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(B) . Tetrahedral $NiCl_2Br_2^{2-}$: Tetrahedral complexes do not exhibit geometric isomerism because all positions are equivalent.$B$. Square planar

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$B, C, E$

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(B) . Tetrahedral $NiCl_2Br_2^{2-}$: Tetrahedral complexes do not exhibit geometric isomerism because all positions are equivalent.$B$. Square planar $Pt(NH_3)_2Cl_2$: This is of the type $Ma_2b_2$,which exhibits cis-trans geometric isomerism.$C$. Octahedral $Co(NH_3)_3Cl_3$: This is of the type $Ma_3b_3$,which exhibits facial (fac) and meridional (mer) geometric isomerism.$D$. Square planar $Pd(NH_3)_3Br^{+}$: This is of the type $Ma_3b$,which does not exhibit geometric isomerism.$E$. Octahedral $Co(en)_3^{3+}$: This is of the type $M(AA)_3$,which exhibits optical isomerism (enantiomers).Therefore,$B, C,$ and $E$ exhibit isomerism.

Which of the following molecules/ions can exhibit isomerism:
$A$. Tetrahedral $NiCl_2Br_2^{2-}$ $B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$ $D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Here,$en = 1,2-\text{diaminoethane}$.

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Number of complexes from the following with an even number of unpaired $d$ electrons is . . . . . . . . . .
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$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Which of the following molecules/ions can exhibit isomerism:$A$. Tetrahedral $NiCl_2Br_2^{2-}$$B$. Square planar $Pt(NH_3)_2Cl_2$$C$. Octahedral $Co(NH_3)_3Cl_3$$D$. Square planar $Pd(NH_3)_3Br^{+}$$E$. Octahedral $Co(en)_3^{3+}$Here,$en = 1,2-\text{diaminoethane}$.