NEET Test Series from KOTA - 10 Papers In MS WORD
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COORDINATION COMPOUNDS
274354
The coordination number and the oxidation state of the element ' $E$ ' in the complex $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ (where (en) is ethylene diamine ) are respectively,
1 6 and 2
2 4 and 2
3 4 and 3
4 6 and 3
Explanation:
(D) : [E (en $\left.)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ As we know that 'en' and $\mathrm{C}_{2} \mathrm{O}_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 . $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{E}$. $x+2(0)+1(-2)-1=0$ $\mathrm{x}=3$ Co- ordination number and oxidation state of the compound are +6 and +3 respectively.
AIIMS-2014
COORDINATION COMPOUNDS
274355
What is the oxidation state of iron in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}^{2+}\right.$ ?
1 0
2 +1
3 +2
4 +3
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $\mathrm{x}+5(0)+1(+1)=+2(\because \mathrm{NO}$ is linear $)$ $x=+1$ So, the oxidation state of $\mathrm{Fe}$ is +1
BCECE-2013
COORDINATION COMPOUNDS
274356
The coordination number and oxidation state of $\mathrm{Cr}$ in $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$ are respectively
1 6 and +3
2 3 and 0
3 4 and +2
4 3 and +3
Explanation:
(A) : Coordination number of $\mathrm{Cr}$ is 6 (oxatate is bidentate ligand) $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \text { is calculated as }$ $3(1)+\mathrm{x}+3(-2)=0$ $3+\mathrm{x}+(-6)=0$ $\mathrm{x}=6-3$ $\mathrm{x}=+3$
CG PET- 2011
COORDINATION COMPOUNDS
274357
The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are respectively
1 $+3,3 \mathrm{~d}$ and 4
2 $+3,4 \mathrm{~d}$ and 6
3 $+3,3 \mathrm{~d}$ and 6
4 $+2,3 \mathrm{~d}$ and 6
Explanation:
(C) : In the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ is - Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Cr}$. $x+2(0)+2(-1)-1=0$ $\mathrm{x}=+3$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}$ The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\left(\mathrm{Cr}(\text { en })_{2}\right) \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are $+3,3 \mathrm{~d}$ and 6 .
274354
The coordination number and the oxidation state of the element ' $E$ ' in the complex $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ (where (en) is ethylene diamine ) are respectively,
1 6 and 2
2 4 and 2
3 4 and 3
4 6 and 3
Explanation:
(D) : [E (en $\left.)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ As we know that 'en' and $\mathrm{C}_{2} \mathrm{O}_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 . $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{E}$. $x+2(0)+1(-2)-1=0$ $\mathrm{x}=3$ Co- ordination number and oxidation state of the compound are +6 and +3 respectively.
AIIMS-2014
COORDINATION COMPOUNDS
274355
What is the oxidation state of iron in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}^{2+}\right.$ ?
1 0
2 +1
3 +2
4 +3
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $\mathrm{x}+5(0)+1(+1)=+2(\because \mathrm{NO}$ is linear $)$ $x=+1$ So, the oxidation state of $\mathrm{Fe}$ is +1
BCECE-2013
COORDINATION COMPOUNDS
274356
The coordination number and oxidation state of $\mathrm{Cr}$ in $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$ are respectively
1 6 and +3
2 3 and 0
3 4 and +2
4 3 and +3
Explanation:
(A) : Coordination number of $\mathrm{Cr}$ is 6 (oxatate is bidentate ligand) $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \text { is calculated as }$ $3(1)+\mathrm{x}+3(-2)=0$ $3+\mathrm{x}+(-6)=0$ $\mathrm{x}=6-3$ $\mathrm{x}=+3$
CG PET- 2011
COORDINATION COMPOUNDS
274357
The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are respectively
1 $+3,3 \mathrm{~d}$ and 4
2 $+3,4 \mathrm{~d}$ and 6
3 $+3,3 \mathrm{~d}$ and 6
4 $+2,3 \mathrm{~d}$ and 6
Explanation:
(C) : In the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ is - Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Cr}$. $x+2(0)+2(-1)-1=0$ $\mathrm{x}=+3$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}$ The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\left(\mathrm{Cr}(\text { en })_{2}\right) \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are $+3,3 \mathrm{~d}$ and 6 .
