JEE Main 2023 — Chemistry Inorganic Chemistry

d-orbital is degenerate which split into two levels ${e}_{g}$eg​ and ${t}_{2}g$t2​g in the presence of ligands. This splitting is due to the presence of ligands. This is called the crystal-field splitting and the energy difference between the two levels is called the crystal-field splitting energy, $\Delta 0.$Δ0.

(A) ${\mathrm{Ti}}^{2+}:4{s}^{0}3{d}^{2}$Ti2+:4s03d2

$${t}_{2g}{e}^{–}=2 {e}_{g}{e}^{–}=0$$t2g​e–=2eg​e–=0

$${\Delta }_{0}=2(-.4{\Delta }_{0})=-.8{\Delta }_{0}$$Δ0​=2(−.4Δ0​)=−.8Δ0​

(B) ${V}^{2+}:4{s}^{0}3{d}^{3}$V2+:4s03d3

$${t}_{2g}{e}^{–}=3 {e}_{g}{e}^{–}=0$$t2g​e–=3eg​e–=0

$${\Delta }_{0}=3(-.4{\Delta }_{0})$$Δ0​=3(−.4Δ0​)

$$=-1.2{\Delta }_{0}$$=−1.2Δ0​

(C) ${\mathrm{Mn}}^{3+}=4{s}^{0}3{d}^{4}$Mn3+=4s03d4

$${t}_{2g}{e}^{–}=3 {e}_{g}{e}^{–}=1$$t2g​e–=3eg​e–=1

$${\Delta }_{0}=3(-.4{\Delta }_{0})+1(.6{\Delta }_{0})$$Δ0​=3(−.4Δ0​)+1(.6Δ0​)

$$=-1.2{\Delta }_{0}+0.6{\Delta }_{0}$$=−1.2Δ0​+0.6Δ0​

$$=-0.6{\Delta }_{0}$$=−0.6Δ0​

(D) ${\mathrm{Fe}}^{3+}=4{s}^{0}3{d}^{5}$Fe3+=4s03d5

$${t}_{2g}{e}^{–}=3 {e}_{g}{e}^{–}=2$$t2g​e–=3eg​e–=2

$${\Delta }_{0}=3(-.4{\Delta }_{0})+2(.6{\Delta }_{0})$$Δ0​=3(−.4Δ0​)+2(.6Δ0​)

$$=-1.2{\Delta }_{0}+1.2{\Delta }_{0}=0$$=−1.2Δ0​+1.2Δ0​=0