274334
The oxidation states of $\mathrm{Pt}$ in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is
1 +1 and +1
2 +1 and +2
3 +2 and +2
4 +2 and +1
Explanation:
(C) : The co- ordination compound is $\left[\mathrm{P}+\left(\mathrm{NH}_{3}\right)_{4}\right]$ $\left[\mathrm{P}+\mathrm{Cl}_{4}\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{P}+$ $x+4(0)+x+4(-1)=0$ $2 \mathrm{x}-4=0$ $\mathrm{x}=+2$ So, oxidation state of $\mathrm{P}+$ is +2
AMU-2019
COORDINATION COMPOUNDS
274328
In complex $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, the correct primary valency, secondary valency, and coordination number are respectively as $x, y$ and $z$. What are $x, y$ and $z$ ?
1 $x=6 \quad y=6 \quad z=3$
2 $x=6 \quad y=3 \quad z=6$
3 $x=3 \quad y=6 \quad z=6$
4 $x=3 \quad y=6 \quad z=4$
Explanation:
(C) : Primary valency $=$ Oxidation state Secondary valency $=$ No. of molecule co-ordinate to central atom. In the molecule $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - $\mathrm{Cr}$ is +3 State. Hence Primary valency $(\mathrm{x})=3$ $6 \mathrm{H}_{2} \mathrm{O}$ is $\mathrm{CO}$-ordinate with central atom. Hence, secondary valency $(\mathrm{y})=6$ And, coordination number of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ is $(\mathrm{z})=6$
AP EAPCET-12.07.2022
COORDINATION COMPOUNDS
274329
Which of the following has least oxidation state of $\mathrm{Fe}$ ?
4 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
Explanation:
(C) : (a) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ $3(+1)+x+6(-1)=0$ $\mathrm{x}=+3$ (b) $\quad \mathrm{K}_{2}\left[\mathrm{FeO}_{4}\right]$ $2(+1)+x+4(-2)=0$ $\mathrm{x}=+6$ (c) $\quad \mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{x}+1(-2)+2(1)+1(-2)=0$ $\mathrm{x}-2+2-2=0$ (d.) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
UPTU/UPSEE-2018
COORDINATION COMPOUNDS
274330
Which of the following is wrong statement?
1 $\mathrm{Ni}(\mathrm{CO})_{4}$ has oxidation number +4 for $\mathrm{Ni}$
2 $\mathrm{Ni}(\mathrm{CO})_{4}$ has zero oxidation number for $\mathrm{Ni}$
3 $\mathrm{Ni}$ is metal
4 $\mathrm{CO}$ is gas
Explanation:
(A) : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ Let, $x$ be the oxidation sate of $\mathrm{Ni}$. $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\therefore$ oxidation state of $\mathrm{Ni}$ is 0 (c) $\mathrm{Ni}$ is a metal which is belongs to the $3 \mathrm{~d}$-series of transition metal. (d) CO is a poisons gas. The name of the gas is carbon monoxide. So, statement (a) is false.
BITSAT-2008]
COORDINATION COMPOUNDS
274331
The oxidation number of central metal in $\left[\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \quad$ and $\quad\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{\oplus}$, respectively, are
1 $+2 ;+1$
2 $+2 ;+2$
3 $+2 ;+3$
4 $+3 ;+2$
Explanation:
(C) : The oxidation no. of central metal Pt and Co are- | $\left[\mathbf{P t}\left(\mathbf{N H}_3\right)_{\mathbf{2}} \mathbf{C l}\left(\mathbf{N O}_2\right)\right]$ | $\left[\mathbf{C o C l}_{\mathbf{2}}(\mathbf{e n})_{\mathbf{2}}\right]^{+}$ | | :--- | :--- | | $\mathrm{x}+2(0)+(-1)+(-1)$ | $\mathrm{x}+2(-1)+2(0)=+1$ | | $\mathrm{x}+(-1)+(-1)=0$ | $\mathrm{x}-2+0=+1$ | | $\mathrm{x}=+2$ | $\mathrm{x}=+3$ |
274334
The oxidation states of $\mathrm{Pt}$ in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is
1 +1 and +1
2 +1 and +2
3 +2 and +2
4 +2 and +1
Explanation:
(C) : The co- ordination compound is $\left[\mathrm{P}+\left(\mathrm{NH}_{3}\right)_{4}\right]$ $\left[\mathrm{P}+\mathrm{Cl}_{4}\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{P}+$ $x+4(0)+x+4(-1)=0$ $2 \mathrm{x}-4=0$ $\mathrm{x}=+2$ So, oxidation state of $\mathrm{P}+$ is +2
AMU-2019
COORDINATION COMPOUNDS
274328
In complex $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, the correct primary valency, secondary valency, and coordination number are respectively as $x, y$ and $z$. What are $x, y$ and $z$ ?
