Chemistry Q74 – JEE-MAIN 2025

Home Chemistry Coordination CompoundsQuestion #341

A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (

M^{3 +} / M^{2 +}

). It forms a metal complex of the type

[M (C N)_{6}]^{4 -}

. The number of electrons present in the

e_{g}

orbital of the complex is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Identify the Metal

Determine which of the given transition metals (Mn, Cr, Co, Fe) has the highest standard electrode potential for the $M^{3 +} / M^{2 +}$ couple.

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● Video Walkthrough

Step 1: Identify the metal M✦ Active

Compare the standard electrode potentials ( $E^{\circ}$ ) for the $M^{3 +} / M^{2 +}$ couple for Mn, Cr, Co, and Fe. The relevant standard electrode potentials are:

E^{\circ} (M n^{3 +} / M n^{2 +}) = + 1.57 V

E^{\circ} (C r^{3 +} / C r^{2 +}) = - 0.41 V

E^{\circ} (C o^{3 +} / C o^{2 +}) = + 1.81 V

E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77 V

Cobalt (Co) has the highest $E^{\circ}$ value ( $+ 1.81 V$ ). Therefore, M = Co.

Step 2: Determine the oxidation state and

d

-electron configuration of M in the complex○ Expand

The complex is $[C o (C N)_{6}]^{4 -}$ . Let the oxidation state of Co be $x$ . The cyanide ligand ( $C N^{-}$ ) has a charge of -1. Setting up the charge balance equation:

x + 6 (- 1) = - 4

x - 6 = - 4

x = + 2

So, the metal is $C o^{2 +}$ . The atomic number of Co is 27, with an electronic configuration of $[A r] 3 d^{7} 4 s^{2}$ . Thus, $C o^{2 +}$ has a $3 d^{7}$ electronic configuration.

Step 3: Apply Crystal Field Theory (CFT) to determine

e_{g}

electrons○ Expand

The complex $[C o (C N)_{6}]^{4 -}$ is an octahedral complex. $C N^{-}$ is a strong field ligand, which causes a large crystal field splitting ( $Δ_{o} > P$ , where P is pairing energy). For a $d^{7}$ ion in a strong octahedral field, electrons will first fill the lower energy $t_{2 g}$ orbitals before pairing up and then occupying the higher energy $e_{g}$ orbitals. The filling order is $t_{2 g}$ first, then $e_{g}$ . Since it's a strong field, electrons pair up in $t_{2 g}$ before moving to $e_{g}$ .

t_{2 g}^{6} e_{g}^{1}

Therefore, there is 1 electron present in the $e_{g}$ orbital of the complex.

💡 Teacher's Secret Hint

Remember that strong field ligands cause maximum pairing in the lower energy orbitals before electrons occupy higher energy orbitals.

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Chemistry Question 74 – JEE-MAIN 2025

Clue 1 of 3: Identify the Metal

🧠 Full Solution Path