question_answer

In nitroprusside ion the iron and NO exist as \[{{\operatorname{Fe}}^{\operatorname{II}}}\] and \[{{\operatorname{NO}}^{+}}\] rather than \[{{\operatorname{Fe}}^{III}}\] and \[\operatorname{NO}\]. These forms can be differentiated by

A) Estimating the concentration of iron

B) Measuring the concentration of CN

C) Measuring the solid state magnetic moment

D) Thermally decomposing the compound.

Correct Answer: C

Solution :

The magnetic moment \[(\mu )\] of a species is related to its number of unpaired electrons (n) in form of following expressions.

\[\mu =\sqrt{n(n+2)}B.M\]

The number of unpaired electrons in the given pairs are as follows:

\[F{{e}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}\]

Or

Thus, \[n=4.\]

\[F{{e}^{3+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}\]

Or

Thus, \[n=5.\]

\[\overset{+}{\mathop{N}}\,O\]or  \[\overset{+}{\mathop{{\underset{\scriptscriptstyle\centerdot\centerdot}{N}}}}\,=\ddot{O}:\]                       \[n=0\]

NO or \[:\dot{N}=\ddot{O}:\]                             \[n=1\]

The given combinations differ in the number of unpaired electrons. Hence these can be differentiated by the measurement on the solid state magnetic moment of nitroprusside ion.