Question

Column 1: Compound Column

Column 2: Oxidation state Column

Column 3: Type of linkage

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \left(I\right)& N{a}_2{S}_2{O}_3& \left(i\right)& +5& \left(P\right)& S-O-S\\ \left(II\right)& N{a}_2{S}_4{O}_6& \left(ii\right)& -2& \left(Q\right)& -S-S-\\ \left(III\right)& H_2{S}_2{O}_7& \left(iii\right)& +6& \left(R\right)& -O-O-\\ \left(I\right)& H_2{S}_2{O}_8& \left(iv\right)& +4& \left(S\right)& S=S\end{array}$

The only correct combination in which the hybridization of one of the S atom is $s{p}^2$:

• A. (I) (ii) (Q)
• B. (II) (i) (Q)
• C. (I) (ii) (S)
• D. (II) (i) (S)

Answer 1

The correct option is C

(I) (ii) (S)