Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Compare the relative stability of the following species and indicate their magnetic properties; O_2, O^+ _2, O^+_ 2- ( superoxide)

4.36     Compare the relative stability of the following species and indicate their magnetic properties;

                 O_{2},O^{+}_{2},O^{-}_{2}-(superoxide),\ O_{2}^{2-} (peroxide)

Answers (1)

best_answer

The electronic configuration of O_{2} molecule can be written as:

(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^1 \equiv\pi^*2p_{y}^1)

Here the number of bonding electrons is N_{b} = 10 and the number of antibonding electrons is N_{a} = 6.

Therefore, 

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

= \frac{1}{2}(10-6) = 2

The electronic configuration of O_{2}^+ molecule can be written as:

(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^1 )

Here the number of bonding electrons is N_{b} = 10 and the number of antibonding electrons is N_{a} = 5.

Therefore, 

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

= \frac{1}{2}(10-5) = 2.5

The electronic configuration of O_{2}^- molecule can be written as:

(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv\pi^*2p_{y}^1)

Here the number of bonding electrons is N_{b} = 10 and the number of antibonding electrons is N_{a} = 7.

Therefore, 

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

= \frac{1}{2}(10-7) = 1.5

The electronic configuration of O_{2}^{2-} molecule can be written as:

(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv\pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv\pi^*2p_{y}^2)

Here the number of bonding electrons is N_{b} = 10 and the number of antibonding electrons is N_{a} = 8.

Therefore, 

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

= \frac{1}{2}(10-8) = 1

Therefore, the bond dissociation energy is directly proportional to the bond order.

Thus, the higher the bond order, the greater will be the stability.

We get this order of stability:

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

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE date sheet 2025 for Classes 10, 12 likely in November last week: Reports
anam-khan

CBSE allows Class 10, 12 students from disaffiliated schools to appear for board exam 2025 from same school
anam-khan

CBSE withdraws affiliation of 21 schools, downgrades 6 schools in two states; most in Delhi

Latest Question