274132
Both $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}$ and diamagnetic. The hybridization of Nickel in these complexes are respectively
1 $\mathrm{sp}^{3}, \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3}, \mathrm{dsp}^{2}$
3 $\mathrm{dsp}^{2}, \mathrm{sp}^{3}$
4 $\mathrm{dsp}^{2}, \mathrm{dsp}^{2}$
Explanation:
(B) : Hybridisation of $\mathrm{Ni}$ atom in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\mathrm{Ni}(\mathrm{CO})_{4}$ are $\mathrm{dsp}^{2}$ and $\mathrm{sp}^{3}$ respectively. In $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$, there is $\mathrm{Ni}^{2+}$ ion for which the electronic configuration in the valence shell is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$. In presence of strong field $\mathrm{CN}^{-}$ions, all the electrons are paired up. The empty orbitals (One $3 \mathrm{~d}$, one $4 \mathrm{~s}$ and two $4 \mathrm{p}$ orbitals) undergo $\mathrm{dsp}^{2}$ hybridization to make bonds with four $\mathrm{CN}$ ligands which result in square planar geometry. Since all the electrons are paired, $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ is diamagnetic. In $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$, the valence shell electronic configuration of ground state $\mathrm{Ni}$ atom is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$. All of these 10 electrons one pushed into $3 \mathrm{~d}$ orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three $4 \mathrm{p}$ orbitals are undergo $\mathrm{sp}^{3}$ hybridization and form bonds with four $\mathrm{CO}$ ligands to give $\mathrm{Ni}(\mathrm{CO})_{4}$ with tetrahedral geometry, since all the electrons are paired, $\mathrm{Ni}\left(\mathrm{CO}_{4}\right)$ is diamagnetic.
Assam CEE-2018
COORDINATION COMPOUNDS
274146
Aluminium chloride in acidified aqueous solution forms a complex ' $A$ ', in which hybridisation state of $\mathrm{Al}$ is ' $\mathrm{B}$ '. What are ' $\mathrm{A}$ ' and ' $B$ ', respectively?
(A) : Alumunium chloride is a salt and when treated with water forms a complex. The complex formed will be $\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$. The oxidation state of $\mathrm{Al}$ is +3 . The hybridization of $\mathrm{Al}$ in complex is $\mathrm{sp}^{3} \mathrm{~d}^{2}$.
Odisha NEET 2019)
COORDINATION COMPOUNDS
274160
Which of the following has square planar geometry?
1 $\left[\mathrm{PtCl}_4\right]^{2-}$
2 $\left[\mathrm{NiCl}_4\right]^{2-}$
3 $\left[\mathrm{ZnCl}_4\right]^{2-}$
4 $\left[\mathrm{CoCl}_4\right]^{2-}$
Explanation:
(A) : All the compounds have four co-ordination number but $\left[\mathrm{PtCl}_4\right]^{2-}$ which contains the $5 \mathrm{~d}$-series metal rather than the other compound. Thus $\left[\mathrm{PtCl}_4\right]^{2-}$ form the square planar complex.
BITSAT-2010
COORDINATION COMPOUNDS
274162
Potassium ferrocyanide is an example of
1 tetrahedral
2 octahedral
3 square planar
4 linear
Explanation:
(B) : The chemical formula of potassium ferrocyanide is $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. Here, the central metal attached with the six cyanide ligand, hence it form the octahedral geometry.
274132
Both $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}$ and diamagnetic. The hybridization of Nickel in these complexes are respectively
1 $\mathrm{sp}^{3}, \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3}, \mathrm{dsp}^{2}$
3 $\mathrm{dsp}^{2}, \mathrm{sp}^{3}$
4 $\mathrm{dsp}^{2}, \mathrm{dsp}^{2}$
Explanation:
(B) : Hybridisation of $\mathrm{Ni}$ atom in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\mathrm{Ni}(\mathrm{CO})_{4}$ are $\mathrm{dsp}^{2}$ and $\mathrm{sp}^{3}$ respectively. In $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$, there is $\mathrm{Ni}^{2+}$ ion for which the electronic configuration in the valence shell is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$. In presence of strong field $\mathrm{CN}^{-}$ions, all the electrons are paired up. The empty orbitals (One $3 \mathrm{~d}$, one $4 \mathrm{~s}$ and two $4 \mathrm{p}$ orbitals) undergo $\mathrm{dsp}^{2}$ hybridization to make bonds with four $\mathrm{CN}$ ligands which result in square planar geometry. Since all the electrons are paired, $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ is diamagnetic. In $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$, the valence shell electronic configuration of ground state $\mathrm{Ni}$ atom is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$. All of these 10 electrons one pushed into $3 \mathrm{~d}$ orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three $4 \mathrm{p}$ orbitals are undergo $\mathrm{sp}^{3}$ hybridization and form bonds with four $\mathrm{CO}$ ligands to give $\mathrm{Ni}(\mathrm{CO})_{4}$ with tetrahedral geometry, since all the electrons are paired, $\mathrm{Ni}\left(\mathrm{CO}_{4}\right)$ is diamagnetic.
