Question

Column -$I$	Column -$II$
$\left(I\right)N_2{O}_5\left(s\right)$	$\left(P\right)$ Cation has $s{p}^3{d}^2$ hybridised central atom
$\left(II\right)IC{l}_3\text{(Molten)}$	$\left(Q\right)$ Anion has square planar shape
$\left(III\right)PB{r}_5\left(s\right)$	$\left(R\right)$ Anion has trigonal planar shape
$\left(IV\right)Xe{F}_6\left(s\right)$	$\left(S\right)$ Cation is linear
	$\left(T\right)$ Cation has tetrahedral shape

Which of the following option has the correct combination considering List-$I$ and List-$II$.

• A. $I,P$
• B. $II,Q$
• C. $III,S$
• D. $IV,T$

Answer 1

The correct option is B $II,Q$

$N_2{O}_5$ in ionic form exist as $N{O}_2^{+}$ and $N{O}_3^{-}$.
$N{O}_2^{+}$ has $sp$ hybridisation. So, it is linear.
$N{O}_3^{-}$ has $s{p}^2$ hybridisation. So, it has trigonal planar shape.


On molten state $IC{l}_3$ dissociates as follows;
$2IC{l}_3\to IC{l}_2^{+}+IC{l}_4^{-}$
$IC{l}_2^{+}$ is $s{p}^3$ hybridised and it's shape is bent.
$IC{l}_4^{-}$ is $s{p}^{3}d$ hybridised and its shape is square planar.


$PB{r}_5$ exists as $[PB{r}_4{]}^{+}B{r}^{-}$.
$[PB{r}_4{]}^{+}$ is $s{p}^3$ hybridised and its shape is tetrahedral

$Xe{F}_6$ exist as $Xe{F}_5^{+}$ and $F^{-}$.
$Xe{F}_5^{+}$ has $s{p}^3{d}^2$ hybridisation and has square pyramidal structure.