1 $x=6 \quad y=6 \quad z=3$
2 $x=6 \quad y=3 \quad z=6$
3 $x=3 \quad y=6 \quad z=6$
4 $x=3 \quad y=6 \quad z=4$
Explanation:
(C) : Primary valency $=$ Oxidation state Secondary valency $=$ No. of molecule co-ordinate to central atom. In the molecule $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - $\mathrm{Cr}$ is +3 State. Hence Primary valency $(\mathrm{x})=3$ $6 \mathrm{H}_{2} \mathrm{O}$ is $\mathrm{CO}$-ordinate with central atom. Hence, secondary valency $(\mathrm{y})=6$ And, coordination number of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ is $(\mathrm{z})=6$
AP EAPCET-12.07.2022
COORDINATION COMPOUNDS
274329
Which of the following has least oxidation state of $\mathrm{Fe}$ ?
4 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
Explanation:
(C) : (a) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ $3(+1)+x+6(-1)=0$ $\mathrm{x}=+3$ (b) $\quad \mathrm{K}_{2}\left[\mathrm{FeO}_{4}\right]$ $2(+1)+x+4(-2)=0$ $\mathrm{x}=+6$ (c) $\quad \mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{x}+1(-2)+2(1)+1(-2)=0$ $\mathrm{x}-2+2-2=0$ (d.) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
UPTU/UPSEE-2018
COORDINATION COMPOUNDS
274330
Which of the following is wrong statement?
1 $\mathrm{Ni}(\mathrm{CO})_{4}$ has oxidation number +4 for $\mathrm{Ni}$
2 $\mathrm{Ni}(\mathrm{CO})_{4}$ has zero oxidation number for $\mathrm{Ni}$
3 $\mathrm{Ni}$ is metal
4 $\mathrm{CO}$ is gas
Explanation:
(A) : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ Let, $x$ be the oxidation sate of $\mathrm{Ni}$. $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\therefore$ oxidation state of $\mathrm{Ni}$ is 0 (c) $\mathrm{Ni}$ is a metal which is belongs to the $3 \mathrm{~d}$-series of transition metal. (d) CO is a poisons gas. The name of the gas is carbon monoxide. So, statement (a) is false.
BITSAT-2008]
COORDINATION COMPOUNDS
274331
The oxidation number of central metal in $\left[\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \quad$ and $\quad\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{\oplus}$, respectively, are
1 $+2 ;+1$
2 $+2 ;+2$
3 $+2 ;+3$
4 $+3 ;+2$
Explanation:
(C) : The oxidation no. of central metal Pt and Co are- | $\left[\mathbf{P t}\left(\mathbf{N H}_3\right)_{\mathbf{2}} \mathbf{C l}\left(\mathbf{N O}_2\right)\right]$ | $\left[\mathbf{C o C l}_{\mathbf{2}}(\mathbf{e n})_{\mathbf{2}}\right]^{+}$ | | :--- | :--- | | $\mathrm{x}+2(0)+(-1)+(-1)$ | $\mathrm{x}+2(-1)+2(0)=+1$ | | $\mathrm{x}+(-1)+(-1)=0$ | $\mathrm{x}-2+0=+1$ | | $\mathrm{x}=+2$ | $\mathrm{x}=+3$ |
274334
The oxidation states of $\mathrm{Pt}$ in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is
1 +1 and +1
2 +1 and +2
3 +2 and +2
4 +2 and +1
Explanation:
(C) : The co- ordination compound is $\left[\mathrm{P}+\left(\mathrm{NH}_{3}\right)_{4}\right]$ $\left[\mathrm{P}+\mathrm{Cl}_{4}\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{P}+$ $x+4(0)+x+4(-1)=0$ $2 \mathrm{x}-4=0$ $\mathrm{x}=+2$ So, oxidation state of $\mathrm{P}+$ is +2
AMU-2019
COORDINATION COMPOUNDS
274328
In complex $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, the correct primary valency, secondary valency, and coordination number are respectively as $x, y$ and $z$. What are $x, y$ and $z$ ?
