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Consider the complexes.
(i) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{ClBr}\right]$
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) [Pd(en) $\mathrm{ClBr}]$
(v) $\left[\mathrm{Pd}(\mathrm{en})_2\right] \mathrm{Cl}_2 \quad$ (en = ethylenediamine)
The total number of geometrical isomers of (A) is same as the total number of geometrical isomers of
Options:
(i) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{ClBr}\right]$
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) [Pd(en) $\mathrm{ClBr}]$
(v) $\left[\mathrm{Pd}(\mathrm{en})_2\right] \mathrm{Cl}_2 \quad$ (en = ethylenediamine)
The total number of geometrical isomers of (A) is same as the total number of geometrical isomers of
Solution:
2182 Upvotes
Verified Answer
The correct answer is:
(ii)
(i) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{ClBr}\right]$
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
$M a_2 b c$
$M a_2 b_2$
G.I $=2$
G.I $=2$
cis and trans
cis and trans
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) $[\mathrm{Pd}(\mathrm{en}) \mathrm{ClBr}]$
$\mathrm{Maab}_2$
$M a b c$
$\mathrm{G} . \mathrm{I}=3$
G.I $=3$
(v) $\mathrm{Pd}(\mathrm{en})_2 \mathrm{Cl}_2$
$\mathrm{Ma}_2 \mathrm{~b}_2$
$\mathrm{G} . \mathrm{I}=4$
So, only (i) and (ii) have two, i.e. same number of geometrical isomers.
(ii) $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
$M a_2 b c$
$M a_2 b_2$
G.I $=2$
G.I $=2$
cis and trans
cis and trans
(iii) $\left[\mathrm{Pd}(\mathrm{en}) \mathrm{Cl}_2\right]$
(iv) $[\mathrm{Pd}(\mathrm{en}) \mathrm{ClBr}]$
$\mathrm{Maab}_2$
$M a b c$
$\mathrm{G} . \mathrm{I}=3$
G.I $=3$
(v) $\mathrm{Pd}(\mathrm{en})_2 \mathrm{Cl}_2$
$\mathrm{Ma}_2 \mathrm{~b}_2$
$\mathrm{G} . \mathrm{I}=4$
So, only (i) and (ii) have two, i.e. same number of geometrical isomers.
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