274152
Among $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$ species, the hybridization state of the $\mathrm{Fe}$ atom are, respectively.
274154
The number of unpaired electrons calculated in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathbf{F}_6\right)\right]^{3-}$
1 4 and 4
2 0 and 2
3 2 and 4
4 0 and 4
Explanation:
(D) : In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{F}_6\right)\right]^{3-}$ both the oxidation state of cobalt ion is +3 . In first case $\mathrm{NH}_3$ is the neutral ligand which is a strong field ligand. Hence, the electrons in Co $(+3)$ i.e. $4 \mathrm{~s}^0 3 \mathrm{~d}^6$ get paired to form inner orbital complex. Hence, no unpaired electron. On the other hand $\mathrm{F}^{-}$is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.
VITEEE- 2008
COORDINATION COMPOUNDS
274155
In $\mathrm{TeCl}_4$ the central atom tellurium involves
1 $\mathrm{sp}^3$ hybridization
2 $\mathrm{sp}^3 \mathrm{~d}$ hybridization
3 $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization
4 $\mathrm{dsp}^2$ hybridization
Explanation:
(B) : Hybridisation $=\frac{1}{2}[$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any] $=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^3 \mathrm{~d}$ ${\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4\right]}$
VITEEE- 2007
COORDINATION COMPOUNDS
274156
The hybridization of nickel in nickel tetracarbonyl is
1 $\mathrm{sp}^3$
2 $\mathrm{sp}^2$
3 $\mathrm{dsp}^2$
4 $d^2 s^3$
Explanation:
(A) : The chemical formula of nickeltetracarbonyl is $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$. The oxidation number of the complex is 0 . ${ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
274152
Among $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$ species, the hybridization state of the $\mathrm{Fe}$ atom are, respectively.
274154
The number of unpaired electrons calculated in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathbf{F}_6\right)\right]^{3-}$
1 4 and 4
2 0 and 2
3 2 and 4
4 0 and 4
Explanation:
(D) : In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{F}_6\right)\right]^{3-}$ both the oxidation state of cobalt ion is +3 . In first case $\mathrm{NH}_3$ is the neutral ligand which is a strong field ligand. Hence, the electrons in Co $(+3)$ i.e. $4 \mathrm{~s}^0 3 \mathrm{~d}^6$ get paired to form inner orbital complex. Hence, no unpaired electron. On the other hand $\mathrm{F}^{-}$is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.
VITEEE- 2008
COORDINATION COMPOUNDS
274155
In $\mathrm{TeCl}_4$ the central atom tellurium involves
1 $\mathrm{sp}^3$ hybridization
2 $\mathrm{sp}^3 \mathrm{~d}$ hybridization
3 $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization
4 $\mathrm{dsp}^2$ hybridization
Explanation:
(B) : Hybridisation $=\frac{1}{2}[$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any] $=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^3 \mathrm{~d}$ ${\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4\right]}$
VITEEE- 2007
COORDINATION COMPOUNDS
274156
The hybridization of nickel in nickel tetracarbonyl is
1 $\mathrm{sp}^3$
2 $\mathrm{sp}^2$
3 $\mathrm{dsp}^2$
4 $d^2 s^3$
Explanation:
(A) : The chemical formula of nickeltetracarbonyl is $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$. The oxidation number of the complex is 0 . ${ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
274152
Among $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$ species, the hybridization state of the $\mathrm{Fe}$ atom are, respectively.
274154
The number of unpaired electrons calculated in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathbf{F}_6\right)\right]^{3-}$
1 4 and 4
2 0 and 2
3 2 and 4
4 0 and 4
Explanation:
(D) : In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{F}_6\right)\right]^{3-}$ both the oxidation state of cobalt ion is +3 . In first case $\mathrm{NH}_3$ is the neutral ligand which is a strong field ligand. Hence, the electrons in Co $(+3)$ i.e. $4 \mathrm{~s}^0 3 \mathrm{~d}^6$ get paired to form inner orbital complex. Hence, no unpaired electron. On the other hand $\mathrm{F}^{-}$is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.
VITEEE- 2008
COORDINATION COMPOUNDS
274155
In $\mathrm{TeCl}_4$ the central atom tellurium involves
1 $\mathrm{sp}^3$ hybridization
2 $\mathrm{sp}^3 \mathrm{~d}$ hybridization
3 $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization
4 $\mathrm{dsp}^2$ hybridization
Explanation:
(B) : Hybridisation $=\frac{1}{2}[$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any] $=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^3 \mathrm{~d}$ ${\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4\right]}$
VITEEE- 2007
COORDINATION COMPOUNDS
274156
The hybridization of nickel in nickel tetracarbonyl is
1 $\mathrm{sp}^3$
2 $\mathrm{sp}^2$
3 $\mathrm{dsp}^2$
4 $d^2 s^3$
Explanation:
(A) : The chemical formula of nickeltetracarbonyl is $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$. The oxidation number of the complex is 0 . ${ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
274152
Among $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$ species, the hybridization state of the $\mathrm{Fe}$ atom are, respectively.
274154
The number of unpaired electrons calculated in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathbf{F}_6\right)\right]^{3-}$
1 4 and 4
2 0 and 2
3 2 and 4
4 0 and 4
Explanation:
(D) : In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{F}_6\right)\right]^{3-}$ both the oxidation state of cobalt ion is +3 . In first case $\mathrm{NH}_3$ is the neutral ligand which is a strong field ligand. Hence, the electrons in Co $(+3)$ i.e. $4 \mathrm{~s}^0 3 \mathrm{~d}^6$ get paired to form inner orbital complex. Hence, no unpaired electron. On the other hand $\mathrm{F}^{-}$is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.
VITEEE- 2008
COORDINATION COMPOUNDS
274155
In $\mathrm{TeCl}_4$ the central atom tellurium involves
1 $\mathrm{sp}^3$ hybridization
2 $\mathrm{sp}^3 \mathrm{~d}$ hybridization
3 $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization
4 $\mathrm{dsp}^2$ hybridization
Explanation:
(B) : Hybridisation $=\frac{1}{2}[$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any] $=\frac{1}{2}[6+4+0-0]=5=\mathrm{sp}^3 \mathrm{~d}$ ${\left[\because \mathrm{Te}(52)=[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^4\right]}$
VITEEE- 2007
COORDINATION COMPOUNDS
274156
The hybridization of nickel in nickel tetracarbonyl is
1 $\mathrm{sp}^3$
2 $\mathrm{sp}^2$
3 $\mathrm{dsp}^2$
4 $d^2 s^3$
Explanation:
(A) : The chemical formula of nickeltetracarbonyl is $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$. The oxidation number of the complex is 0 . ${ }_{28} \mathrm{Ni}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
