274157
The shape of $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$and the hybridisation of $\mathrm{Pt}$ respectively are
1 tetrahedral, $\mathrm{sp}^3$
2 trigonal pyramidal, $\mathrm{sp}^3$
3 square planar, $\mathrm{dsp}^2$
4 square planar, $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(C) : The complex is $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$. The platinum metal connected with the four ligand. The complex formed the square planar geometry because $\mathrm{Pt}$ belongs to the $5 \mathrm{~d}$ series. $\begin{gathered}{\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}} \\ \mathrm{x}+3(-1)+0=-1 \\ \mathrm{x}=+2 \\ \mathrm{Pt}^{2+}=[\mathrm{Xe}] 5 \mathrm{~d}^8 6 \mathrm{~s}^0 6 \mathrm{p}^0 \text { (Pairing take place) }\end{gathered}$ Hence, shape of compound is square planar with $\mathrm{dsp}^2$ hybridization.
UPTU/ UPSEE-2013
COORDINATION COMPOUNDS
274159
Which of the following is an outer orbital complex ?
(B) : when outer $4 \mathrm{~d}$ orbitals are involved in hybridisation, it is called an outer orbital complex.
AMU - 2010
COORDINATION COMPOUNDS
274164
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{2+}$ is
1 $s p^3$
2 $d^3 s$
3 $s p^2 f$
4 $d s p^2$
Explanation:
(D) : Copper form complex with $\mathrm{NH}_3$ in +2 and +3 oxidation state. $\mathrm{Cu}^{2+}$ make the square planar complex with $\mathrm{NH}_3$ because $\mathrm{NH}_3$ act as strong field ligand. $\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^{\circ} 4 \mathrm{p}^{\circ}$ Hence, the hybridization of $\mathrm{Cu}^{2+}$ is $\mathrm{dsp}^2$.
WB-JEE-2009
COORDINATION COMPOUNDS
274153
Which one of the following complexes is not expected to exhibit isomerism?
(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ has tetrahedral geometry.
VITEEE- 2012
COORDINATION COMPOUNDS
274145
According to Werner's theory the geometry of the complex is determined by
1 only from the primary valence in space
2 number and position of the primary valences in space
3 number and position of the secondary valency
4 only from the position of secondary valence in space
Explanation:
(C) : According to the Werner's theory the primary valency is satisfy by the oxidation state of the complex whereas secondary valency is satisfy by the coordination number of the complex. Thus the geometry of the complex is determined by number and position of the secondary valency.
274157
The shape of $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$and the hybridisation of $\mathrm{Pt}$ respectively are
1 tetrahedral, $\mathrm{sp}^3$
2 trigonal pyramidal, $\mathrm{sp}^3$
3 square planar, $\mathrm{dsp}^2$
4 square planar, $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(C) : The complex is $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$. The platinum metal connected with the four ligand. The complex formed the square planar geometry because $\mathrm{Pt}$ belongs to the $5 \mathrm{~d}$ series. $\begin{gathered}{\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}} \\ \mathrm{x}+3(-1)+0=-1 \\ \mathrm{x}=+2 \\ \mathrm{Pt}^{2+}=[\mathrm{Xe}] 5 \mathrm{~d}^8 6 \mathrm{~s}^0 6 \mathrm{p}^0 \text { (Pairing take place) }\end{gathered}$ Hence, shape of compound is square planar with $\mathrm{dsp}^2$ hybridization.
UPTU/ UPSEE-2013
COORDINATION COMPOUNDS
274159
Which of the following is an outer orbital complex ?
(B) : when outer $4 \mathrm{~d}$ orbitals are involved in hybridisation, it is called an outer orbital complex.
AMU - 2010
COORDINATION COMPOUNDS
274164
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{2+}$ is
1 $s p^3$
2 $d^3 s$
3 $s p^2 f$
4 $d s p^2$
Explanation:
(D) : Copper form complex with $\mathrm{NH}_3$ in +2 and +3 oxidation state. $\mathrm{Cu}^{2+}$ make the square planar complex with $\mathrm{NH}_3$ because $\mathrm{NH}_3$ act as strong field ligand. $\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^{\circ} 4 \mathrm{p}^{\circ}$ Hence, the hybridization of $\mathrm{Cu}^{2+}$ is $\mathrm{dsp}^2$.
