Questions

# Consider the following species : Which one of these will have the highest bond order?

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a
NO
b
CN-
c
CN+
d
CN
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detailed solution

Correct option is B

NO:(σ1s)2,σ⋆1s2,(σ2s)2,σ*2s2,σ2pz2,π2px2=π2py2,π⋆2px1=π⋆2py0 B. O.=10-52=2.5CN-:(σ1s)2,σ⋆1s2,(σ2s)2,σ*2s2,π2px2=π2py2,σ2pz2 B.O. =10-42=3CN:(σ1s)2,σ⋆1s2,(σ2s)2,σ⋆2s2,π2px2=π2py2,σ2pz1B.O·=9-42=2.5CN+:(σ1s)2,σ⋆1s2,(σ2s)2,σ⋆2s2,π2px2=π2py2 B.O. =8-42=2 Hence, CN-has highest bond order.

Similar Questions

The increasing order of energies of various molecular orbitals are

$\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}\right)<\left({\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}={\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}\right)<{\sigma }^{*}2{\mathrm{p}}_{\mathrm{z}}$

$<\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}<\left({\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}={\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}\right)<{\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}$

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