The compound(s) that exhibit(s) geometrical isomerism or optical isomerism or both is(are):

Question

The compound(s) that exhibit(s) geometrical isomerism or optical isomerism or both is(are): (A) [ Pt ( en )3] Cl 4 (B) [ Pt ( NH 3)2( CN )2] (C) K 2[ NiCl 2 Br 2] (D) Both (a) and (b)

Tardigrade Admin · Accepted Answer

(B) [ Pt ( NH 3)2( CN )2]: Pt is in +2 oxidation state having 5 d 8 configuration. Hence the hybridisation of complex is dsp 2 and geometry is square planar.

(C) [ NiCl 2 Br 2]2-= tetrahedral, no GI, no optical isomerism

Q. The compound(s) that exhibit(s) geometrical isomerism or optical isomerism or both is(are):

$[Pt (e n)_{3}] C l_{4}$

$[Pt (N H_{3})_{2} (CN)_{2}]$

$K_{2} [N i C l_{2} B r_{2}]$

Both (a) and (b)

(B) $[Pt (N H_{3})_{2} (CN)_{2}] : Pt$ is in +2 oxidation state having $5 d^{8}$ configuration. Hence the hybridisation of complex is $d s p^{2}$ and geometry is square planar.

(C) $[N i C l_{2} B r_{2}]^{2 -} =$ tetrahedral, no GI, no optical isomerism

Q. The compound(s) that exhibit(s) geometrical isomerism or optical isomerism or both is(are):

[Pt(en)3​]Cl4​

[Pt(NH3​)2​(CN)2​]

K2​[NiCl2​Br2​]

Both (a) and (b)

(B) [Pt(NH3​)2​(CN)2​]:Pt is in +2 oxidation state having 5d8 configuration. Hence the hybridisation of complex is dsp2 and geometry is square planar. (C) [NiCl2​Br2​]2−= tetrahedral, no GI, no optical isomerism

$[Pt (e n)_{3}] C l_{4}$

$[Pt (N H_{3})_{2} (CN)_{2}]$

$K_{2} [N i C l_{2} B r_{2}]$

(B) $[Pt (N H_{3})_{2} (CN)_{2}] : Pt$ is in +2 oxidation state having $5 d^{8}$ configuration. Hence the hybridisation of complex is $d s p^{2}$ and geometry is square planar.

(C) $[N i C l_{2} B r_{2}]^{2 -} =$ tetrahedral, no GI, no optical isomerism