KCET · Chemistry · Classification of Elements and Periodicity in Properties
A pair of isoelectronic species having bond order of one is
- A \(\mathrm{N}_2, \mathrm{CO}\)
- B \(\mathrm{N}_2, \mathrm{NO}^{+}\)
- C \(\mathrm{O}_2^{2-}, \mathrm{F}_2\)
- D \(\mathrm{CO}, \mathrm{NO}^{+}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{O}_2^{2-}, \mathrm{F}_2\)
Step-by-step Solution
Detailed explanation
Among the given isoelectronic species, \(\mathrm{O}_2^{2-}\) and \(\mathrm{F}_2\) contains \(18 e^{-}\)each and their bond order is 1.
For \(\mathrm{O}_2^{2-} \mathrm{B} \cdot \mathrm{O}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)
For \(\mathrm{F}_2\)
B.O \(=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)
For \(\mathrm{O}_2^{2-} \mathrm{B} \cdot \mathrm{O}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)
For \(\mathrm{F}_2\)
B.O \(=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)
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