Question

Which of the following order of Bond strength is not correct?

• A. Diamond $<$ Graphite ( $C-C$ bond)
• B. $S{O}_2>S{O}_3^{-2}>S{O}_4^{-2}$ ( $S-O$ bond)
• C. $C{O}_3^{-2}<HC{O}_3^{-}$ ( $C-O$ bond)
• D. $O_2^{-}<O_2<O_2^{+}$ ( $O-O$ bond)

Answer 1

Answer: (B)

We know that bond length is inversely proportional to the $s$ -character. In diamond each carbon is $s{p}^3$ hybridised, while in graphite each carbon is $s{p}^2$ hybridised. Therefore, graphite has higher $s$ -character, so lower bond length than diamond.Lower the bond length, higher will be the bond strength. hence, Diamond $<$ Graphite ( $C-C$ bond).
Bond strength also depends on bond order. The higher the order of the bond the greater the pull between the two atoms and the shorter the length of the bond.
Bond order $(BO)=\frac{1}{2}\times$ [Number of an electron in anti bonding molecular orbitals] – [Number of electrons in bonding molecular orbitals]
$B.Ofor{O}_2=1/2\times [10-6]=2B.Ofor{O}_2^{-}=1/2\times [10-7]=1.5B.Ofor{O}_2^{+}=1/2\times [10-5]=2.5$
Hence, the order of bond strength $O_2^{-}<O_2<O_2^{+}$
In bicarbonate and carbonate, the structure and resonating structure are

Bond order = $\frac{totalno.ofbonds}{totalno.ofresonatingstructures}$

For $BOforC{O}_3^{-2}=\frac{4}{3}=1.33BOforHC{O}_3^{-2}=\frac{4}{2}=2$
Hence bond strength order will be $C{O}_3^{-2}<HC{O}_3^{-}$
For $S{O}_2$


$BOforS{O}_2^{}=\frac{4}{2}=1.33BOforS{O}_3^{-2}=\frac{4}{3}=1.33BOforS{O}_4^{-2}=\frac{6}{4}=1.5$
Hence $S{O}_4^{-2}$ should have the highest bond strength.