AP EAMCET202125 Aug 2021Evening ShiftChemistryChemical Bonding and Molecular StructureActual
Arrange the following species in the correct order of their stabilities N ₂⁻, C ₂, Ne ₂, O ₂²⁻
Options
- ANe ₂ < O ₂²⁻ < C ₂ < N ₂⁻
- BNe ₂ < C ₂ < O ₂²⁻ < N ₂⁻
- CNe ₂ < N ₂⁻ < O ₂²⁻ < C ₂
- DNe ₂ < O ₂²⁻ < N ₂⁻ < C ₂
Correct answer
A. Ne ₂ < O ₂²⁻ < C ₂ < N ₂⁻
Step-by-step solution
Higher the bond order, more is the stability. The bond order of the given species can be calculated as (i) N ₂⁻= Diatomic molecule Number of electrons =(7+7+1)=15 e⁻ i.e. B.O. = 2.5 (ii) C ₂= Diatomic molecule Number of electrons =(6+6)=12 e⁻ i.e. B.O. =2.0 (iii) Ne ₂= Number of bonding and antibonding electrons are equal i.e. B.O. =0 Ne ₂ does not exist. (iv) O ₂²⁻= Diatomic molecule Number of electrons =(8+8+2) e⁻=18 e⁻ i.e. B.O. =1.0 Hence, correct sequence of B.O. is Hence, correct sequence of B.O. is 2.5 ~N ₂⁻