SATHEE: Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure - Result Question 97

####99. Decreasing order of stability of $O_{2}, O_{2}^{-}, O_{2}^{+}$ and $O_{2}^{2 -}$ is :

[2015 RS]

(a) $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$

(b) $O_{2}^{2 -} > O_{2}^{-} > O_{2} > O_{2}^{+}$

(d) $O_{2}^{-} > O_{2}^{2 -} > O_{2}^{+} > O_{2}$

Show Answer

Solution:

(a) According to molecular orbital theory as bond order decreases stability of the molecule decreases

Bond order $= \frac{1}{2} (N_{b} - N_{a})$

Bond order for $O_{2}^{+} = \frac{1}{2} (10 - 5) = 2.5$

Bond order for $O_{2} = \frac{1}{2} (10 - 6) = 2$

Bond order for $O_{2}^{-} = \frac{1}{2} (10 - 7) = 1.5$

Bond order for $O_{2}^{2 -} = \frac{1}{2} (10 - 8) = 1.0$

hence the correct order is

$O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Chemical Bonding and Molecular Structure - Result Question 97

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu