How many unpaired electrons are present in the ground state of Fe³⁺ (z=26), Mn²⁺ (z=25) and argon (z=18)?
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in case of Fe³⁺ :- first , we have to write electronic configuration of Fe,
e.g., [Ar]4s²3d⁶ , now remove three electrons from outermost energy level to form Fe³⁺ . e.g., configuration of Fe³⁺ : [Ar]3d⁵
But we know, d orbital has 10 electron so, number of unpaired electrons = 10 - 5 = 5
in case of Mn²⁺ :- similarly we have to write configuration of Mn.
e.g., [Ar]4s²3d⁵ , now remove two electrons from outermost energy level of it
e.g., electronic configuration of Mn²⁺ : [Ar]3d⁵ , hence number of unpaired electrons = 5
in case of Argon :- Argon is inert gas . So, number of unpaired electrons = 0
e.g., [Ar]4s²3d⁶ , now remove three electrons from outermost energy level to form Fe³⁺ . e.g., configuration of Fe³⁺ : [Ar]3d⁵
But we know, d orbital has 10 electron so, number of unpaired electrons = 10 - 5 = 5
in case of Mn²⁺ :- similarly we have to write configuration of Mn.
e.g., [Ar]4s²3d⁵ , now remove two electrons from outermost energy level of it
e.g., electronic configuration of Mn²⁺ : [Ar]3d⁵ , hence number of unpaired electrons = 5
in case of Argon :- Argon is inert gas . So, number of unpaired electrons = 0
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