Chemistry Q60 – JEE-MAIN 2026

Home Chemistry Periodicity & ElementsQuestion #502

Given below are two statements : Statement I : Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc. Statement II : The correct order of the following complexes in terms of CFSE is

[C o (H_{2} O)_{6}]^{2 +} < [C o (H_{2} O)_{6}]^{3 +} < [C o (e n)_{3}]^{3 +}

. In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: Ionization Energy Trends

Consider the electronic configurations of the elements and the stability of the resulting ions after removing the first, second, and third electrons.

🧠 Full Solution Path

Step 1: Evaluate Statement I (Third Ionization Energy)✦ Active

We analyze the electronic configurations and the stability of $M^{2 +}$ ions for Sc, Mn, Cu, and Zn to determine the energy required to remove the third electron ( $I E_{3}$ ):

- Sc ( $[A r] 3 d^{1} 4 s^{2}$ ): $S c^{2 +}$ is $[A r] 3 d^{1}$ . Removing the third electron means removing the $3 d^{1}$ electron, which is relatively easy.

- Mn ( $[A r] 3 d^{5} 4 s^{2}$ ): $M n^{2 +}$ is $[A r] 3 d^{5}$ . Removing the third electron means removing an electron from a stable half-filled $3 d^{5}$ configuration, requiring high energy.

- Cu ( $[A r] 3 d^{10} 4 s^{1}$ ): $C u^{2 +}$ is $[A r] 3 d^{9}$ . Removing the third electron means removing a $3 d^{9}$ electron.

- Zn ( $[A r] 3 d^{10} 4 s^{2}$ ): $Z n^{2 +}$ is $[A r] 3 d^{10}$ . Removing the third electron means removing an electron from a very stable fully-filled $3 d^{10}$ configuration, requiring extremely high energy.

The order of third ionization energy is $S c < C u < M n < Z n$ . Thus, the highest $I E_{3}$ is for Zn and the lowest is for Sc. Statement I is true.

Step 2: Evaluate Statement II (CFSE Order)○ Expand

We calculate the CFSE for each complex:

- ** $[C o (H_{2} O)_{6}]^{2 +}$ :** $C o^{2 +}$ is $d^{7}$ . $H_{2} O$ is a weak field ligand, so it's high spin: $t_{2 g}^{5} e_{g}^{2}$ . $C F S E = (- 0.4 \times 5 + 0.6 \times 2) Δ_{o} = - 0.8 Δ_{o} (C o^{2 +})$ . The magnitude is $0.8 Δ_{o} (C o^{2 +})$ . - ** $[C o (H_{2} O)_{6}]^{3 +}$ :** $C o^{3 +}$ is $d^{6}$ . For $C o^{3 +}$ complexes, even $H_{2} O$ can induce low spin due to the higher charge: $t_{2 g}^{6} e_{g}^{0}$ . $C F S E = (- 0.4 \times 6 + 0.6 \times 0) Δ_{o} = - 2.4 Δ_{o} (C o^{3 +})$ . The magnitude is $2.4 Δ_{o} (C o^{3 +})$ . Since $Δ_{o} (C o^{3 +}) > Δ_{o} (C o^{2 +})$ , the magnitude of CFSE for $[C o (H_{2} O)_{6}]^{3 +}$ is greater than for $[C o (H_{2} O)_{6}]^{2 +}$ . - ** $[C o (e n)_{3}]^{3 +}$ :** $C o^{3 +}$ is $d^{6}$ . $e n$ (ethylenediamine) is a strong field ligand, so it's low spin: $t_{2 g}^{6} e_{g}^{0}$ . $C F S E = (- 0.4 \times 6 + 0.6 \times 0) Δ_{o}^{'} = - 2.4 Δ_{o}^{'}$ . The magnitude is $2.4 Δ_{o}^{'}$ . Since $e n$ is a stronger ligand than $H_{2} O$ , $Δ_{o}^{'} > Δ_{o} (C o^{3 +})$ . Thus, the magnitude of CFSE for $[C o (e n)_{3}]^{3 +}$ is greater than for $[C o (H_{2} O)_{6}]^{3 +}$ .

The correct order of CFSE magnitude is $[C o (H_{2} O)_{6}]^{2 +} < [C o (H_{2} O)_{6}]^{3 +} < [C o (e n)_{3}]^{3 +}$ . Statement II is true.

💡 Teacher's Secret Hint

Remember that $Δ_{o}$ increases with increasing oxidation state of the metal ion and with stronger field ligands.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are true, the correct option is 1.

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Chemistry Question 60 – JEE-MAIN 2026

Clue 1 of 3: Ionization Energy Trends

🧠 Full Solution Path