QBManager

COORDINATION COMPOUNDS

274251 Which of the following facts about the complex $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ is wrong?

1 The complex involves

d^{2} {sp}^{3}

hybridization and is octahedral in shape

2 The complex is paramagnetic

3 The complex is an outer orbital complex

4 The complex gives white precipitate with silver nitrate solution

Explanation:

(C) : $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ complex, $Cr$ is present in +3 oxidation.
In ${Cr}^{3 +}$ always inner d-orbital i.e., $(n - 1) d$ is used for hybridisation and forms $d^{2} {sp}^{3}$ hybridisation with octahedral geometry.
original image
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior.
$[Cr {({NH}_{3})}_{6}] \cdot {Cl}_{3} + {AgNO}_{3} ⟶$
$(ppt.) 3$

AIEEE 2011

COORDINATION COMPOUNDS

274253 The effective atomic number for ${[Rh {(H_{2} O)}_{6}]}^{3 +}$ (atomic number of $R h$ is 45 ) is

1 42

2 45

3 48

4 54

Explanation:

(D) : Atomic Number of $Rh = 45$
Oxidation state of $Rh = + 3$
No. of donor site $= 6$
EAN of ${[Rh {(H_{2} O)}_{6}]}^{3 +} =$ Atomic no. - Oxidation $+ 2 \times$
Number of donor atom attached with metal. $= 45 - 3 + 2 \times 6 = 54$

J and K CET-(2012)

COORDINATION COMPOUNDS

274257 The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is
(At. $No$ of $Cr = 24, Fe = 26, Pd = 46$ )

1

[pd {({NH}_{3})}_{6}] {Cl}_{4}

2

[Cr {({NH}_{3})}_{5} Cl] {SO}_{4}

3

{Na}_{4} [Fe (CN)_{6}]

4

K_{3} [Fe (CN)_{6}]

Explanation:

(C) : Atomic number of krypton $= 36$
E.AN of complex A = Atomic No of Krypton
Atomic number of $Fe = 26$
Oxidation state of $Fe$ in ${Na}_{4} [Fe (CN)_{6}]$ complex
$x + 6 (- 1) = - 4$
$x = 2$
Number of donor sites attached with $Fe = 6$
E.AN of ${[Fe (CN)_{6}]}^{4 -} = Z - (ON) + 2 (CN)$
Where, $ON =$ Oxidation no.
$CN =$ coordination number
$Z =$ atomic number
$= 26 - 2 + 2 (6)$
$= 36$
Hence, complex A is ${Na}_{4} [Fe (CN)]_{6}$

J and K CET-(2006)

COORDINATION COMPOUNDS

274189 What is the value of $x$ on the ${[Ni (CN)_{4}]}^{x}$ complex ion?

1 +2

2 -2

3 0 (zero)

4 +4

Explanation:

(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 .
${[Ni (CN)_{4}]}^{x}$
$2 + 4 \times (- 1) = x$
$2 - 4 = x$
$x = - 2$
The complex is ${[Ni (CN)_{4}]}^{2 -}$

Manipal-2020

COORDINATION COMPOUNDS

274165 ${Fe}^{3 +}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

1

{[Fe {(H_{2} O)}_{6}]}_{2} [F_{e} (CN)_{6}]

2

{Fe}_{2} {[Fe (CN)_{6}]}_{2}

3

{Fe}_{3} {[Fe (OH)_{2} (CN)_{4}]}_{2}

4

{Fe}_{4} {[Fe (CN)_{6}]}_{3}

Explanation:

(D) : The reaction is-
$4 {Fe}^{3 +} + 3 [Fe (CN)_{6}] ⟶ {Fe}_{4} {[Fe (CN)_{6}]}_{3}$
$Potassium$
$Ferrocynide$
$(PrussianBlue)$

JEE Main 27.07.2022

COORDINATION COMPOUNDS

274251 Which of the following facts about the complex $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ is wrong?

1 The complex involves

d^{2} {sp}^{3}

hybridization and is octahedral in shape

2 The complex is paramagnetic

3 The complex is an outer orbital complex

4 The complex gives white precipitate with silver nitrate solution

Explanation:

(C) : $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ complex, $Cr$ is present in +3 oxidation.
In ${Cr}^{3 +}$ always inner d-orbital i.e., $(n - 1) d$ is used for hybridisation and forms $d^{2} {sp}^{3}$ hybridisation with octahedral geometry.
original image
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior.
$[Cr {({NH}_{3})}_{6}] \cdot {Cl}_{3} + {AgNO}_{3} ⟶$
$(ppt.) 3$

AIEEE 2011

COORDINATION COMPOUNDS

274253 The effective atomic number for ${[Rh {(H_{2} O)}_{6}]}^{3 +}$ (atomic number of $R h$ is 45 ) is

1 42

2 45

3 48

4 54

Explanation:

