigonal pyramidal, hence, 'A' belongs to nitrogen family.

⇒ DCl₃+ ACl₃→ [DCl₂]⁺[ACl₄]^–

or  [ACl₂]⁺[DCl₄]^–

if anionic part is [ACl₄]^–which has See - saw structure then cationic part will be [DCl₂]⁺where hybridization of D will be 'sp' and hence, shape will be linear.

⇒  If anionic part is${\text{DCl}}_4^{-}$i.e, it is of tetrahedral shape the cation part will be${\text{ACl}}_2^{+}$which will be bent with sp²hybridisation.