JEE Advanced 2023 Paper 1 · Q15 · Crystal Field Theory

Question

Match the electronic configurations in List-I with appropriate metal complex ions in List-II. [Atomic numbers: Fe = 26, Mn = 25, Co = 27]
List-I:
(P) $t_{2g}^6 e_g^0$
(Q) $t_{2g}^3 e_g^2$
(R) $e^2 t_2^3$
(S) $t_{2g}^4 e_g^2$
List-II: (1) $[Fe(H_2O)_6]^{2+}$, (2) $[Mn(H_2O)_6]^{2+}$, (3) $[Co(NH_3)_6]^{3+}$, (4) $[FeCl_4]^-$, (5) $[CoCl_4]^{2-}$

SolveKar AI · Accepted Answer

$[Mn(H_2O)_6]^{2+}$ is $d^5$ with weak ligand → high-spin octahedral: $t_{2g}^3 e_g^2$ — (Q)→(2). $[Fe(H_2O)_6]^{2+}$ is $d^6$ high-spin: $t_{2g}^4 e_g^2$ — (S)→(1). $[Co(NH_3)_6]^{3+}$ is $d^6$ low-spin (strong ligand): $t_{2g}^6 e_g^0$ — (P)→(3). $[FeCl_4]^-$ is $d^5$ tetrahedral, $e^2 t_2^3$ — (R)→(4).

JEE Advanced 2023 Paper 1 · Q15 · Crystal Field Theory

Solution