(D) : Atomic Number of $Rh = 45$
Oxidation state of $Rh = + 3$
No. of donor site $= 6$
EAN of ${[Rh {(H_{2} O)}_{6}]}^{3 +} =$ Atomic no. - Oxidation $+ 2 \times$
Number of donor atom attached with metal. $= 45 - 3 + 2 \times 6 = 54$

J and K CET-(2012)

COORDINATION COMPOUNDS

274257 The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is
(At. $No$ of $Cr = 24, Fe = 26, Pd = 46$ )

1

[pd {({NH}_{3})}_{6}] {Cl}_{4}

2

[Cr {({NH}_{3})}_{5} Cl] {SO}_{4}

3

{Na}_{4} [Fe (CN)_{6}]

4

K_{3} [Fe (CN)_{6}]

Explanation:

(C) : Atomic number of krypton $= 36$
E.AN of complex A = Atomic No of Krypton
Atomic number of $Fe = 26$
Oxidation state of $Fe$ in ${Na}_{4} [Fe (CN)_{6}]$ complex
$x + 6 (- 1) = - 4$
$x = 2$
Number of donor sites attached with $Fe = 6$
E.AN of ${[Fe (CN)_{6}]}^{4 -} = Z - (ON) + 2 (CN)$
Where, $ON =$ Oxidation no.
$CN =$ coordination number
$Z =$ atomic number
$= 26 - 2 + 2 (6)$
$= 36$
Hence, complex A is ${Na}_{4} [Fe (CN)]_{6}$

J and K CET-(2006)

COORDINATION COMPOUNDS

274189 What is the value of $x$ on the ${[Ni (CN)_{4}]}^{x}$ complex ion?

1 +2

2 -2

3 0 (zero)

4 +4

Explanation:

(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 .
${[Ni (CN)_{4}]}^{x}$
$2 + 4 \times (- 1) = x$
$2 - 4 = x$
$x = - 2$
The complex is ${[Ni (CN)_{4}]}^{2 -}$

Manipal-2020

COORDINATION COMPOUNDS

274165 ${Fe}^{3 +}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

1

{[Fe {(H_{2} O)}_{6}]}_{2} [F_{e} (CN)_{6}]

2

{Fe}_{2} {[Fe (CN)_{6}]}_{2}

3

{Fe}_{3} {[Fe (OH)_{2} (CN)_{4}]}_{2}

4

{Fe}_{4} {[Fe (CN)_{6}]}_{3}

Explanation:

(D) : The reaction is-
$4 {Fe}^{3 +} + 3 [Fe (CN)_{6}] ⟶ {Fe}_{4} {[Fe (CN)_{6}]}_{3}$
$Potassium$
$Ferrocynide$
$(PrussianBlue)$

JEE Main 27.07.2022

COORDINATION COMPOUNDS

274251 Which of the following facts about the complex $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ is wrong?

1 The complex involves

d^{2} {sp}^{3}

hybridization and is octahedral in shape

2 The complex is paramagnetic

3 The complex is an outer orbital complex

4 The complex gives white precipitate with silver nitrate solution

Explanation:

(C) : $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ complex, $Cr$ is present in +3 oxidation.
In ${Cr}^{3 +}$ always inner d-orbital i.e., $(n - 1) d$ is used for hybridisation and forms $d^{2} {sp}^{3}$ hybridisation with octahedral geometry.
original image
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior.
$[Cr {({NH}_{3})}_{6}] \cdot {Cl}_{3} + {AgNO}_{3} ⟶$
$(ppt.) 3$

AIEEE 2011

COORDINATION COMPOUNDS

274253 The effective atomic number for ${[Rh {(H_{2} O)}_{6}]}^{3 +}$ (atomic number of $R h$ is 45 ) is

1 42

2 45

3 48

4 54

Explanation:

(D) : Atomic Number of $Rh = 45$
Oxidation state of $Rh = + 3$
No. of donor site $= 6$
EAN of ${[Rh {(H_{2} O)}_{6}]}^{3 +} =$ Atomic no. - Oxidation $+ 2 \times$
Number of donor atom attached with metal. $= 45 - 3 + 2 \times 6 = 54$

J and K CET-(2012)

COORDINATION COMPOUNDS

274257 The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is
(At. $No$ of $Cr = 24, Fe = 26, Pd = 46$ )

1

[pd {({NH}_{3})}_{6}] {Cl}_{4}

2

[Cr {({NH}_{3})}_{5} Cl] {SO}_{4}

3

{Na}_{4} [Fe (CN)_{6}]

4

K_{3} [Fe (CN)_{6}]

Explanation:

