Question

The correct order of $Cl-O$ bond order is :

• A. ${ClO}_3^{-}<{ClO}_4^{-}<{ClO}_2^{-}<{ClO}^{-}$
• B. ${ClO}^{-}<{ClO}_4^{-}<{ClO}_3^{-}<{ClO}_2^{-}$
• C. ${ClO}^{-}<{ClO}_2^{-}<{ClO}_3^{-}<{ClO}_4^{-}$
• D. ${ClO}_4^{-}<{ClO}_3^{-}<{ClO}_2^{-}<{ClO}^{-}$

Answer 1

The correct option is C ${ClO}^{-}<{ClO}_2^{-}<{ClO}_3^{-}<{ClO}_4^{-}$

All of these are oxide acid radicals and for such acid radicals, bond order can be calculated using the given formula.

$Bondorder=Valencyofoneperipheralatom$

$+$

$\left(Chargeonacidradical/Totalnumberofperipheralatoms\right)$

For $Cl{O}^{-}$, $B.O$

$=2+(-1/1)$

$=2-1$

$=1$

For $Cl{O}_2^{-}$, $B.O$

$=2+(-1/2)$

$=2-0.5$

$=1.5$

For $Cl{O}_3^{-}$, $B.O$

$=2+(-1/3)$

$=2-0.33$
$=1.66$

For $Cl{O}_4^{-}$, $B.O$

$=2+(-1/4)$

$=2-0.25$

$=1.75$

Hence,

${ClO}^{-}<{ClO}_2^{-}<{ClO}_3^{-}<{ClO}_4^{-}$