274354
The coordination number and the oxidation state of the element ' $E$ ' in the complex $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ (where (en) is ethylene diamine ) are respectively,
1 6 and 2
2 4 and 2
3 4 and 3
4 6 and 3
Explanation:
(D) : [E (en $\left.)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ As we know that 'en' and $\mathrm{C}_{2} \mathrm{O}_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 . $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{E}$. $x+2(0)+1(-2)-1=0$ $\mathrm{x}=3$ Co- ordination number and oxidation state of the compound are +6 and +3 respectively.
AIIMS-2014
COORDINATION COMPOUNDS
274355
What is the oxidation state of iron in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}^{2+}\right.$ ?
1 0
2 +1
3 +2
4 +3
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $\mathrm{x}+5(0)+1(+1)=+2(\because \mathrm{NO}$ is linear $)$ $x=+1$ So, the oxidation state of $\mathrm{Fe}$ is +1
BCECE-2013
COORDINATION COMPOUNDS
274356
The coordination number and oxidation state of $\mathrm{Cr}$ in $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$ are respectively
1 6 and +3
2 3 and 0
3 4 and +2
4 3 and +3
Explanation:
(A) : Coordination number of $\mathrm{Cr}$ is 6 (oxatate is bidentate ligand) $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \text { is calculated as }$ $3(1)+\mathrm{x}+3(-2)=0$ $3+\mathrm{x}+(-6)=0$ $\mathrm{x}=6-3$ $\mathrm{x}=+3$
CG PET- 2011
COORDINATION COMPOUNDS
274357
The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are respectively
1 $+3,3 \mathrm{~d}$ and 4
2 $+3,4 \mathrm{~d}$ and 6
3 $+3,3 \mathrm{~d}$ and 6
4 $+2,3 \mathrm{~d}$ and 6
Explanation:
(C) : In the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ is - Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Cr}$. $x+2(0)+2(-1)-1=0$ $\mathrm{x}=+3$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}$ The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\left(\mathrm{Cr}(\text { en })_{2}\right) \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are $+3,3 \mathrm{~d}$ and 6 .
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274354
The coordination number and the oxidation state of the element ' $E$ ' in the complex $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ (where (en) is ethylene diamine ) are respectively,
1 6 and 2
2 4 and 2
3 4 and 3
4 6 and 3
Explanation:
(D) : [E (en $\left.)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ As we know that 'en' and $\mathrm{C}_{2} \mathrm{O}_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 . $\left[\mathrm{E}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{NO}_{2}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{E}$. $x+2(0)+1(-2)-1=0$ $\mathrm{x}=3$ Co- ordination number and oxidation state of the compound are +6 and +3 respectively.
AIIMS-2014
COORDINATION COMPOUNDS
274355
What is the oxidation state of iron in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}^{2+}\right.$ ?
1 0
2 +1
3 +2
4 +3
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Fe}$. $\mathrm{x}+5(0)+1(+1)=+2(\because \mathrm{NO}$ is linear $)$ $x=+1$ So, the oxidation state of $\mathrm{Fe}$ is +1
BCECE-2013
COORDINATION COMPOUNDS
274356
The coordination number and oxidation state of $\mathrm{Cr}$ in $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$ are respectively
1 6 and +3
2 3 and 0
3 4 and +2
4 3 and +3
Explanation:
(A) : Coordination number of $\mathrm{Cr}$ is 6 (oxatate is bidentate ligand) $\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \text { is calculated as }$ $3(1)+\mathrm{x}+3(-2)=0$ $3+\mathrm{x}+(-6)=0$ $\mathrm{x}=6-3$ $\mathrm{x}=+3$
CG PET- 2011
COORDINATION COMPOUNDS
274357
The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are respectively
1 $+3,3 \mathrm{~d}$ and 4
2 $+3,4 \mathrm{~d}$ and 6
3 $+3,3 \mathrm{~d}$ and 6
4 $+2,3 \mathrm{~d}$ and 6
Explanation:
(C) : In the complex cis $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ is - Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Cr}$. $x+2(0)+2(-1)-1=0$ $\mathrm{x}=+3$ The oxidation state of $\mathrm{Cr}$ is +3 . $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}$ The oxidation number, d-orbital occupation and coordination number of $\mathrm{Cr}$ in the complex cis $\left[\left(\mathrm{Cr}(\text { en })_{2}\right) \mathrm{Cl}_{2}\right] \mathrm{Cl}$ are $+3,3 \mathrm{~d}$ and 6 .