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Calculate the bond order for an ion with this configuration: (σ2s)2(σ2s*)2(σ2px)2(π2py, π2pz)4(π2py*, π2pz*)3. a) 1 b) 2 c) 3 d) 4
Calculate the bond order for an ion with this configuration: (σ2s)2(σ2s*)2(σ2px)2(π2py, π2pz)4(π2py*, π2pz*)3. a) 1 b) 2 c) 3 d) 4
Final answer:Thebond orderof an ion with the given electron configuration is determined by applying the formula (bonding electrons - antibonding electrons) / 2. This results in a bond order of 1.5, implying a stable yet partial bond.Explanation:The bond order of a molecule is defined as one-half the difference between the number of bonding electrons and the number of antibondingelectrons.For the ion with electron configurations (σ2s)2(σ2s*)2(σ2px)2(π2py, π2pz)4(π2py*, π2pz*)3, we can count the number of bonding and antibonding electrons.Bonding electrons are found in (σ2s)2 (σ2px)2 (π2py, π2pz)4, totaling 8. The antibonding electrons are in (σ2s*)2 (π2py*, π2pz*)3, totaling 5.Use the bond order formula, which is (bonding electrons - antibonding electrons) / 2. In this case, this equals (8 - 5) / 2 = 1.5.So, none of the options given (a) 1, (b) 2, (c) 3, (d) 4 are correct. The bond order would actually be 1.5, denoting a stable yetpartial bond.Learn more aboutBond Orderhere:brainly.com/question/12447843#SPJ11...