(C) : Atomic number of krypton $= 36$
E.AN of complex A = Atomic No of Krypton
Atomic number of $Fe = 26$
Oxidation state of $Fe$ in ${Na}_{4} [Fe (CN)_{6}]$ complex
$x + 6 (- 1) = - 4$
$x = 2$
Number of donor sites attached with $Fe = 6$
E.AN of ${[Fe (CN)_{6}]}^{4 -} = Z - (ON) + 2 (CN)$
Where, $ON =$ Oxidation no.
$CN =$ coordination number
$Z =$ atomic number
$= 26 - 2 + 2 (6)$
$= 36$
Hence, complex A is ${Na}_{4} [Fe (CN)]_{6}$

J and K CET-(2006)

COORDINATION COMPOUNDS

274189 What is the value of $x$ on the ${[Ni (CN)_{4}]}^{x}$ complex ion?

1 +2

2 -2

3 0 (zero)

4 +4

Explanation:

(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 .
${[Ni (CN)_{4}]}^{x}$
$2 + 4 \times (- 1) = x$
$2 - 4 = x$
$x = - 2$
The complex is ${[Ni (CN)_{4}]}^{2 -}$

Manipal-2020

COORDINATION COMPOUNDS

274165 ${Fe}^{3 +}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

1

{[Fe {(H_{2} O)}_{6}]}_{2} [F_{e} (CN)_{6}]

2

{Fe}_{2} {[Fe (CN)_{6}]}_{2}

3

{Fe}_{3} {[Fe (OH)_{2} (CN)_{4}]}_{2}

4

{Fe}_{4} {[Fe (CN)_{6}]}_{3}

Explanation:

(D) : The reaction is-
$4 {Fe}^{3 +} + 3 [Fe (CN)_{6}] ⟶ {Fe}_{4} {[Fe (CN)_{6}]}_{3}$
$Potassium$
$Ferrocynide$
$(PrussianBlue)$

JEE Main 27.07.2022

COORDINATION COMPOUNDS

274251 Which of the following facts about the complex $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ is wrong?

1 The complex involves

d^{2} {sp}^{3}

hybridization and is octahedral in shape

2 The complex is paramagnetic

3 The complex is an outer orbital complex

4 The complex gives white precipitate with silver nitrate solution

Explanation:

(C) : $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ complex, $Cr$ is present in +3 oxidation.
In ${Cr}^{3 +}$ always inner d-orbital i.e., $(n - 1) d$ is used for hybridisation and forms $d^{2} {sp}^{3}$ hybridisation with octahedral geometry.
original image
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior.
$[Cr {({NH}_{3})}_{6}] \cdot {Cl}_{3} + {AgNO}_{3} ⟶$
$(ppt.) 3$

AIEEE 2011

COORDINATION COMPOUNDS

274253 The effective atomic number for ${[Rh {(H_{2} O)}_{6}]}^{3 +}$ (atomic number of $R h$ is 45 ) is

1 42

2 45

3 48

4 54

Explanation:

(D) : Atomic Number of $Rh = 45$
Oxidation state of $Rh = + 3$
No. of donor site $= 6$
EAN of ${[Rh {(H_{2} O)}_{6}]}^{3 +} =$ Atomic no. - Oxidation $+ 2 \times$
Number of donor atom attached with metal. $= 45 - 3 + 2 \times 6 = 54$

J and K CET-(2012)

COORDINATION COMPOUNDS

274257 The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is
(At. $No$ of $Cr = 24, Fe = 26, Pd = 46$ )

1

[pd {({NH}_{3})}_{6}] {Cl}_{4}

2

[Cr {({NH}_{3})}_{5} Cl] {SO}_{4}

3

{Na}_{4} [Fe (CN)_{6}]

4

K_{3} [Fe (CN)_{6}]

Explanation:

(C) : Atomic number of krypton $= 36$
E.AN of complex A = Atomic No of Krypton
Atomic number of $Fe = 26$
Oxidation state of $Fe$ in ${Na}_{4} [Fe (CN)_{6}]$ complex
$x + 6 (- 1) = - 4$
$x = 2$
Number of donor sites attached with $Fe = 6$
E.AN of ${[Fe (CN)_{6}]}^{4 -} = Z - (ON) + 2 (CN)$
Where, $ON =$ Oxidation no.
$CN =$ coordination number
$Z =$ atomic number
$= 26 - 2 + 2 (6)$
$= 36$
Hence, complex A is ${Na}_{4} [Fe (CN)]_{6}$

J and K CET-(2006)

COORDINATION COMPOUNDS

274189 What is the value of $x$ on the ${[Ni (CN)_{4}]}^{x}$ complex ion?

1 +2

2 -2

3 0 (zero)

4 +4

Explanation:

(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 .
${[Ni (CN)_{4}]}^{x}$
$2 + 4 \times (- 1) = x$
$2 - 4 = x$
$x = - 2$
The complex is ${[Ni (CN)_{4}]}^{2 -}$