WB-JEE-2009
COORDINATION COMPOUNDS
274153
Which one of the following complexes is not expected to exhibit isomerism?
(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ has tetrahedral geometry.
VITEEE- 2012
COORDINATION COMPOUNDS
274145
According to Werner's theory the geometry of the complex is determined by
1 only from the primary valence in space
2 number and position of the primary valences in space
3 number and position of the secondary valency
4 only from the position of secondary valence in space
Explanation:
(C) : According to the Werner's theory the primary valency is satisfy by the oxidation state of the complex whereas secondary valency is satisfy by the coordination number of the complex. Thus the geometry of the complex is determined by number and position of the secondary valency.
274157
The shape of $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$and the hybridisation of $\mathrm{Pt}$ respectively are
1 tetrahedral, $\mathrm{sp}^3$
2 trigonal pyramidal, $\mathrm{sp}^3$
3 square planar, $\mathrm{dsp}^2$
4 square planar, $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(C) : The complex is $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$. The platinum metal connected with the four ligand. The complex formed the square planar geometry because $\mathrm{Pt}$ belongs to the $5 \mathrm{~d}$ series. $\begin{gathered}{\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}} \\ \mathrm{x}+3(-1)+0=-1 \\ \mathrm{x}=+2 \\ \mathrm{Pt}^{2+}=[\mathrm{Xe}] 5 \mathrm{~d}^8 6 \mathrm{~s}^0 6 \mathrm{p}^0 \text { (Pairing take place) }\end{gathered}$ Hence, shape of compound is square planar with $\mathrm{dsp}^2$ hybridization.
UPTU/ UPSEE-2013
COORDINATION COMPOUNDS
274159
Which of the following is an outer orbital complex ?
(B) : when outer $4 \mathrm{~d}$ orbitals are involved in hybridisation, it is called an outer orbital complex.
AMU - 2010
COORDINATION COMPOUNDS
274164
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{2+}$ is
1 $s p^3$
2 $d^3 s$
3 $s p^2 f$
4 $d s p^2$
Explanation:
(D) : Copper form complex with $\mathrm{NH}_3$ in +2 and +3 oxidation state. $\mathrm{Cu}^{2+}$ make the square planar complex with $\mathrm{NH}_3$ because $\mathrm{NH}_3$ act as strong field ligand. $\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^{\circ} 4 \mathrm{p}^{\circ}$ Hence, the hybridization of $\mathrm{Cu}^{2+}$ is $\mathrm{dsp}^2$.
WB-JEE-2009
COORDINATION COMPOUNDS
274153
Which one of the following complexes is not expected to exhibit isomerism?
(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ has tetrahedral geometry.
VITEEE- 2012
COORDINATION COMPOUNDS
274145
According to Werner's theory the geometry of the complex is determined by
1 only from the primary valence in space
2 number and position of the primary valences in space
3 number and position of the secondary valency
4 only from the position of secondary valence in space
Explanation:
(C) : According to the Werner's theory the primary valency is satisfy by the oxidation state of the complex whereas secondary valency is satisfy by the coordination number of the complex. Thus the geometry of the complex is determined by number and position of the secondary valency.
NEET Test Series from KOTA - 10 Papers In MS WORD
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COORDINATION COMPOUNDS
274157
The shape of $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$and the hybridisation of $\mathrm{Pt}$ respectively are
1 tetrahedral, $\mathrm{sp}^3$
2 trigonal pyramidal, $\mathrm{sp}^3$
3 square planar, $\mathrm{dsp}^2$
4 square planar, $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(C) : The complex is $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$. The platinum metal connected with the four ligand. The complex formed the square planar geometry because $\mathrm{Pt}$ belongs to the $5 \mathrm{~d}$ series. $\begin{gathered}{\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}} \\ \mathrm{x}+3(-1)+0=-1 \\ \mathrm{x}=+2 \\ \mathrm{Pt}^{2+}=[\mathrm{Xe}] 5 \mathrm{~d}^8 6 \mathrm{~s}^0 6 \mathrm{p}^0 \text { (Pairing take place) }\end{gathered}$ Hence, shape of compound is square planar with $\mathrm{dsp}^2$ hybridization.
