(B) : $\left[\mathrm{RhCl}_6\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Rh}$. $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Rh}=[\mathrm{Kr}] 4 \mathrm{~d}^7 5 \mathrm{~s}^2$ $\mathrm{Rh}^{3+}=[\mathrm{Kr}] 4 \mathrm{~d}^6 \quad(\because$ Pairing occur because $\mathrm{Rh}$ belongs to $4 \mathrm{~d}$ series) It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $\operatorname{spin} t t_{2 g}^6 e_g^0$
J and K CET-(2018)
COORDINATION COMPOUNDS
274150
The number of d-electrons in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is
1 2
2 3
3 4
4 5
Explanation:
(B) : $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5$ $\mathrm{Cr}^{+3}=[\mathrm{Ar}] 4 \mathrm{~s}^0 3 \mathrm{~d}^3$ $\therefore$ The d-electron in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}$ is 3 .
MPPET- 2009
COORDINATION COMPOUNDS
274151
The hybridization of $\mathrm{Fe}$ in $\mathrm{K}_3 \mathrm{Fe}(\mathrm{CN})_6$ is
1 $\mathrm{sp}^3$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3 \mathrm{~d}^2$
4 $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(D) : $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $3(+1)+x+6(-1)=0$ $x=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5(\because$ Pairing occur because $\mathrm{CN}$ is $\mathrm{SFL})$ Hence, the hybridization of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ is $\mathrm{d}^2 \mathrm{sp}^3$.
BITSAT-2016
COORDINATION COMPOUNDS
274158
In $\left[\mathrm{NiCl}_4\right]^{2-}$, the type of hybridization for $\mathrm{Ni}$ is
1 $\mathrm{sp}^3 \mathrm{~d}^2$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3$
4 $\mathrm{dsp}^2$
Explanation:
(C) : For the complex- $\left[\mathrm{NiCl}_4\right]^{2-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ni}$ $\therefore \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8$ $\mathrm{Cl}$ is a weak field ligand thus the pairing is not occur. Hence, the hybridization of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is $\mathrm{sp}^3$.
(B) : $\left[\mathrm{RhCl}_6\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Rh}$. $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Rh}=[\mathrm{Kr}] 4 \mathrm{~d}^7 5 \mathrm{~s}^2$ $\mathrm{Rh}^{3+}=[\mathrm{Kr}] 4 \mathrm{~d}^6 \quad(\because$ Pairing occur because $\mathrm{Rh}$ belongs to $4 \mathrm{~d}$ series) It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $\operatorname{spin} t t_{2 g}^6 e_g^0$
J and K CET-(2018)
COORDINATION COMPOUNDS
274150
The number of d-electrons in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is
1 2
2 3
3 4
4 5
Explanation:
(B) : $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5$ $\mathrm{Cr}^{+3}=[\mathrm{Ar}] 4 \mathrm{~s}^0 3 \mathrm{~d}^3$ $\therefore$ The d-electron in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}$ is 3 .
MPPET- 2009
COORDINATION COMPOUNDS
274151
The hybridization of $\mathrm{Fe}$ in $\mathrm{K}_3 \mathrm{Fe}(\mathrm{CN})_6$ is
1 $\mathrm{sp}^3$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3 \mathrm{~d}^2$
4 $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(D) : $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $3(+1)+x+6(-1)=0$ $x=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5(\because$ Pairing occur because $\mathrm{CN}$ is $\mathrm{SFL})$ Hence, the hybridization of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ is $\mathrm{d}^2 \mathrm{sp}^3$.
BITSAT-2016
COORDINATION COMPOUNDS
274158
In $\left[\mathrm{NiCl}_4\right]^{2-}$, the type of hybridization for $\mathrm{Ni}$ is
1 $\mathrm{sp}^3 \mathrm{~d}^2$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3$
4 $\mathrm{dsp}^2$
Explanation:
(C) : For the complex- $\left[\mathrm{NiCl}_4\right]^{2-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ni}$ $\therefore \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8$ $\mathrm{Cl}$ is a weak field ligand thus the pairing is not occur. Hence, the hybridization of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is $\mathrm{sp}^3$.
