Question

Identify the molecule / ion in which the ratio of $\sigma$ to $\pi$ bonds is 3:2

• A. HCO$_3^-$
• B. CH$_2$(CN)$_2$
• C. HClO$_4$
• D. XeO$_3$

Hint:

To count bonds quickly: count all single connections between atoms for the total sigma bonds. Then, add one pi bond for every double bond and two pi bonds for every triple bond. Drawing the correct Lewis structure is the crucial first step.

Answer 1

The Correct Option is B

We are asked to identify the molecule/ion in which the ratio of $\sigma$ to $\pi$ bonds is 3:2. Step 1: Recall bond counting rules

• Single bond = 1 $\sigma$ bond
• Double bond = 1 $\sigma$ + 1 $\pi$ bond
• Triple bond = 1 $\sigma$ + 2 $\pi$ bonds

Step 2: Analyze each option (A) HCO$_3^-$ (Bicarbonate ion): Structure: O=C(OH)-O$^-$.

• $\sigma$ bonds: C=O (1) + C-O (1) + C-OH (1) + O-H (1) = 4 $\sigma$
• $\pi$ bonds: C=O (1) = 1 $\pi$
• Ratio $\sigma:\pi = 4:1$

(B) CH$_2$(CN)$_2$ (Malononitrile): Structure: H$_2$C-C$\equiv$N (two CN groups).

• C-H bonds: 2 ($\sigma$)
• C-C bonds: 2 ($\sigma$)
• C$\equiv$N bonds: each has 1 $\sigma$ + 2 $\pi$; two CN groups give 2 $\sigma$ + 4 $\pi$
• Total $\sigma$ = 2 + 2 + 2 = 6
• Total $\pi$ = 4
• Ratio $\sigma:\pi = 6:4 = 3:2$

(C) HClO$_4$ (Perchloric acid): Structure: O=Cl(=O)(=O)-OH

• $\sigma$ bonds: 4 (Cl-O) + 1 (O-H) = 5
• $\pi$ bonds: 3 (from three Cl=O double bonds) = 3
• Ratio $\sigma:\pi = 5:3$

(D) XeO$_3$ (Xenon trioxide): Structure: Xe(=O)(=O)=O

• $\sigma$ bonds: 3 (Xe-O) = 3
• $\pi$ bonds: 3 (Xe=O double bonds) = 3
• Ratio $\sigma:\pi = 3:3 = 1:1$

Step 3: Conclusion The molecule with $\sigma:\pi$ ratio of 3:2 is: \[ \text{CH}_2(\text{CN})_2 \] \[ \text{Correct Answer: (B) CH}_2(\text{CN})_2 \]