Statement I `[Fe(H_(2)O)_(5)NO]SO_(4)` is paramagnetic
Statement II The Fe in `[Fe(H_(2)O)_(5)NO]SO_(4)` has three unpaired electrons .

Options

A. Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for Statement-6
B. Statement-1 is True, Statement is 2 is True , Statement-2 is NOT a correct explanation for Statement-6
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

A

Explanation

In the compound `[Fe(H_(2)O)_(5)NO]SO_(4)`, NO is present as `NO^(+)`. Hence, oxidation state of Fe is `x+5xx0+1=+2` or x=+1. In fact, odd `e^(-)` on N is transferred from NO to Fe, changing NO to `NO^(+)` and `Fe^(2+)` to `Fe^(+)`).
E.C. of `Fe^(+)=1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(1)`
The electron present in 4s shifts to 3d giving `3d^(7)` configuration, i.e.,

`H_(2)O` and `NO^(+)` both are weak field ligands and cannot cause pairing of 3d electrons. Thus, there are 3 unpaired electrons and hybridisation is `sp^(3)d^(2)` (an outer orbital complex).