Assam CEE-2018
COORDINATION COMPOUNDS
274146
Aluminium chloride in acidified aqueous solution forms a complex ' $A$ ', in which hybridisation state of $\mathrm{Al}$ is ' $\mathrm{B}$ '. What are ' $\mathrm{A}$ ' and ' $B$ ', respectively?
(A) : Alumunium chloride is a salt and when treated with water forms a complex. The complex formed will be $\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$. The oxidation state of $\mathrm{Al}$ is +3 . The hybridization of $\mathrm{Al}$ in complex is $\mathrm{sp}^{3} \mathrm{~d}^{2}$.
Odisha NEET 2019)
COORDINATION COMPOUNDS
274160
Which of the following has square planar geometry?
1 $\left[\mathrm{PtCl}_4\right]^{2-}$
2 $\left[\mathrm{NiCl}_4\right]^{2-}$
3 $\left[\mathrm{ZnCl}_4\right]^{2-}$
4 $\left[\mathrm{CoCl}_4\right]^{2-}$
Explanation:
(A) : All the compounds have four co-ordination number but $\left[\mathrm{PtCl}_4\right]^{2-}$ which contains the $5 \mathrm{~d}$-series metal rather than the other compound. Thus $\left[\mathrm{PtCl}_4\right]^{2-}$ form the square planar complex.
BITSAT-2010
COORDINATION COMPOUNDS
274162
Potassium ferrocyanide is an example of
1 tetrahedral
2 octahedral
3 square planar
4 linear
Explanation:
(B) : The chemical formula of potassium ferrocyanide is $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. Here, the central metal attached with the six cyanide ligand, hence it form the octahedral geometry.
274132
Both $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}$ and diamagnetic. The hybridization of Nickel in these complexes are respectively
1 $\mathrm{sp}^{3}, \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3}, \mathrm{dsp}^{2}$
3 $\mathrm{dsp}^{2}, \mathrm{sp}^{3}$
4 $\mathrm{dsp}^{2}, \mathrm{dsp}^{2}$
Explanation:
(B) : Hybridisation of $\mathrm{Ni}$ atom in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\mathrm{Ni}(\mathrm{CO})_{4}$ are $\mathrm{dsp}^{2}$ and $\mathrm{sp}^{3}$ respectively. In $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$, there is $\mathrm{Ni}^{2+}$ ion for which the electronic configuration in the valence shell is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$. In presence of strong field $\mathrm{CN}^{-}$ions, all the electrons are paired up. The empty orbitals (One $3 \mathrm{~d}$, one $4 \mathrm{~s}$ and two $4 \mathrm{p}$ orbitals) undergo $\mathrm{dsp}^{2}$ hybridization to make bonds with four $\mathrm{CN}$ ligands which result in square planar geometry. Since all the electrons are paired, $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ is diamagnetic. In $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$, the valence shell electronic configuration of ground state $\mathrm{Ni}$ atom is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$. All of these 10 electrons one pushed into $3 \mathrm{~d}$ orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three $4 \mathrm{p}$ orbitals are undergo $\mathrm{sp}^{3}$ hybridization and form bonds with four $\mathrm{CO}$ ligands to give $\mathrm{Ni}(\mathrm{CO})_{4}$ with tetrahedral geometry, since all the electrons are paired, $\mathrm{Ni}\left(\mathrm{CO}_{4}\right)$ is diamagnetic.
Assam CEE-2018
COORDINATION COMPOUNDS
274146
Aluminium chloride in acidified aqueous solution forms a complex ' $A$ ', in which hybridisation state of $\mathrm{Al}$ is ' $\mathrm{B}$ '. What are ' $\mathrm{A}$ ' and ' $B$ ', respectively?
(A) : Alumunium chloride is a salt and when treated with water forms a complex. The complex formed will be $\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$. The oxidation state of $\mathrm{Al}$ is +3 . The hybridization of $\mathrm{Al}$ in complex is $\mathrm{sp}^{3} \mathrm{~d}^{2}$.
