The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25 and 26 Which one of these may be expected to have the highest second ionization enthalpy?

Question

The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25 and 26 Which one of these may be expected to have the highest second ionization enthalpy? (A) V (B) Cr (C) Mn (D) Fe

Tardigrade Admin · Accepted Answer

(by first IP) This is stable EC hence formation of Cr 2+ by second IP requires maximum enthalpy

Q. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25 and $26$ Which one of these may be expected to have the highest second ionization enthalpy?

$V$

$C r$

$M n$

$F e$

(by first IP)

This is stable EC hence formation of $C r^{2 +}$ by second IP requires maximum enthalpy

Q. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25 and 26 Which one of these may be expected to have the highest second ionization enthalpy?

V

Cr

Mn

Fe

(by first IP) This is stable EC hence formation of Cr2+ by second IP requires maximum enthalpy

Q. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25 and $26$ Which one of these may be expected to have the highest second ionization enthalpy?

$V$

$C r$

$M n$

$F e$

(by first IP)

This is stable EC hence formation of $C r^{2 +}$ by second IP requires maximum enthalpy