How many geometrical isomers are possible in the following coordination entities?

(i) [Cr(C₂O₄)₃]³⁻ (ii) [Co(NH₃)₃Cl₃]

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Solution

(i) For [Cr(C₂O₄)₃]³⁻, no geometric isomer is possible as it is a bidentate ligand.

(ii) [Co(NH₃)₃Cl₃]

Two geometrical isomers are possible.

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Similar questions

How many geometrical isomers are possible in the following coordination entities?
(1). ${[C r {(C_{2} O_{4})}_{3}]}_{3}^{3 -}$

(2). $[C o {(N H_{3})}_{3} C l_{3}]$

[C o {({N H}_{3})}_{3} {C l}_{3}]

[C r (O X)_{3}]^{3 -}

[C o C l_{3} (N H_{3})_{3}]

[C o (e n)_{2} C l_{2}]^{+} (e n = e t h a n e - 1, 2 - d i a m i n e)

[C r (C_{2} O_{4})_{3}]^{3 -}

[C o (N H_{3})_{3} C l_{3}]

C r = 24, C o = 27

(i) {C H}_{3} - C H = C H - C H = C H - {C H}_{3}

(i i) {C H}_{3} - C H = C H - C H = C H - C l

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