1 $x=6 \quad y=6 \quad z=3$
2 $x=6 \quad y=3 \quad z=6$
3 $x=3 \quad y=6 \quad z=6$
4 $x=3 \quad y=6 \quad z=4$
Explanation:
(C) : Primary valency $=$ Oxidation state Secondary valency $=$ No. of molecule co-ordinate to central atom. In the molecule $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - $\mathrm{Cr}$ is +3 State. Hence Primary valency $(\mathrm{x})=3$ $6 \mathrm{H}_{2} \mathrm{O}$ is $\mathrm{CO}$-ordinate with central atom. Hence, secondary valency $(\mathrm{y})=6$ And, coordination number of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ is $(\mathrm{z})=6$
AP EAPCET-12.07.2022
COORDINATION COMPOUNDS
274329
Which of the following has least oxidation state of $\mathrm{Fe}$ ?
4 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
Explanation:
(C) : (a) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ $3(+1)+x+6(-1)=0$ $\mathrm{x}=+3$ (b) $\quad \mathrm{K}_{2}\left[\mathrm{FeO}_{4}\right]$ $2(+1)+x+4(-2)=0$ $\mathrm{x}=+6$ (c) $\quad \mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{x}+1(-2)+2(1)+1(-2)=0$ $\mathrm{x}-2+2-2=0$ (d.) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
UPTU/UPSEE-2018
COORDINATION COMPOUNDS
274330
Which of the following is wrong statement?
1 $\mathrm{Ni}(\mathrm{CO})_{4}$ has oxidation number +4 for $\mathrm{Ni}$
2 $\mathrm{Ni}(\mathrm{CO})_{4}$ has zero oxidation number for $\mathrm{Ni}$
3 $\mathrm{Ni}$ is metal
4 $\mathrm{CO}$ is gas
Explanation:
(A) : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ Let, $x$ be the oxidation sate of $\mathrm{Ni}$. $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\therefore$ oxidation state of $\mathrm{Ni}$ is 0 (c) $\mathrm{Ni}$ is a metal which is belongs to the $3 \mathrm{~d}$-series of transition metal. (d) CO is a poisons gas. The name of the gas is carbon monoxide. So, statement (a) is false.
BITSAT-2008]
COORDINATION COMPOUNDS
274331
The oxidation number of central metal in $\left[\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \quad$ and $\quad\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{\oplus}$, respectively, are
1 $+2 ;+1$
2 $+2 ;+2$
3 $+2 ;+3$
4 $+3 ;+2$
Explanation:
(C) : The oxidation no. of central metal Pt and Co are- | $\left[\mathbf{P t}\left(\mathbf{N H}_3\right)_{\mathbf{2}} \mathbf{C l}\left(\mathbf{N O}_2\right)\right]$ | $\left[\mathbf{C o C l}_{\mathbf{2}}(\mathbf{e n})_{\mathbf{2}}\right]^{+}$ | | :--- | :--- | | $\mathrm{x}+2(0)+(-1)+(-1)$ | $\mathrm{x}+2(-1)+2(0)=+1$ | | $\mathrm{x}+(-1)+(-1)=0$ | $\mathrm{x}-2+0=+1$ | | $\mathrm{x}=+2$ | $\mathrm{x}=+3$ |
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274334
The oxidation states of $\mathrm{Pt}$ in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is
1 +1 and +1
2 +1 and +2
3 +2 and +2
4 +2 and +1
Explanation:
(C) : The co- ordination compound is $\left[\mathrm{P}+\left(\mathrm{NH}_{3}\right)_{4}\right]$ $\left[\mathrm{P}+\mathrm{Cl}_{4}\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{P}+$ $x+4(0)+x+4(-1)=0$ $2 \mathrm{x}-4=0$ $\mathrm{x}=+2$ So, oxidation state of $\mathrm{P}+$ is +2
AMU-2019
COORDINATION COMPOUNDS
274328
In complex $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, the correct primary valency, secondary valency, and coordination number are respectively as $x, y$ and $z$. What are $x, y$ and $z$ ?
