The correct order of the following complexes in terms of their crystal field stabilization energies is : [2025]

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Solution

Q.
The correct order of the following complexes in terms of their crystal field stabilization energies is : [2025]

1 ${[Co {({NH}_{3})}_{4}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co {(en)}_{3}]}^{3 +}$

2 ${[Co {({NH}_{3})}_{4}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {(en)}_{3}]}^{3 +} < {[Co {({NH}_{3})}_{6}]}^{3 +}$

3 ${[Co {(en)}_{3}]}^{3 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {({NH}_{3})}_{4}]}^{2 +}$

4 ${[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co {({NH}_{3})}_{4}]}^{2 +} < {[Co {(en)}_{3}]}^{3 +}$

Ans.
(1)

Complex Electronic configuration of central ion before octahedral splitting Electronic configuration of central ion after octahedral splitting CFSE

${[Co {({NH}_{3})}_{6}]}^{2 +}$ ${Co}^{2 +} =_{18} [Ar]$ $3 d^{7} 4 s^{0}$ $t_{2 g}^{2, 2, 2} e_{g}^{1, 0}$ $6 \times (- 0.4 Δ_{0}) + 1 \times (0.6 Δ_{0}) + P = - 1.8 Δ_{0} + P$

${[{Co(NH}_{3})_{6}]}^{3 +}$ ${Co}^{3 +} =_{18} [Ar]$ $3 d^{6} 4 s^{0}$ $t_{2 g}^{2, 2, 2} e_{g}^{0, 0}$ $6 \times (- 0.4 Δ_{0}) + 0 \times (0.6 Δ_{0}) + 2 P = - 2.4 Δ_{0} + 2 P$

${[Co(en)_{3}]}^{3 +}$ ${Co}^{3 +} =_{18} [Ar]$ $3 d^{6} 4 s^{0}$ $t_{2 g}^{2, 2, 2} e_{g}^{0, 0}$ $6 \times (- 0.4 Δ_{0}) + 0 \times (0.6 Δ_{0}) + 2 P = - 2.4 Δ_{0} + 2 P$

As en is a stronger ligand than $N H_{3}$ , CFSE of ${[Co {(en)}_{3}]}^{3 +}$ is more than that of ${[Co {({NH}_{3})}_{6}]}^{3 +}$ . For octahedral complexes, order of CFSE is:

${[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co(en)_{3}]}^{3 +} . {[Co {({NH}_{3})}_{4}]}^{2 +}$ is a square planar complex and we need not to find its CFSE to attempt this question.

The only option where order of CFSE given is ${[Co {({NH}_{3})}_{6}]}^{2 +} < {[Co {({NH}_{3})}_{6}]}^{3 +} < {[Co(en)_{3}]}^{3 +}$ is option 1.

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