4 cis \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\); trans \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\)
Explanation:
\({\mathrm{\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}}}\) can not show cis/trans isomerism. It shows only optical isomerism.
CHXII09:COORDINATION COMPOUNDS
321953
A square planar complex represented below shows
1 Optical isomerism
2 Linkage isomerism
3 Geometrical isomerism
4 None of these
Explanation:
All donating sides are same thus it doesn't show geometrical and optical isomerism in square planar complex.
CHXII09:COORDINATION COMPOUNDS
321954
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
1 Relative position of unidentate ligands on central atom is not same wrt each other.
2 Relative position of unidentate ligands on central atom is same wrt each other.
3 Tetrahedral complexes are not considered for isomerism.
4 Since it is 3 dimensional structure.
Explanation:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
CHXII09:COORDINATION COMPOUNDS
321955
Give the ratio of trans-isomers in \({\mathrm{\left[\mathrm{M}(\mathrm{AA}) \mathrm{b}_{2} \mathrm{c}_{2}\right](\mathrm{A})}}\) and \({\mathrm{\left[\mathrm{Ma}_{4} \mathrm{~b}_{2}\right]}}\), (B) respectively.
1 1
2 2
3 3
4 4
Explanation:
Trans-isomer in \({\mathrm{\mathrm{A}=2}}\) Trans-isomer in \({\mathrm{\mathrm{B}=1}}\) Ratio \({\mathrm{=\dfrac{2}{1}=2}}\)
CHXII09:COORDINATION COMPOUNDS
321956
A complex exists in two geometrical isomeric forms. One of the geometrical isomer is given.
Select ligand for missing position ‘‘O”.
1 \(\mathrm{NH}_{3}\)
2 \(\mathrm{Cr}^{-}\)
3 \(\mathrm{Br}^{-}\)
4 Either \(\mathrm{Cl}^{-}\) or \({\rm{B}}{{\rm{r}}^ - }.\)
Explanation:
\({\mathrm{\mathrm{Ma}_{4} \mathrm{~b}_{2}}}\) and \({\mathrm{\mathrm{Ma}_{4} \mathrm{bc}}}\) can show geometrical isomersim. So, 'O' can be \({\mathrm{\mathrm{Cl}^{-}}}\)or \({\rm{B}}{{\rm{r}}^ - }.\)
4 cis \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\); trans \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\)
Explanation:
\({\mathrm{\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}}}\) can not show cis/trans isomerism. It shows only optical isomerism.
CHXII09:COORDINATION COMPOUNDS
321953
A square planar complex represented below shows
1 Optical isomerism
2 Linkage isomerism
3 Geometrical isomerism
4 None of these
Explanation:
All donating sides are same thus it doesn't show geometrical and optical isomerism in square planar complex.
CHXII09:COORDINATION COMPOUNDS
321954
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
1 Relative position of unidentate ligands on central atom is not same wrt each other.
2 Relative position of unidentate ligands on central atom is same wrt each other.
3 Tetrahedral complexes are not considered for isomerism.
4 Since it is 3 dimensional structure.
Explanation:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
CHXII09:COORDINATION COMPOUNDS
321955
Give the ratio of trans-isomers in \({\mathrm{\left[\mathrm{M}(\mathrm{AA}) \mathrm{b}_{2} \mathrm{c}_{2}\right](\mathrm{A})}}\) and \({\mathrm{\left[\mathrm{Ma}_{4} \mathrm{~b}_{2}\right]}}\), (B) respectively.
1 1
2 2
3 3
4 4
Explanation:
Trans-isomer in \({\mathrm{\mathrm{A}=2}}\) Trans-isomer in \({\mathrm{\mathrm{B}=1}}\) Ratio \({\mathrm{=\dfrac{2}{1}=2}}\)
CHXII09:COORDINATION COMPOUNDS
321956
A complex exists in two geometrical isomeric forms. One of the geometrical isomer is given.
Select ligand for missing position ‘‘O”.
1 \(\mathrm{NH}_{3}\)
2 \(\mathrm{Cr}^{-}\)
3 \(\mathrm{Br}^{-}\)
4 Either \(\mathrm{Cl}^{-}\) or \({\rm{B}}{{\rm{r}}^ - }.\)
Explanation:
\({\mathrm{\mathrm{Ma}_{4} \mathrm{~b}_{2}}}\) and \({\mathrm{\mathrm{Ma}_{4} \mathrm{bc}}}\) can show geometrical isomersim. So, 'O' can be \({\mathrm{\mathrm{Cl}^{-}}}\)or \({\rm{B}}{{\rm{r}}^ - }.\)
4 cis \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\); trans \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\)
Explanation:
\({\mathrm{\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}}}\) can not show cis/trans isomerism. It shows only optical isomerism.
CHXII09:COORDINATION COMPOUNDS
321953
A square planar complex represented below shows
1 Optical isomerism
2 Linkage isomerism
3 Geometrical isomerism
4 None of these
Explanation:
All donating sides are same thus it doesn't show geometrical and optical isomerism in square planar complex.
CHXII09:COORDINATION COMPOUNDS
321954
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
1 Relative position of unidentate ligands on central atom is not same wrt each other.
2 Relative position of unidentate ligands on central atom is same wrt each other.
3 Tetrahedral complexes are not considered for isomerism.
4 Since it is 3 dimensional structure.
Explanation:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
CHXII09:COORDINATION COMPOUNDS
321955
Give the ratio of trans-isomers in \({\mathrm{\left[\mathrm{M}(\mathrm{AA}) \mathrm{b}_{2} \mathrm{c}_{2}\right](\mathrm{A})}}\) and \({\mathrm{\left[\mathrm{Ma}_{4} \mathrm{~b}_{2}\right]}}\), (B) respectively.
