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# Shape of $DCl_4$ is

$\begin{array}{1 1} See-Saw \\ Perfect\;tetrahedral \\ square\;planar \\ none\;of\;these \end{array}$

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Solution :
$DCl_3$ is trigonal planar, hence 'D' belongs to Boron family
$ACl_3$ is trigonal pyramidal, hence 'A' belongs to Nitrogen family
$DCl_3 + ACl_3 [DCl_2]^{+} [ACl_4]^{–}$ or $[ACl_2]^{+} [DCl_4]^{–}$
If anionic part is $[ACl_4]^{–}$ which has Sea-saw structure then cationic part will be $[DCl_2]^{+}$ where Hybridization of D will be 'sp' and hence shape will be linear.
If anionic part is $DCl_{4^{–}}$ i.e. it is of tetrahedral shape then cation part will be $ACl^{+}2$ which will be bent]
Answer :$Perfect\;tetrahedral$