STATEMENT-1: [Fe(H_2O)_5NO]SO_4 is paramagnet | vedclass

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(A) In the complex $[Fe(H_2O)_5NO]SO_4$,the $NO$ ligand exists as $NO^+$.Thus,the oxidation state of $Fe$ is calculated as: $x + 5(0) + 1 = +2$ (since

Vedclass · Accepted Answer

$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is correct explanation for $STATEMENT-1$

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(A) In the complex $[Fe(H_2O)_5NO]SO_4$,the $NO$ ligand exists as $NO^+$.Thus,the oxidation state of $Fe$ is calculated as: $x + 5(0) + 1 = +2$ (since $SO_4$ is $-2$,the complex ion is $[Fe(H_2O)_5NO]^{2+}$). Wait,for $[Fe(H_2O)_5NO]SO_4$,$x + 0 + 1 = 2$,so $x = +1$.The electronic configuration of $Fe^+$ $(Z=26)$ is $[Ar] 3d^6 4s^1$.In the presence of the strong field ligand $NO^+$,the $4s$ electron pairs up into the $3d$ orbital.The configuration becomes $3d^7$,which has $3$ unpaired electrons as shown in the orbital diagram.Since it has unpaired electrons,the complex is paramagnetic.Therefore,$STATEMENT-1$ is True and $STATEMENT-2$ is True,and $STATEMENT-2$ correctly explains $STATEMENT-1$.

$STATEMENT-1$: $[Fe(H_2O)_5NO]SO_4$ is paramagnetic.
$STATEMENT-2$: The $Fe$ in $[Fe(H_2O)_5NO]SO_4$ has three unpaired electrons.

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$STATEMENT-1$: $[Fe(H_2O)_5NO]SO_4$ is paramagnetic.$STATEMENT-2$: The $Fe$ in $[Fe(H_2O)_5NO]SO_4$ has three unpaired electrons.