1 1
2 2
3 3
4 4
Explanation:
Trans-isomer in \({\mathrm{\mathrm{A}=2}}\) Trans-isomer in \({\mathrm{\mathrm{B}=1}}\) Ratio \({\mathrm{=\dfrac{2}{1}=2}}\)
CHXII09:COORDINATION COMPOUNDS
321956
A complex exists in two geometrical isomeric forms. One of the geometrical isomer is given.
Select ligand for missing position ‘‘O”.
1 \(\mathrm{NH}_{3}\)
2 \(\mathrm{Cr}^{-}\)
3 \(\mathrm{Br}^{-}\)
4 Either \(\mathrm{Cl}^{-}\) or \({\rm{B}}{{\rm{r}}^ - }.\)
Explanation:
\({\mathrm{\mathrm{Ma}_{4} \mathrm{~b}_{2}}}\) and \({\mathrm{\mathrm{Ma}_{4} \mathrm{bc}}}\) can show geometrical isomersim. So, 'O' can be \({\mathrm{\mathrm{Cl}^{-}}}\)or \({\rm{B}}{{\rm{r}}^ - }.\)
4 cis \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\); trans \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\)
Explanation:
\({\mathrm{\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}}}\) can not show cis/trans isomerism. It shows only optical isomerism.
CHXII09:COORDINATION COMPOUNDS
321953
A square planar complex represented below shows
1 Optical isomerism
2 Linkage isomerism
3 Geometrical isomerism
4 None of these
Explanation:
All donating sides are same thus it doesn't show geometrical and optical isomerism in square planar complex.
CHXII09:COORDINATION COMPOUNDS
321954
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
1 Relative position of unidentate ligands on central atom is not same wrt each other.
2 Relative position of unidentate ligands on central atom is same wrt each other.
3 Tetrahedral complexes are not considered for isomerism.
4 Since it is 3 dimensional structure.
Explanation:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
CHXII09:COORDINATION COMPOUNDS
321955
Give the ratio of trans-isomers in \({\mathrm{\left[\mathrm{M}(\mathrm{AA}) \mathrm{b}_{2} \mathrm{c}_{2}\right](\mathrm{A})}}\) and \({\mathrm{\left[\mathrm{Ma}_{4} \mathrm{~b}_{2}\right]}}\), (B) respectively.
1 1
2 2
3 3
4 4
Explanation:
Trans-isomer in \({\mathrm{\mathrm{A}=2}}\) Trans-isomer in \({\mathrm{\mathrm{B}=1}}\) Ratio \({\mathrm{=\dfrac{2}{1}=2}}\)
CHXII09:COORDINATION COMPOUNDS
321956
A complex exists in two geometrical isomeric forms. One of the geometrical isomer is given.
Select ligand for missing position ‘‘O”.
1 \(\mathrm{NH}_{3}\)
2 \(\mathrm{Cr}^{-}\)
3 \(\mathrm{Br}^{-}\)
4 Either \(\mathrm{Cl}^{-}\) or \({\rm{B}}{{\rm{r}}^ - }.\)
Explanation:
\({\mathrm{\mathrm{Ma}_{4} \mathrm{~b}_{2}}}\) and \({\mathrm{\mathrm{Ma}_{4} \mathrm{bc}}}\) can show geometrical isomersim. So, 'O' can be \({\mathrm{\mathrm{Cl}^{-}}}\)or \({\rm{B}}{{\rm{r}}^ - }.\)
4 cis \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\); trans \(-\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\)
Explanation:
\({\mathrm{\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{+3}}}\) can not show cis/trans isomerism. It shows only optical isomerism.
CHXII09:COORDINATION COMPOUNDS
321953
A square planar complex represented below shows
1 Optical isomerism
2 Linkage isomerism
3 Geometrical isomerism
4 None of these
Explanation:
All donating sides are same thus it doesn't show geometrical and optical isomerism in square planar complex.
CHXII09:COORDINATION COMPOUNDS
321954
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
1 Relative position of unidentate ligands on central atom is not same wrt each other.
2 Relative position of unidentate ligands on central atom is same wrt each other.
3 Tetrahedral complexes are not considered for isomerism.
4 Since it is 3 dimensional structure.
Explanation:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.
CHXII09:COORDINATION COMPOUNDS
321955
Give the ratio of trans-isomers in \({\mathrm{\left[\mathrm{M}(\mathrm{AA}) \mathrm{b}_{2} \mathrm{c}_{2}\right](\mathrm{A})}}\) and \({\mathrm{\left[\mathrm{Ma}_{4} \mathrm{~b}_{2}\right]}}\), (B) respectively.
1 1
2 2
3 3
4 4
Explanation:
Trans-isomer in \({\mathrm{\mathrm{A}=2}}\) Trans-isomer in \({\mathrm{\mathrm{B}=1}}\) Ratio \({\mathrm{=\dfrac{2}{1}=2}}\)
CHXII09:COORDINATION COMPOUNDS
321956
A complex exists in two geometrical isomeric forms. One of the geometrical isomer is given.
Select ligand for missing position ‘‘O”.
1 \(\mathrm{NH}_{3}\)
2 \(\mathrm{Cr}^{-}\)
3 \(\mathrm{Br}^{-}\)
4 Either \(\mathrm{Cl}^{-}\) or \({\rm{B}}{{\rm{r}}^ - }.\)
Explanation:
\({\mathrm{\mathrm{Ma}_{4} \mathrm{~b}_{2}}}\) and \({\mathrm{\mathrm{Ma}_{4} \mathrm{bc}}}\) can show geometrical isomersim. So, 'O' can be \({\mathrm{\mathrm{Cl}^{-}}}\)or \({\rm{B}}{{\rm{r}}^ - }.\)
