QBManager

COORDINATION COMPOUNDS

274354 The coordination number and the oxidation state of the element ' $E$ ' in the complex $[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$ (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 $)_{2} (C_{2} O_{4})] {NO}_{2}$
As we know that 'en' and $C_{2} O_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 .
$[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$
Let, $x$ be the oxidation state of $E$ .
$x + 2 (0) + 1 (- 2) - 1 = 0$
$x = 3$
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 $[Fe {(H_{2} O)}_{5} {NO}^{2 +}$ ?

1 0

2 +1

3 +2

4 +3

Explanation:

(B) : ${[Fe {(H_{2} O)}_{5} NO]}^{2 +}$
Let, $x$ be the oxidation state of $Fe$ .
$x + 5 (0) + 1 (+ 1) = + 2 (∵ NO$ is linear $)$
$x = + 1$
So, the oxidation state of $Fe$ is +1

BCECE-2013

COORDINATION COMPOUNDS

274356 The coordination number and oxidation state of $Cr$ in $K_{3} [Cr {(C_{2} O_{4})}_{3}]$ are respectively

1 6 and +3

2 3 and 0

3 4 and +2

4 3 and +3

Explanation:

(A) : Coordination number of $Cr$ is 6 (oxatate is bidentate ligand)
$K_{3} [Cr (C_{2} O_{4})] is calculated as$
$3 (1) + x + 3 (- 2) = 0$
$3 + x + (- 6) = 0$
$x = 6 - 3$
$x = + 3$

CG PET- 2011

COORDINATION COMPOUNDS

274357 The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ are respectively

1

+ 3, 3 d

and 4

2

+ 3, 4 d

and 6

3

+ 3, 3 d

and 6

4

+ 2, 3 d

and 6

Explanation:

(C) : In the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ is -
Let, $x$ be the oxidation state of $Cr$ .
$x + 2 (0) + 2 (- 1) - 1 = 0$
$x = + 3$
The oxidation state of $Cr$ is +3 .
${Cr}^{3 +} = [Ar] 3 d^{3}$
The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[(Cr (en)_{2}) {Cl}_{2}] Cl$ are $+ 3, 3 d$ and 6 .

J and K CET-(2010)

COORDINATION COMPOUNDS

274354 The coordination number and the oxidation state of the element ' $E$ ' in the complex $[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$ (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 $)_{2} (C_{2} O_{4})] {NO}_{2}$
As we know that 'en' and $C_{2} O_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 .
$[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$
Let, $x$ be the oxidation state of $E$ .
$x + 2 (0) + 1 (- 2) - 1 = 0$
$x = 3$
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 $[Fe {(H_{2} O)}_{5} {NO}^{2 +}$ ?

1 0

2 +1

3 +2

4 +3

Explanation:

(B) : ${[Fe {(H_{2} O)}_{5} NO]}^{2 +}$
Let, $x$ be the oxidation state of $Fe$ .
$x + 5 (0) + 1 (+ 1) = + 2 (∵ NO$ is linear $)$
$x = + 1$
So, the oxidation state of $Fe$ is +1

BCECE-2013

COORDINATION COMPOUNDS

274356 The coordination number and oxidation state of $Cr$ in $K_{3} [Cr {(C_{2} O_{4})}_{3}]$ are respectively

1 6 and +3

2 3 and 0

3 4 and +2

4 3 and +3

Explanation:

(A) : Coordination number of $Cr$ is 6 (oxatate is bidentate ligand)
$K_{3} [Cr (C_{2} O_{4})] is calculated as$
$3 (1) + x + 3 (- 2) = 0$
$3 + x + (- 6) = 0$
$x = 6 - 3$
$x = + 3$

CG PET- 2011

COORDINATION COMPOUNDS

274357 The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ are respectively

1

+ 3, 3 d

and 4

2

+ 3, 4 d

and 6

3

+ 3, 3 d

and 6

4

+ 2, 3 d

and 6

Explanation:

(C) : In the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ is -
Let, $x$ be the oxidation state of $Cr$ .
$x + 2 (0) + 2 (- 1) - 1 = 0$
$x = + 3$
The oxidation state of $Cr$ is +3 .
${Cr}^{3 +} = [Ar] 3 d^{3}$
The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[(Cr (en)_{2}) {Cl}_{2}] Cl$ are $+ 3, 3 d$ and 6 .