Odisha NEET 2019)
COORDINATION COMPOUNDS
274160
Which of the following has square planar geometry?
1 $\left[\mathrm{PtCl}_4\right]^{2-}$
2 $\left[\mathrm{NiCl}_4\right]^{2-}$
3 $\left[\mathrm{ZnCl}_4\right]^{2-}$
4 $\left[\mathrm{CoCl}_4\right]^{2-}$
Explanation:
(A) : All the compounds have four co-ordination number but $\left[\mathrm{PtCl}_4\right]^{2-}$ which contains the $5 \mathrm{~d}$-series metal rather than the other compound. Thus $\left[\mathrm{PtCl}_4\right]^{2-}$ form the square planar complex.
BITSAT-2010
COORDINATION COMPOUNDS
274162
Potassium ferrocyanide is an example of
1 tetrahedral
2 octahedral
3 square planar
4 linear
Explanation:
(B) : The chemical formula of potassium ferrocyanide is $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. Here, the central metal attached with the six cyanide ligand, hence it form the octahedral geometry.
274132
Both $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}$ and diamagnetic. The hybridization of Nickel in these complexes are respectively
1 $\mathrm{sp}^{3}, \mathrm{sp}^{3}$
2 $\mathrm{sp}^{3}, \mathrm{dsp}^{2}$
3 $\mathrm{dsp}^{2}, \mathrm{sp}^{3}$
4 $\mathrm{dsp}^{2}, \mathrm{dsp}^{2}$
Explanation:
(B) : Hybridisation of $\mathrm{Ni}$ atom in $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\mathrm{Ni}(\mathrm{CO})_{4}$ are $\mathrm{dsp}^{2}$ and $\mathrm{sp}^{3}$ respectively. In $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$, there is $\mathrm{Ni}^{2+}$ ion for which the electronic configuration in the valence shell is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$. In presence of strong field $\mathrm{CN}^{-}$ions, all the electrons are paired up. The empty orbitals (One $3 \mathrm{~d}$, one $4 \mathrm{~s}$ and two $4 \mathrm{p}$ orbitals) undergo $\mathrm{dsp}^{2}$ hybridization to make bonds with four $\mathrm{CN}$ ligands which result in square planar geometry. Since all the electrons are paired, $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ is diamagnetic. In $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$, the valence shell electronic configuration of ground state $\mathrm{Ni}$ atom is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$. All of these 10 electrons one pushed into $3 \mathrm{~d}$ orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three $4 \mathrm{p}$ orbitals are undergo $\mathrm{sp}^{3}$ hybridization and form bonds with four $\mathrm{CO}$ ligands to give $\mathrm{Ni}(\mathrm{CO})_{4}$ with tetrahedral geometry, since all the electrons are paired, $\mathrm{Ni}\left(\mathrm{CO}_{4}\right)$ is diamagnetic.
Assam CEE-2018
COORDINATION COMPOUNDS
274146
Aluminium chloride in acidified aqueous solution forms a complex ' $A$ ', in which hybridisation state of $\mathrm{Al}$ is ' $\mathrm{B}$ '. What are ' $\mathrm{A}$ ' and ' $B$ ', respectively?
(A) : Alumunium chloride is a salt and when treated with water forms a complex. The complex formed will be $\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$. The oxidation state of $\mathrm{Al}$ is +3 . The hybridization of $\mathrm{Al}$ in complex is $\mathrm{sp}^{3} \mathrm{~d}^{2}$.
Odisha NEET 2019)
COORDINATION COMPOUNDS
274160
Which of the following has square planar geometry?
1 $\left[\mathrm{PtCl}_4\right]^{2-}$
2 $\left[\mathrm{NiCl}_4\right]^{2-}$
3 $\left[\mathrm{ZnCl}_4\right]^{2-}$
4 $\left[\mathrm{CoCl}_4\right]^{2-}$
Explanation:
(A) : All the compounds have four co-ordination number but $\left[\mathrm{PtCl}_4\right]^{2-}$ which contains the $5 \mathrm{~d}$-series metal rather than the other compound. Thus $\left[\mathrm{PtCl}_4\right]^{2-}$ form the square planar complex.
BITSAT-2010
COORDINATION COMPOUNDS
274162
Potassium ferrocyanide is an example of
1 tetrahedral
2 octahedral
3 square planar
4 linear
Explanation:
(B) : The chemical formula of potassium ferrocyanide is $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. Here, the central metal attached with the six cyanide ligand, hence it form the octahedral geometry.
