Chemistry Q60 – JEE-MAIN 2026

Home Chemistry Periodicity & ElementsQuestion #512

The correct order of first (

Δ H_{1}

) and second (

Δ H_{2}

) ionisation enthalpy values of Cr and Mn are : A.

Δ H_{1}

: Cr > Mn B.

Δ H_{2}

: Cr > Mn C.

Δ H_{1}

: Mn > Cr D.

Δ H_{2}

: Mn > Cr Choose the correct answer from the options given below :

Clue 1 of 3: Electronic Configuration Stability

The stability of half-filled or fully-filled subshells significantly influences ionization enthalpies.

🧠 Full Solution Path

Step 1: Determine Electronic Configurations✦ Active

First, write down the electronic configurations of Cr and Mn to understand their electron arrangements and subshell stabilities.

Cr (Z=24): [A r] 3 d^{5} 4 s^{1}

Mn (Z=25): [A r] 3 d^{5} 4 s^{2}

Step 2: Analyze First Ionization Enthalpy (

Δ H_{1}

)○ Expand

For Cr, the first electron is removed from the $4 s^{1}$ orbital, leading to a stable half-filled $3 d^{5}$ configuration ( $C r \to C r^{+}$ , $3 d^{5} 4 s^{1} \to 3 d^{5}$ ). This process is relatively easy. For Mn, the first electron is removed from the $4 s^{2}$ orbital ( $M n \to M n^{+}$ , $3 d^{5} 4 s^{2} \to 3 d^{5} 4 s^{1}$ ). Generally, ionization enthalpy increases across a period, but the stability gained by Cr makes its $Δ H_{1}$ lower than Mn. Thus, $Δ H_{1} (M n) > Δ H_{1} (C r)$ . This means statement C is correct.

💡 Teacher's Secret Hint

Remember that exceptional stability of half-filled or fully-filled subshells can override general periodic trends.

Step 3: Analyze Second Ionization Enthalpy (

Δ H_{2}

)○ Expand

For Cr, the second electron is removed from the highly stable half-filled $3 d^{5}$ configuration of $C r^{+}$ ( $C r^{+} \to C r^{2 +}$ , $3 d^{5} \to 3 d^{4}$ ). This requires a very high amount of energy. For Mn, the second electron is removed from the $4 s^{1}$ orbital of $M n^{+}$ to achieve a stable half-filled $3 d^{5}$ configuration ( $M n^{+} \to M n^{2 +}$ , $3 d^{5} 4 s^{1} \to 3 d^{5}$ ). This process is relatively easy. Therefore, $Δ H_{2} (C r)$ is significantly higher than $Δ H_{2} (M n)$ . Thus, $Δ H_{2} (C r) > Δ H_{2} (M n)$ . This means statement B is correct.

💡 Teacher's Secret Hint

Consider the electronic configuration of the *ion* from which the second electron is being removed.

Step 4: Identify the Correct Option○ Expand

Based on the analysis, statements B ( $Δ H_{2}$ : Cr > Mn) and C ( $Δ H_{1}$ : Mn > Cr) are correct. Therefore, the option 'B and C only' is the correct answer.

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Chemistry Question 60 – JEE-MAIN 2026

Clue 1 of 3: Electronic Configuration Stability

🧠 Full Solution Path