The correct structures of trans-[NiBr_{2}(PPh | vedclass

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(A) $1$. For the square planar complex $trans-[NiBr_{2}(PPh_{3})_{2}]$,the two identical ligands ($Br$ or $PPh_{3}$) must be positioned opposite to ea

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(A) $1$. For the square planar complex $trans-[NiBr_{2}(PPh_{3})_{2}]$,the two identical ligands ($Br$ or $PPh_{3}$) must be positioned opposite to each other (at $180^{\circ}$ angle). In the structure shown in option $A$,the $Br$ atoms are trans to each other,and the $PPh_{3}$ groups are trans to each other.$2$. For the octahedral complex $meridional-[Co(NH_{3})_{3}(NO_{2})_{3}]$,the three identical ligands must lie on the same plane (meridian). In the structure shown in option $A$,the three $NH_{3}$ ligands are in a meridional arrangement,and the three $NO_{2}$ ligands are also in a meridional arrangement.$3$. Therefore,the structure in option $A$ correctly represents both the trans-isomer of the nickel complex and the meridional-isomer of the cobalt complex.

The correct structures of trans-$[NiBr_{2}(PPh_{3})_{2}]$ and meridional-$[Co(NH_{3})_{3}(NO_{2})_{3}]$,respectively,are

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