A) \[MnS{{O}_{4}}\cdot 4{{H}_{2}}O\]
B) \[NiS{{O}_{4}}\cdot 6{{H}_{2}}O\]
C) \[FeS{{O}_{4}}\cdot 6{{H}_{2}}O\]
D) \[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\]
Correct Answer: D
Solution :
\[M{{n}^{2+}}=\underset{(unpaired\,\,electrons\,\,=\,\,5)}{\mathop{\overset{3{{d}^{5}}}{\mathop{}}\,\overset{4{{s}^{0}}}{\mathop{}}\,}}\,\] In \[CuS{{O}_{4}}\cdot 5{{H}_{2}}O,Cu\]is present as\[C{{u}^{2+}}.\] \[C{{u}^{2+}}=\underset{(unpaired\,\,electrons\,\,=\,\,1)}{\mathop{\overset{3{{d}^{9}}}{\mathop{}}\,\overset{4{{s}^{0}}}{\mathop{}}\,}}\,\] In \[FeS{{O}_{4}}\cdot 6{{H}_{2}}O\,,\]\[Fe\]is present as\[F{{e}^{2+}}\] \[F{{e}^{2+}}=\underset{(unpaired\,\,electrons\,\,=\,\,4)}{\mathop{\overset{3{{d}^{6}}}{\mathop{}}\,\overset{4{{s}^{0}}}{\mathop{}}\,}}\,\] In \[NiS{{O}_{4}}\cdot 6{{H}_{2}}O,\,Ni\] is present as\[N{{i}^{2+}}\] \[N{{i}^{2+}}=\underset{(unpaired\,\,electrons\,\,=\,\,2)}{\mathop{\overset{3{{d}^{8}}}{\mathop{}}\,\overset{4{{s}^{0}}}{\mathop{}}\,}}\,\] Since, paramagnetic character\[\propto \]unpaired electrons. Thus, \[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\]has the lowest degree of paramagnetism among the given at 298 K.
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