In the brown ring complex,if NO is considered | vedclass

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(A) The brown ring complex is $[Fe(H_2O)_5(NO)]SO_4$. Case $1$: When $NO$ is considered as $NO^+$ ($2e^-$ donor): $Fe$ is in $+1$ oxidation state. $EA

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(A) The brown ring complex is $[Fe(H_2O)_5(NO)]SO_4$. Case $1$: When $NO$ is considered as $NO^+$ ($2e^-$ donor): $Fe$ is in $+1$ oxidation state. $EAN = Z - 	ext{oxidation state} + 2 	imes (	ext{coordination number}) = 26 - 1 + 2 	imes 6 = 26 - 1 + 12 = 37$. Case $2$: When $NO$ is considered as a neutral ligand ($NO$,$3e^-$ donor): $Fe$ is in $0$ oxidation state. $EAN = Z - 	ext{oxidation state} + 2 	imes (	ext{number of } H_2O) + 3 	imes (	ext{number of } NO) = 26 - 0 + 2 	imes 5 + 3 	imes 1 = 26 + 10 + 3 = 39$. Wait,re-evaluating the standard convention: In the brown ring complex $[Fe(H_2O)_5(NO)]^{+2}$,$Fe$ is $+1$ and $NO$ is $NO^+$. If we treat $NO$ as neutral,the oxidation state of $Fe$ becomes $0$. $EAN = 26 - 0 + 10 + 3 = 39$. Comparing $37$ and $39$,the $EAN$ increases.

In the brown ring complex,if $NO$ is considered as a neutral ligand,then the $EAN$ would:

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