A) \[{{O}_{2}}\]
B) \[{{H}_{2}}O\]
C) \[Cr{{O}_{2}}\]
D) \[F{{e}_{3}}{{O}_{4}}\]
E) \[ZnF{{e}_{2}}{{O}_{4}}\]
Correct Answer: B
Solution :
\[{{O}_{2}}\]molecule is paramagnetic and the presence of two unpaired electrons in the\[\overset{*}{\mathop{\pi }}\,(2{{p}_{x}})\]or\[\overset{*}{\mathop{\pi }}\,(2{{p}_{y}})\]molecular orbitals accounts for its paramagnetic nature. Similary in \[Cr{{O}_{2}},\] \[F{{e}_{3}}{{O}_{4}}\]and\[ZnF{{e}_{2}}{{O}_{4}},\]the d-orbitals of transition elements are not fulfilled and have unpaired electrons and hence, are paramagnetic. Whereas,\[{{H}_{2}}O\]is diamagnetic because of the presence of all paired electrons and hence repelled by magnetic field.
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