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(B) For triamminetrinitrocobalt $(III)$,the formula is $[Co(NH_{3})_{3}(NO_{2})_{3}]$. This complex exhibits facial $(fac)$ and meridional $(mer)$ iso

Vedclass · Accepted Answer

$2$

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(B) For triamminetrinitrocobalt $(III)$,the formula is $[Co(NH_{3})_{3}(NO_{2})_{3}]$. This complex exhibits facial $(fac)$ and meridional $(mer)$ isomers,so $X = 2$.For trioxalatochromate $(III)$,the formula is $[Cr(C_{2}O_{4})_{3}]^{3-}$. This is a tris-chelated complex with identical bidentate ligands. Such complexes do not show geometrical isomerism,so $Y = 0$.Therefore,$X+Y = 2+0 = 2$.

The number of geometrical isomers possible in triamminetrinitrocobalt $(III)$ is $X$ and in trioxalatochromate $(III)$ is $Y$. Then the value of $X+Y$ is $....$

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