UPTU/ UPSEE-2013
COORDINATION COMPOUNDS
274159
Which of the following is an outer orbital complex ?
(B) : when outer $4 \mathrm{~d}$ orbitals are involved in hybridisation, it is called an outer orbital complex.
AMU - 2010
COORDINATION COMPOUNDS
274164
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{2+}$ is
1 $s p^3$
2 $d^3 s$
3 $s p^2 f$
4 $d s p^2$
Explanation:
(D) : Copper form complex with $\mathrm{NH}_3$ in +2 and +3 oxidation state. $\mathrm{Cu}^{2+}$ make the square planar complex with $\mathrm{NH}_3$ because $\mathrm{NH}_3$ act as strong field ligand. $\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^{\circ} 4 \mathrm{p}^{\circ}$ Hence, the hybridization of $\mathrm{Cu}^{2+}$ is $\mathrm{dsp}^2$.
WB-JEE-2009
COORDINATION COMPOUNDS
274153
Which one of the following complexes is not expected to exhibit isomerism?
(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ has tetrahedral geometry.
VITEEE- 2012
COORDINATION COMPOUNDS
274145
According to Werner's theory the geometry of the complex is determined by
1 only from the primary valence in space
2 number and position of the primary valences in space
3 number and position of the secondary valency
4 only from the position of secondary valence in space
Explanation:
(C) : According to the Werner's theory the primary valency is satisfy by the oxidation state of the complex whereas secondary valency is satisfy by the coordination number of the complex. Thus the geometry of the complex is determined by number and position of the secondary valency.
274157
The shape of $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$and the hybridisation of $\mathrm{Pt}$ respectively are
1 tetrahedral, $\mathrm{sp}^3$
2 trigonal pyramidal, $\mathrm{sp}^3$
3 square planar, $\mathrm{dsp}^2$
4 square planar, $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(C) : The complex is $\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}$. The platinum metal connected with the four ligand. The complex formed the square planar geometry because $\mathrm{Pt}$ belongs to the $5 \mathrm{~d}$ series. $\begin{gathered}{\left[\mathrm{PtCl}_3\left(\mathrm{C}_2 \mathrm{H}_4\right)\right]^{-}} \\ \mathrm{x}+3(-1)+0=-1 \\ \mathrm{x}=+2 \\ \mathrm{Pt}^{2+}=[\mathrm{Xe}] 5 \mathrm{~d}^8 6 \mathrm{~s}^0 6 \mathrm{p}^0 \text { (Pairing take place) }\end{gathered}$ Hence, shape of compound is square planar with $\mathrm{dsp}^2$ hybridization.
UPTU/ UPSEE-2013
COORDINATION COMPOUNDS
274159
Which of the following is an outer orbital complex ?
(B) : when outer $4 \mathrm{~d}$ orbitals are involved in hybridisation, it is called an outer orbital complex.
AMU - 2010
COORDINATION COMPOUNDS
274164
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{2+}$ is
1 $s p^3$
2 $d^3 s$
3 $s p^2 f$
4 $d s p^2$
Explanation:
(D) : Copper form complex with $\mathrm{NH}_3$ in +2 and +3 oxidation state. $\mathrm{Cu}^{2+}$ make the square planar complex with $\mathrm{NH}_3$ because $\mathrm{NH}_3$ act as strong field ligand. $\mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^{\circ} 4 \mathrm{p}^{\circ}$ Hence, the hybridization of $\mathrm{Cu}^{2+}$ is $\mathrm{dsp}^2$.
WB-JEE-2009
COORDINATION COMPOUNDS
274153
Which one of the following complexes is not expected to exhibit isomerism?
(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ has tetrahedral geometry.
VITEEE- 2012
COORDINATION COMPOUNDS
274145
According to Werner's theory the geometry of the complex is determined by
1 only from the primary valence in space
2 number and position of the primary valences in space
3 number and position of the secondary valency
4 only from the position of secondary valence in space
Explanation:
(C) : According to the Werner's theory the primary valency is satisfy by the oxidation state of the complex whereas secondary valency is satisfy by the coordination number of the complex. Thus the geometry of the complex is determined by number and position of the secondary valency.
