Question

If true enter 1, else enter 0.
Brown coloured ring formed in the nitrate test has formula $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$. The magnetic moment in the complex is 3.87

Answer 1

(a) The iron is present in +1 oxidation state. Its electronic confguration in the complex is $\left[Ar\right]4s^{o}3d^7\begin{array}{ccc}\uparrow \downarrow & \uparrow \downarrow & \uparrow \uparrow \uparrow \end{array}$.
Thus three unpaired electrons are present. This is also confirmed by magnetic moment of 3.87 B.M
$\sqrt{N(N+2)}=\sqrt{15}=3.87$