The correct order of first ($$\Delta_i H_1$$) and second ($$\Delta_i H_2$$) ionisation enthalpy values of Cr and Mn are :
A. $$\Delta_i H_1$$ : Cr > Mn
B. $$\Delta_i H_2$$ : Cr > Mn
C. $$\Delta_i H_1$$ : Mn > Cr
D. $$\Delta_i H_2$$ : Mn > Cr
Choose the correct answer from the options given below :

The electronic configurations of the two elements are

$$Cr:[Ar],3d^54s^1,$$

$$Mn:[Ar],3d^54s^2.$$

The first ionization enthalpy corresponds to the removal of the outermost (4s) electron.

For chromium,

$$Cr\rightarrow Cr^++e^-,$$

giving

$$Cr^+:[Ar],3d^5.$$

For manganese,

$$Mn\rightarrow Mn^++e^-,$$

giving

$$Mn^+:[Ar],3d^54s^1.$$

Since manganese has a higher effective nuclear charge than chromium, it holds its outer electron more strongly. Therefore,

$$\Delta_iH_1(Mn)>\Delta_iH_1(Cr).$$

For the second ionization enthalpy, chromium loses an electron from the stable half-filled (3d^5) configuration of (Cr^+), whereas manganese loses the remaining (4s) electron from (Mn^+):

$$Cr^+:[Ar],3d^5\rightarrow Cr^{2+}:[Ar],3d^4,$$

$$Mn^+:[Ar],3d^54s^1\rightarrow Mn^{2+}:[Ar],3d^5.$$

Removing an electron from the exceptionally stable half-filled (3d^5) configuration of (Cr^+) requires more energy than removing the (4s) electron from (Mn^+). Hence,

$$\Delta_iH_2(Cr)>\Delta_iH_2(Mn).$$

Therefore,

$$\Delta_iH_1: Mn>Cr,$$

$$\Delta_iH_2: Cr>Mn.$$

Hence, the correct statements are B and C only.