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The smaller the ligand, the closer it comes to the metal ion and thus the greater the repulsion between ligand and the electron present in the d-orbital of central atom that is why 1- < Br- < ci- < F- are present in spectrochemical series as order.

Aminoothanethiolnte complex of Pt(II), $[Pt {({NH}_{2} {CH}_{2} {CH}_{2} S)}_{2}]$ has no enantiomers but has geometrical isomers while aminoethanethiolate complex of cobalt(II), $[Co {({NH}_{2} {CH}_{2} {CH}_{2} S)}_{2}]$ has no enantiomers and no geometrical isomers.

${[Ni {(H_{2} O)}_{6}]}^{2 +}$ is green, ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ is deep blue and ${[Ni (CN)_{4}]}^{2 -}$ ,is yellow because ${[Ni {(H_{2} O)}_{6}]}^{2 +}$ has absorbed red colour, ${[Ni {({NH}_{3})}_{6}]}^{2 +}$ has absorbed orange colour and ${[Ni (CN)_{4}]}^{2 -}$ has absorbed violet indigo colour.

${[Fe {(C_{2} O_{4})}_{3}]}^{3 -}$ complex ion has $d^{2} s p^{3}$ hybridisation, is inner orbital complex and low spin.

answer is A.

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Detailed Solution

When the ligand is smaller, the more closer it comes to the metal ion. This results in the greater repulsion between ligand and the electron present in the d-orbital of central atom. Thus,

$I^{-} < {Br}^{-} < {Cl}^{-} < F^{2}$ are present in spectrochemical series as order.

Also the color of complexes,

${[Ni {(H_{2} O)}_{6}]}^{2 +} \to$ green,

${[Ni {({NH}_{3})}_{6}]}^{2 +} \to$ deep blue

${[Ni (CN)_{4}]}^{2 -} \to$ yellow

${[Ni {(H_{2} O)}_{6}]}^{2 +} \to$ red colour

${[Ni {({NH}_{3})}_{6}]}^{2 +} \to$ orange colour

${[Ni (CN)_{4}]}^{2 -} \to$ violet indigo colour.

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