Q18 / 75 · JEE Main 2026 (22 January Shift 2) · Chemistry
Given below are two statements :
Statement I: The first ionization enthalpy of Cr is lower than that of Mn .
Statement II: The second and third ionization enthalpies of Cr are higher than those of Mn.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: The first ionization enthalpy of Cr is lower than that of Mn .
Statement II: The second and third ionization enthalpies of Cr are higher than those of Mn.
In the light of the above statements, choose the correct answer from the options given below :
- ABoth Statement I and Statement II are true
- BBoth Statement I and Statement II are false
- CStatement I is false but Statement II is true
- Correct answerDStatement I is true but Statement II is false
Solution
Statement I:
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$ and $Mn$ is $[Ar] 3d^5 4s^2$.
For $Cr$, the first electron is removed from $4s^1$, while for $Mn$, it is removed from a stable fully filled $4s^2$ subshell.
Additionally, $Mn$ has a higher nuclear charge.
Thus, the first ionization enthalpy ($IE_1$) of $Cr$ ($653 \text{ kJ/mol}$) is lower than that of $Mn$ ($717 \text{ kJ/mol}$).
Statement I is true.
Statement II:
For the second ionization enthalpy ($IE_2$), $Cr^+$ has configuration $[Ar] 3d^5$ (stable half-filled) and $Mn^+$ has $[Ar] 3d^5 4s^1$.
Removing an electron from the stable $3d^5$ configuration of $Cr^+$ requires more energy than removing it from the $4s^1$ of $Mn^+$.
Thus, $IE_2$ of $Cr$ ($1592 \text{ kJ/mol}$) is higher than $Mn$ ($1509 \text{ kJ/mol}$).
For the third ionization enthalpy ($IE_3$), $Cr^{2+}$ is $[Ar] 3d^4$ and $Mn^{2+}$ is $[Ar] 3d^5$.
Removing an electron from the stable half-filled $3d^5$ configuration of $Mn^{2+}$ requires significantly more energy than from the $3d^4$ of $Cr^{2+}$.
Thus, $IE_3$ of $Mn$ ($3248 \text{ kJ/mol}$) is higher than $Cr$ ($2987 \text{ kJ/mol}$).
Since $IE_3$ of $Cr$ is lower than $Mn$, Statement II is false.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$ and $Mn$ is $[Ar] 3d^5 4s^2$.
For $Cr$, the first electron is removed from $4s^1$, while for $Mn$, it is removed from a stable fully filled $4s^2$ subshell.
Additionally, $Mn$ has a higher nuclear charge.
Thus, the first ionization enthalpy ($IE_1$) of $Cr$ ($653 \text{ kJ/mol}$) is lower than that of $Mn$ ($717 \text{ kJ/mol}$).
Statement I is true.
Statement II:
For the second ionization enthalpy ($IE_2$), $Cr^+$ has configuration $[Ar] 3d^5$ (stable half-filled) and $Mn^+$ has $[Ar] 3d^5 4s^1$.
Removing an electron from the stable $3d^5$ configuration of $Cr^+$ requires more energy than removing it from the $4s^1$ of $Mn^+$.
Thus, $IE_2$ of $Cr$ ($1592 \text{ kJ/mol}$) is higher than $Mn$ ($1509 \text{ kJ/mol}$).
For the third ionization enthalpy ($IE_3$), $Cr^{2+}$ is $[Ar] 3d^4$ and $Mn^{2+}$ is $[Ar] 3d^5$.
Removing an electron from the stable half-filled $3d^5$ configuration of $Mn^{2+}$ requires significantly more energy than from the $3d^4$ of $Cr^{2+}$.
Thus, $IE_3$ of $Mn$ ($3248 \text{ kJ/mol}$) is higher than $Cr$ ($2987 \text{ kJ/mol}$).
Since $IE_3$ of $Cr$ is lower than $Mn$, Statement II is false.
Hence, the correct answer is option: 'D'.