A) \[Al,Al{{(OH)}_{3}},AlC{{l}_{3}}\]
B) \[Al,AlC{{l}_{3}},NaAl{{O}_{2}}\]
C) \[Zn,N{{a}_{2}}Zn{{O}_{2}},Al{{(OH)}_{2}}\]
D) \[\begin{matrix} {{R}_{3}}\overline{C} \\ P \\ \end{matrix}\,\,\,\,\,\,\,\,\,\begin{matrix} {{R}_{2}}\overline{C}H \\ Q \\ \end{matrix}\,\,\,\,\,\,\,\,\,\begin{matrix} \overline{C}{{H}_{3}} \\ R \\ \end{matrix}\,\,\,\,\,\,\,\,\,\,\begin{matrix} R-\overline{C}{{H}_{2}} \\ S \\ \end{matrix}\]
Correct Answer: B
Solution :
Fe(II) Magnetic moment \[0.1M\], 4 unpaired electrons Magnetic moment = 0, 0 unpaired electron. To have either four or zero unpaired electron, the configuration must be \[N{{H}_{4}}Cl\].
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