J and K CET-(2010)

COORDINATION COMPOUNDS

274354 The coordination number and the oxidation state of the element ' $E$ ' in the complex $[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$ (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 $)_{2} (C_{2} O_{4})] {NO}_{2}$
As we know that 'en' and $C_{2} O_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 .
$[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$
Let, $x$ be the oxidation state of $E$ .
$x + 2 (0) + 1 (- 2) - 1 = 0$
$x = 3$
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 $[Fe {(H_{2} O)}_{5} {NO}^{2 +}$ ?

1 0

2 +1

3 +2

4 +3

Explanation:

(B) : ${[Fe {(H_{2} O)}_{5} NO]}^{2 +}$
Let, $x$ be the oxidation state of $Fe$ .
$x + 5 (0) + 1 (+ 1) = + 2 (∵ NO$ is linear $)$
$x = + 1$
So, the oxidation state of $Fe$ is +1

BCECE-2013

COORDINATION COMPOUNDS

274356 The coordination number and oxidation state of $Cr$ in $K_{3} [Cr {(C_{2} O_{4})}_{3}]$ are respectively

1 6 and +3

2 3 and 0

3 4 and +2

4 3 and +3

Explanation:

(A) : Coordination number of $Cr$ is 6 (oxatate is bidentate ligand)
$K_{3} [Cr (C_{2} O_{4})] is calculated as$
$3 (1) + x + 3 (- 2) = 0$
$3 + x + (- 6) = 0$
$x = 6 - 3$
$x = + 3$

CG PET- 2011

COORDINATION COMPOUNDS

274357 The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ are respectively

1

+ 3, 3 d

and 4

2

+ 3, 4 d

and 6

3

+ 3, 3 d

and 6

4

+ 2, 3 d

and 6

Explanation:

(C) : In the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ is -
Let, $x$ be the oxidation state of $Cr$ .
$x + 2 (0) + 2 (- 1) - 1 = 0$
$x = + 3$
The oxidation state of $Cr$ is +3 .
${Cr}^{3 +} = [Ar] 3 d^{3}$
The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[(Cr (en)_{2}) {Cl}_{2}] Cl$ are $+ 3, 3 d$ and 6 .

J and K CET-(2010)

COORDINATION COMPOUNDS

274354 The coordination number and the oxidation state of the element ' $E$ ' in the complex $[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$ (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 $)_{2} (C_{2} O_{4})] {NO}_{2}$
As we know that 'en' and $C_{2} O_{4}$ are bidentate ligand. Thus, central metal connect with the six donor site and Hence coordination number is 6 .
$[E (en)_{2} (C_{2} O_{4})] {NO}_{2}$
Let, $x$ be the oxidation state of $E$ .
$x + 2 (0) + 1 (- 2) - 1 = 0$
$x = 3$
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 $[Fe {(H_{2} O)}_{5} {NO}^{2 +}$ ?

1 0

2 +1

3 +2

4 +3

Explanation:

(B) : ${[Fe {(H_{2} O)}_{5} NO]}^{2 +}$
Let, $x$ be the oxidation state of $Fe$ .
$x + 5 (0) + 1 (+ 1) = + 2 (∵ NO$ is linear $)$
$x = + 1$
So, the oxidation state of $Fe$ is +1

BCECE-2013

COORDINATION COMPOUNDS

274356 The coordination number and oxidation state of $Cr$ in $K_{3} [Cr {(C_{2} O_{4})}_{3}]$ are respectively

1 6 and +3

2 3 and 0

3 4 and +2

4 3 and +3

Explanation:

(A) : Coordination number of $Cr$ is 6 (oxatate is bidentate ligand)
$K_{3} [Cr (C_{2} O_{4})] is calculated as$
$3 (1) + x + 3 (- 2) = 0$
$3 + x + (- 6) = 0$
$x = 6 - 3$
$x = + 3$

CG PET- 2011

COORDINATION COMPOUNDS

274357 The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ are respectively

1

+ 3, 3 d

and 4

2

+ 3, 4 d

and 6

3

+ 3, 3 d

and 6

4

+ 2, 3 d

and 6

Explanation:

(C) : In the complex cis $[Cr (en)_{2} {Cl}_{2}] Cl$ is -
Let, $x$ be the oxidation state of $Cr$ .
$x + 2 (0) + 2 (- 1) - 1 = 0$
$x = + 3$
The oxidation state of $Cr$ is +3 .
${Cr}^{3 +} = [Ar] 3 d^{3}$
The oxidation number, d-orbital occupation and coordination number of $Cr$ in the complex cis $[(Cr (en)_{2}) {Cl}_{2}] Cl$ are $+ 3, 3 d$ and 6 .

J and K CET-(2010)