question_answer

Match the following

List - I	List -II
(A) \[BC{{l}_{3}}\]	(i) Linear
(B) \[Pdr_{4}^{2-}\]	(ii) Planar triangular
(C) \[S{{F}_{6}}\]	(iii) Tetrahedral
(D) \[\text{I}_{3}^{-}\]	(iv) Square planar

The correct match is

A) A-(ii)      B-(iii)     C-(iv)    D-(i)

B) A-(v)      B-(iii)     C-(ii)      D-(i)

C) A-(ii)      B-(v)      C-(iv)    D-(i)

D) A-(v)      B-(iv)     C-(iii)    D-(ii)

Correct Answer: C

Solution :

\[BC{{l}_{3}}=3bp+0lp\] ie, hybridisation is \[s{{p}^{2}}\] and geometry is planar triangular. \[{{[PdB{{r}_{4}}]}^{2-}}\] The outer electronic configuration of Pd \[[Kr]\,4{{d}^{10}},5{{s}^{o}}.\] \[P{{d}^{2+}}=[Kr]4{{d}^{8}},5{{s}^{0}}\] \[S{{F}_{6}}=6bp+olp\]ie, hybridization is \[s{{p}^{3}}{{d}^{2}}\]and the geometry is octahedral. \[\text{I}_{3}^{-}=2bp+3lp\]ie, hybridization is \[s{{p}^{3}}d\]but geometry is linear due to presence of 3 lone pair of electrons. Hence, \[A-(ii),\,B-(v),\,C-(iv),D-(i)\]