(B) : $\left[\mathrm{RhCl}_6\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Rh}$. $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Rh}=[\mathrm{Kr}] 4 \mathrm{~d}^7 5 \mathrm{~s}^2$ $\mathrm{Rh}^{3+}=[\mathrm{Kr}] 4 \mathrm{~d}^6 \quad(\because$ Pairing occur because $\mathrm{Rh}$ belongs to $4 \mathrm{~d}$ series) It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $\operatorname{spin} t t_{2 g}^6 e_g^0$
J and K CET-(2018)
COORDINATION COMPOUNDS
274150
The number of d-electrons in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is
1 2
2 3
3 4
4 5
Explanation:
(B) : $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5$ $\mathrm{Cr}^{+3}=[\mathrm{Ar}] 4 \mathrm{~s}^0 3 \mathrm{~d}^3$ $\therefore$ The d-electron in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}$ is 3 .
MPPET- 2009
COORDINATION COMPOUNDS
274151
The hybridization of $\mathrm{Fe}$ in $\mathrm{K}_3 \mathrm{Fe}(\mathrm{CN})_6$ is
1 $\mathrm{sp}^3$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3 \mathrm{~d}^2$
4 $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(D) : $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $3(+1)+x+6(-1)=0$ $x=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5(\because$ Pairing occur because $\mathrm{CN}$ is $\mathrm{SFL})$ Hence, the hybridization of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ is $\mathrm{d}^2 \mathrm{sp}^3$.
BITSAT-2016
COORDINATION COMPOUNDS
274158
In $\left[\mathrm{NiCl}_4\right]^{2-}$, the type of hybridization for $\mathrm{Ni}$ is
1 $\mathrm{sp}^3 \mathrm{~d}^2$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3$
4 $\mathrm{dsp}^2$
Explanation:
(C) : For the complex- $\left[\mathrm{NiCl}_4\right]^{2-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ni}$ $\therefore \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8$ $\mathrm{Cl}$ is a weak field ligand thus the pairing is not occur. Hence, the hybridization of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is $\mathrm{sp}^3$.
(B) : $\left[\mathrm{RhCl}_6\right]^{3-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Rh}$. $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ $\mathrm{Rh}=[\mathrm{Kr}] 4 \mathrm{~d}^7 5 \mathrm{~s}^2$ $\mathrm{Rh}^{3+}=[\mathrm{Kr}] 4 \mathrm{~d}^6 \quad(\because$ Pairing occur because $\mathrm{Rh}$ belongs to $4 \mathrm{~d}$ series) It forms the low spin complex because electron fill with forcefully in $t_{2 g}$ orbital i.e. low $\operatorname{spin} t t_{2 g}^6 e_g^0$
J and K CET-(2018)
COORDINATION COMPOUNDS
274150
The number of d-electrons in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is
1 2
2 3
3 4
4 5
Explanation:
(B) : $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5$ $\mathrm{Cr}^{+3}=[\mathrm{Ar}] 4 \mathrm{~s}^0 3 \mathrm{~d}^3$ $\therefore$ The d-electron in $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}$ is 3 .
MPPET- 2009
COORDINATION COMPOUNDS
274151
The hybridization of $\mathrm{Fe}$ in $\mathrm{K}_3 \mathrm{Fe}(\mathrm{CN})_6$ is
1 $\mathrm{sp}^3$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3 \mathrm{~d}^2$
4 $\mathrm{d}^2 \mathrm{sp}^3$
Explanation:
(D) : $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $3(+1)+x+6(-1)=0$ $x=+3$ $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5(\because$ Pairing occur because $\mathrm{CN}$ is $\mathrm{SFL})$ Hence, the hybridization of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ is $\mathrm{d}^2 \mathrm{sp}^3$.
BITSAT-2016
COORDINATION COMPOUNDS
274158
In $\left[\mathrm{NiCl}_4\right]^{2-}$, the type of hybridization for $\mathrm{Ni}$ is
1 $\mathrm{sp}^3 \mathrm{~d}^2$
2 $\mathrm{dsp}^3$
3 $\mathrm{sp}^3$
4 $\mathrm{dsp}^2$
Explanation:
(C) : For the complex- $\left[\mathrm{NiCl}_4\right]^{2-}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ni}$ $\therefore \mathrm{x}+4(-1)=-2$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8$ $\mathrm{Cl}$ is a weak field ligand thus the pairing is not occur. Hence, the hybridization of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is $\mathrm{sp}^3$.