1 $x=6 \quad y=6 \quad z=3$
2 $x=6 \quad y=3 \quad z=6$
3 $x=3 \quad y=6 \quad z=6$
4 $x=3 \quad y=6 \quad z=4$
Explanation:
(C) : Primary valency $=$ Oxidation state Secondary valency $=$ No. of molecule co-ordinate to central atom. In the molecule $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - $\mathrm{Cr}$ is +3 State. Hence Primary valency $(\mathrm{x})=3$ $6 \mathrm{H}_{2} \mathrm{O}$ is $\mathrm{CO}$-ordinate with central atom. Hence, secondary valency $(\mathrm{y})=6$ And, coordination number of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ is $(\mathrm{z})=6$
AP EAPCET-12.07.2022
COORDINATION COMPOUNDS
274329
Which of the following has least oxidation state of $\mathrm{Fe}$ ?
4 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
Explanation:
(C) : (a) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ $3(+1)+x+6(-1)=0$ $\mathrm{x}=+3$ (b) $\quad \mathrm{K}_{2}\left[\mathrm{FeO}_{4}\right]$ $2(+1)+x+4(-2)=0$ $\mathrm{x}=+6$ (c) $\quad \mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{x}+1(-2)+2(1)+1(-2)=0$ $\mathrm{x}-2+2-2=0$ (d.) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
UPTU/UPSEE-2018
COORDINATION COMPOUNDS
274330
Which of the following is wrong statement?
1 $\mathrm{Ni}(\mathrm{CO})_{4}$ has oxidation number +4 for $\mathrm{Ni}$
2 $\mathrm{Ni}(\mathrm{CO})_{4}$ has zero oxidation number for $\mathrm{Ni}$
3 $\mathrm{Ni}$ is metal
4 $\mathrm{CO}$ is gas
Explanation:
(A) : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ Let, $x$ be the oxidation sate of $\mathrm{Ni}$. $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\therefore$ oxidation state of $\mathrm{Ni}$ is 0 (c) $\mathrm{Ni}$ is a metal which is belongs to the $3 \mathrm{~d}$-series of transition metal. (d) CO is a poisons gas. The name of the gas is carbon monoxide. So, statement (a) is false.
BITSAT-2008]
COORDINATION COMPOUNDS
274331
The oxidation number of central metal in $\left[\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \quad$ and $\quad\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{\oplus}$, respectively, are
1 $+2 ;+1$
2 $+2 ;+2$
3 $+2 ;+3$
4 $+3 ;+2$
Explanation:
(C) : The oxidation no. of central metal Pt and Co are- | $\left[\mathbf{P t}\left(\mathbf{N H}_3\right)_{\mathbf{2}} \mathbf{C l}\left(\mathbf{N O}_2\right)\right]$ | $\left[\mathbf{C o C l}_{\mathbf{2}}(\mathbf{e n})_{\mathbf{2}}\right]^{+}$ | | :--- | :--- | | $\mathrm{x}+2(0)+(-1)+(-1)$ | $\mathrm{x}+2(-1)+2(0)=+1$ | | $\mathrm{x}+(-1)+(-1)=0$ | $\mathrm{x}-2+0=+1$ | | $\mathrm{x}=+2$ | $\mathrm{x}=+3$ |
274334
The oxidation states of $\mathrm{Pt}$ in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is
1 +1 and +1
2 +1 and +2
3 +2 and +2
4 +2 and +1
Explanation:
(C) : The co- ordination compound is $\left[\mathrm{P}+\left(\mathrm{NH}_{3}\right)_{4}\right]$ $\left[\mathrm{P}+\mathrm{Cl}_{4}\right]$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{P}+$ $x+4(0)+x+4(-1)=0$ $2 \mathrm{x}-4=0$ $\mathrm{x}=+2$ So, oxidation state of $\mathrm{P}+$ is +2
AMU-2019
COORDINATION COMPOUNDS
274328
In complex $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, the correct primary valency, secondary valency, and coordination number are respectively as $x, y$ and $z$. What are $x, y$ and $z$ ?
