The third ionization enthalpy is minimum for

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

$Fe$

Vedclass · Answer

(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.

Column-$I$	Column-$II$
$(a) \ Co^{3+}$	$(i) \ \sqrt{8} \ B.M.$
$(b) \ Cr^{3+}$	$(ii) \ \sqrt{35} \ B.M.$
$(c) \ Fe^{3+}$	$(iii) \ \sqrt{3} \ B.M.$
$(d) \ Ni^{2+}$	$(iv) \ \sqrt{24} \ B.M.$
	$(v) \ \sqrt{15} \ B.M.$

Similar Questions

Which of the following metals forms more than one chloride?

Which of the following electronic configurations would be associated with the highest magnetic moment?

Which compound has a coloured aqueous solution?

The least reactive metal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module