The third ionization enthalpy is minimum for

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Question

(A) The electronic configurations of the given elements are:$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$$Co (Z=27): [Ar] 3d^7 4s^2$$Ni (Z=28

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$Fe$

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(A) The electronic configurations of the given elements are:$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$$Co (Z=27): [Ar] 3d^7 4s^2$$Ni (Z=28): [Ar] 3d^8 4s^2$After the removal of two electrons ($M^{2+}$ ions):$Mn^{2+}: [Ar] 3d^5$$Fe^{2+}: [Ar] 3d^6$$Co^{2+}: [Ar] 3d^7$$Ni^{2+}: [Ar] 3d^8$The third ionization enthalpy involves the removal of an electron from the $3d$ orbital. For $Fe^{2+}$,removing one electron results in the stable half-filled $3d^5$ configuration. Since the removal of an electron from $Fe^{2+}$ leads to a more stable state,the energy required (third ionization enthalpy) is the lowest among the given options.

List-$I$ (Element)	List-$II$ (Block)
$(A)$ $Ra$	$(I)$ $p$-block
$(B)$ $Uuq$	$(II)$ $s$-block
$(C)$ $Ds$	$(III)$ $f$-block
$(D)$ $Fm$	$(IV)$ $d$-block

Similar Questions

Match the following:
List-$I$ (Element) List-$II$ (Block)
$(A)$ $Ra$ $(I)$ $p$-block
$(B)$ $Uuq$ $(II)$ $s$-block
$(C)$ $Ds$ $(III)$ $f$-block
$(D)$ $Fm$ $(IV)$ $d$-block

The correct answer is

Define transition elements and inner transition elements.

Identify the transition element.

Which of the following has the maximum number of unpaired electrons?
(Atomic numbers: $Fe=26, Co=27, Mn=25, Cr=24$)

Out of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one?

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