Question

Match the species in Column I with the shape in Column II.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ i.& OS{F}_2& p.& \text{Tetrahedral}\\ ii.& O_{2}S{F}_2& q.& \text{Linear}\\ iii.& Xe{F}_4& r.& \text{Square planar}\\ iv.& Cl{O}_4^{-}& s.& \text{Trigonal pyramidal}\\ v.& I_3^{-}& t.& \text{See - saw}\end{array}$

• A. $i-s,ii-p,iii-r,iv-p,v-q$
• B. $i-s,ii-p,iii-r,iv-q,v-t$
• C. $i-s,ii-p,iii-r,iv-t,v-q$
• D. $i-q,ii-r,iii-s,iv-t,v-p$

Answer 1

The correct option is A $i-s,ii-p,iii-r,iv-p,v-q$

$\left(i\right)OS{F}_2$

steric number = $(6+2)/2=4$ so, $s{p}^3$ - hybridised $S$ atom, due to the presence of one lone pair shape is trigonal pyramidal. $\left(i\right)\to \left(s\right)$
$\left(ii\right)O_{2}S{F}_2$

steric number = $(6+2)/2=4$ so, $s{p}^3$ - hybridised $S$ atom, the shape and geometry is tetrahedral
$\left(ii\right)\to \left(p\right)$
$\left(iii\right)Xe{F}_4$

steric number = $(8+4)/2=6$ so, $s{p}^3{d}^2$ - hybridised $Xe$ - atom.
Two lone pairs are placed above and below the plane so, shape is square planar $\left(iii\right)\to \left(r\right)$

$\left(iv\right)Cl{O}_4^{-}$

steric number = $(7+1)/2=4$ so, $s{p}^3$ - hybridised Cl atom, tetrahedral. $\left(iv\right)\to \left(p\right)$

$\left(v\right)I_3^{-}$

steric number = $(7+2+1)/2=5$ so, $s{p}^{3}d$ - hybridised $I$, three lone pairs at equatorial position hence linear shape
$\left(v\right)\to \left(q\right)$