1 $x=6 \quad y=6 \quad z=3$
2 $x=6 \quad y=3 \quad z=6$
3 $x=3 \quad y=6 \quad z=6$
4 $x=3 \quad y=6 \quad z=4$
Explanation:
(C) : Primary valency $=$ Oxidation state Secondary valency $=$ No. of molecule co-ordinate to central atom. In the molecule $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ - $\mathrm{Cr}$ is +3 State. Hence Primary valency $(\mathrm{x})=3$ $6 \mathrm{H}_{2} \mathrm{O}$ is $\mathrm{CO}$-ordinate with central atom. Hence, secondary valency $(\mathrm{y})=6$ And, coordination number of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ is $(\mathrm{z})=6$
AP EAPCET-12.07.2022
COORDINATION COMPOUNDS
274329
Which of the following has least oxidation state of $\mathrm{Fe}$ ?
4 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
Explanation:
(C) : (a) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{6}\right]$ $3(+1)+x+6(-1)=0$ $\mathrm{x}=+3$ (b) $\quad \mathrm{K}_{2}\left[\mathrm{FeO}_{4}\right]$ $2(+1)+x+4(-2)=0$ $\mathrm{x}=+6$ (c) $\quad \mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{x}+1(-2)+2(1)+1(-2)=0$ $\mathrm{x}-2+2-2=0$ (d.) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ $\mathrm{x}+6(-1)=-3$ $\mathrm{x}=+3$ So, $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ has the least oxidation state.
UPTU/UPSEE-2018
COORDINATION COMPOUNDS
274330
Which of the following is wrong statement?
1 $\mathrm{Ni}(\mathrm{CO})_{4}$ has oxidation number +4 for $\mathrm{Ni}$
2 $\mathrm{Ni}(\mathrm{CO})_{4}$ has zero oxidation number for $\mathrm{Ni}$
3 $\mathrm{Ni}$ is metal
4 $\mathrm{CO}$ is gas
Explanation:
(A) : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ Let, $x$ be the oxidation sate of $\mathrm{Ni}$. $\mathrm{x}+4(0)=0$ $\mathrm{x}=0$ $\therefore$ oxidation state of $\mathrm{Ni}$ is 0 (c) $\mathrm{Ni}$ is a metal which is belongs to the $3 \mathrm{~d}$-series of transition metal. (d) CO is a poisons gas. The name of the gas is carbon monoxide. So, statement (a) is false.
BITSAT-2008]
COORDINATION COMPOUNDS
274331
The oxidation number of central metal in $\left[\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \quad$ and $\quad\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{\oplus}$, respectively, are
1 $+2 ;+1$
2 $+2 ;+2$
3 $+2 ;+3$
4 $+3 ;+2$
Explanation:
(C) : The oxidation no. of central metal Pt and Co are- | $\left[\mathbf{P t}\left(\mathbf{N H}_3\right)_{\mathbf{2}} \mathbf{C l}\left(\mathbf{N O}_2\right)\right]$ | $\left[\mathbf{C o C l}_{\mathbf{2}}(\mathbf{e n})_{\mathbf{2}}\right]^{+}$ | | :--- | :--- | | $\mathrm{x}+2(0)+(-1)+(-1)$ | $\mathrm{x}+2(-1)+2(0)=+1$ | | $\mathrm{x}+(-1)+(-1)=0$ | $\mathrm{x}-2+0=+1$ | | $\mathrm{x}=+2$ | $\mathrm{x}=+3$ |