Manipal-2020

COORDINATION COMPOUNDS

274165 ${Fe}^{3 +}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

1

{[Fe {(H_{2} O)}_{6}]}_{2} [F_{e} (CN)_{6}]

2

{Fe}_{2} {[Fe (CN)_{6}]}_{2}

3

{Fe}_{3} {[Fe (OH)_{2} (CN)_{4}]}_{2}

4

{Fe}_{4} {[Fe (CN)_{6}]}_{3}

Explanation:

(D) : The reaction is-
$4 {Fe}^{3 +} + 3 [Fe (CN)_{6}] ⟶ {Fe}_{4} {[Fe (CN)_{6}]}_{3}$
$Potassium$
$Ferrocynide$
$(PrussianBlue)$

JEE Main 27.07.2022

COORDINATION COMPOUNDS

274251 Which of the following facts about the complex $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ is wrong?

1 The complex involves

d^{2} {sp}^{3}

hybridization and is octahedral in shape

2 The complex is paramagnetic

3 The complex is an outer orbital complex

4 The complex gives white precipitate with silver nitrate solution

Explanation:

(C) : $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ complex, $Cr$ is present in +3 oxidation.
In ${Cr}^{3 +}$ always inner d-orbital i.e., $(n - 1) d$ is used for hybridisation and forms $d^{2} {sp}^{3}$ hybridisation with octahedral geometry.
original image
$[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ contains three unpaired electrons therefore it shows paramagnetic behavior.
$[Cr {({NH}_{3})}_{6}] \cdot {Cl}_{3} + {AgNO}_{3} ⟶$
$(ppt.) 3$

AIEEE 2011

COORDINATION COMPOUNDS

274253 The effective atomic number for ${[Rh {(H_{2} O)}_{6}]}^{3 +}$ (atomic number of $R h$ is 45 ) is

1 42

2 45

3 48

4 54

Explanation:

(D) : Atomic Number of $Rh = 45$
Oxidation state of $Rh = + 3$
No. of donor site $= 6$
EAN of ${[Rh {(H_{2} O)}_{6}]}^{3 +} =$ Atomic no. - Oxidation $+ 2 \times$
Number of donor atom attached with metal. $= 45 - 3 + 2 \times 6 = 54$

J and K CET-(2012)

COORDINATION COMPOUNDS

274257 The metal ion in complex $A$ has EAN identical to the atomic number of krypton $A$ is
(At. $No$ of $Cr = 24, Fe = 26, Pd = 46$ )

1

[pd {({NH}_{3})}_{6}] {Cl}_{4}

2

[Cr {({NH}_{3})}_{5} Cl] {SO}_{4}

3

{Na}_{4} [Fe (CN)_{6}]

4

K_{3} [Fe (CN)_{6}]

Explanation:

(C) : Atomic number of krypton $= 36$
E.AN of complex A = Atomic No of Krypton
Atomic number of $Fe = 26$
Oxidation state of $Fe$ in ${Na}_{4} [Fe (CN)_{6}]$ complex
$x + 6 (- 1) = - 4$
$x = 2$
Number of donor sites attached with $Fe = 6$
E.AN of ${[Fe (CN)_{6}]}^{4 -} = Z - (ON) + 2 (CN)$
Where, $ON =$ Oxidation no.
$CN =$ coordination number
$Z =$ atomic number
$= 26 - 2 + 2 (6)$
$= 36$
Hence, complex A is ${Na}_{4} [Fe (CN)]_{6}$

J and K CET-(2006)

COORDINATION COMPOUNDS

274189 What is the value of $x$ on the ${[Ni (CN)_{4}]}^{x}$ complex ion?

1 +2

2 -2

3 0 (zero)

4 +4

Explanation:

(B) : Nickel forms square planar complex with coordination number 4 and strong field ligand with oxidation state is +2 .
${[Ni (CN)_{4}]}^{x}$
$2 + 4 \times (- 1) = x$
$2 - 4 = x$
$x = - 2$
The complex is ${[Ni (CN)_{4}]}^{2 -}$

Manipal-2020

COORDINATION COMPOUNDS

274165 ${Fe}^{3 +}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

1

{[Fe {(H_{2} O)}_{6}]}_{2} [F_{e} (CN)_{6}]

2

{Fe}_{2} {[Fe (CN)_{6}]}_{2}

3

{Fe}_{3} {[Fe (OH)_{2} (CN)_{4}]}_{2}

4

{Fe}_{4} {[Fe (CN)_{6}]}_{3}

Explanation:

(D) : The reaction is-
$4 {Fe}^{3 +} + 3 [Fe (CN)_{6}] ⟶ {Fe}_{4} {[Fe (CN)_{6}]}_{3}$
$Potassium$
$Ferrocynide$
$(PrussianBlue)$

JEE Main 27.07.2022

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