Which of the following metal is expected to have the highest third ionization enthalpy.

Question

Which of the following metal is expected to have the highest third ionization enthalpy. (A) Cr(Z = 24) (B) V(.Z=23.) (C) Mn(.Z=25.) (D) Fe(.Z=26.)

Tardigrade Admin · Accepted Answer

Mn had outer electronic configuration as 4s23d5 , first two electrons are removed from 4s- orbital, but the removal of 3r d electron is from half filled 3d- orbital, which requires higher ionization energy then other metal.

Q. Which of the following metal is expected to have the highest third ionization enthalpy.

$C r (Z = 24)$

$V (Z = 23)$

$M n (Z = 25)$

$F e (Z = 26)$

$M n$ had outer electronic configuration as $4 s^{2} 3 d^{5}$ , first two electrons are removed from $4 s -$ orbital, but the removal of $3^{r d}$ electron is from half filled $3 d -$ orbital, which requires higher ionization energy then other metal.

Q. Which of the following metal is expected to have the highest third ionization enthalpy.

Cr(Z=24)

V(Z=23)

Mn(Z=25)

Fe(Z=26)

Mn had outer electronic configuration as 4s23d5 , first two electrons are removed from 4s− orbital, but the removal of 3rd electron is from half filled 3d− orbital, which requires higher ionization energy then other metal.

$C r (Z = 24)$

$V (Z = 23)$

$M n (Z = 25)$

$F e (Z = 26)$

$M n$ had outer electronic configuration as $4 s^{2} 3 d^{5}$ , first two electrons are removed from $4 s -$ orbital, but the removal of $3^{r d}$ electron is from half filled $3 d -$ orbital, which requires higher ionization energy then other metal.