The correct electron affinity order is (EA = | vedclass

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Question

$ \mathrm{N}  \Rightarrow 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{3} \Rightarrow \mathrm{N}^{+} \Rightarrow 1 \mathrm{s}^{2} 2 \mathrm{s}

Vedclass · Accepted Answer

${N^ + } > {O^ + } > N{O^ + }$

Vedclass · Answer

$ \mathrm{N}  \Rightarrow 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{3} \Rightarrow \mathrm{N}^{+} \Rightarrow 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{2}$

$ \mathrm{O} =1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{4} \Rightarrow \mathrm{O}^{+} \Rightarrow 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{3} $

$ \mathrm{NO} \Rightarrow \sigma_{1 \mathrm{s}}^{2}, \sigma_{1 \mathrm{s}}^{* 2}, \sigma_{2 \mathrm{s}}^{2}, \sigma_{2 \mathrm{s}}^{* 2}, \pi_{2 \mathrm{py}}^{2}=\pi_{2 \mathrm{p} z}^{2}, \sigma_{2 \mathrm{px}}^{2}, \pi_{2 \mathrm{py}}^{* 1}=\pi_{2 \mathrm{pz}}^{* 0}$

The correct electron affinity order is $(EA = -\Delta H_{EG})$

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