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The correct statement regarding the observed magnetic properties of NO, O2, B2, and C2 in their ground state is
- a)NO, B2, and C2 are paramagnetic
- b)B2, O2 and NO are paramagnetic
- c)O2, C2 and NO are paramagnetic
- d)O2, B2 and C2 are paramagnetic
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The correct statement regarding the observed magnetic properties of NO...
- In B2, there are two unpaired electrons in the π bonding orbital.
- In O2, there are also two π electrons in antibonding orbitals.
- In NO, there is one unpaired electron in π* antibonding orbital
- Since the paramagnetic nature is based on the presence of unpaired electrons.
- Thus, B2, O2, and NO are paramagnetic.
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The correct statement regarding the observed magnetic properties of NO...
Magnetic Properties of NO, O2, B2, and C2
Observation:
- The magnetic properties of molecules depend on the number of unpaired electrons present in them.
- The ground state electronic configuration of NO, O2, B2, and C2 are as follows:
- NO: σ^2π^4
- O2: σ^2σ*^2π^4σ*^1
- B2: σ^2σ*^2π^2
- C2: σ^2σ*^2π^2
- By analyzing the electronic configurations, we can determine the number of unpaired electrons in each molecule and their magnetic nature.
Explanation:
- NO: In NO molecule, there are two unpaired electrons present in the π orbitals. Therefore, NO is paramagnetic.
- O2: In O2 molecule, there are two unpaired electrons present in the π* orbitals. Therefore, O2 is paramagnetic.
- B2: In B2 molecule, there are two unpaired electrons present in the π orbitals. Therefore, B2 is paramagnetic.
- C2: In C2 molecule, there are no unpaired electrons present. Therefore, C2 is diamagnetic.
Conclusion:
- Based on the above analysis, it is clear that option 'B' is the correct answer as it states that B2, O2, and NO are paramagnetic.
Observation:
- The magnetic properties of molecules depend on the number of unpaired electrons present in them.
- The ground state electronic configuration of NO, O2, B2, and C2 are as follows:
- NO: σ^2π^4
- O2: σ^2σ*^2π^4σ*^1
- B2: σ^2σ*^2π^2
- C2: σ^2σ*^2π^2
- By analyzing the electronic configurations, we can determine the number of unpaired electrons in each molecule and their magnetic nature.
Explanation:
- NO: In NO molecule, there are two unpaired electrons present in the π orbitals. Therefore, NO is paramagnetic.
- O2: In O2 molecule, there are two unpaired electrons present in the π* orbitals. Therefore, O2 is paramagnetic.
- B2: In B2 molecule, there are two unpaired electrons present in the π orbitals. Therefore, B2 is paramagnetic.
- C2: In C2 molecule, there are no unpaired electrons present. Therefore, C2 is diamagnetic.
Conclusion:
- Based on the above analysis, it is clear that option 'B' is the correct answer as it states that B2, O2, and NO are paramagnetic.
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Community Answer
The correct statement regarding the observed magnetic properties of NO...
In B2 there is two unpaired electron in π bonding orbital,, in O2 there is also two π electrons in antibonding MO and there is also one unpaired electron in π* antibonding orbital in NO....so we know that paramagnetic nature is based on the present of unpaired electron so the option is B answer.
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The correct statement regarding the observed magnetic properties of NO, O2, B2, and C2 in their ground state isa)NO, B2, and C2 are paramagneticb)B2, O2 and NO are paramagneticc)O2, C2 and NO are paramagneticd)O2, B2 and C2 are paramagneticCorrect answer is option 'B'. Can you explain this answer?
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