QBManager

COORDINATION COMPOUNDS

274187 For octahedral $Mn (II)$ and tetrahedral $Ni (II)$ complexes, consider the following statements,
(I) Both the complexes can be high spin.
(II) $Ni$
(II) complex can very rarely be low spin.
(III) With strong field ligands, Mn(II) complexes can be low spin.
(IV) Aqueous solution of $Mn$ (II) ions is yellow in colour.
The correct statements is

1 (I) and (II) only

2 (II), (III) and (IV) only

3 (I), (II) and (III) only

4 (I), (III) and (IV) only

Explanation:

(C) : Tetrahedral complexes mostly form high spin complexes because metal form bonds with weak field ligand. Hence, Ni (II) tetrahedral complexes can be high spin.
$Mn (II)$ Octahedral complex form high spin when it form bonds with weak field ligand due to which outer dorbital's are used.
$Mn (II) : d^{5}$

Mn (II) complexes with strong field ligands form low spin complex i.e., pairing of d-electron occurs, and inner d-orbitals are used for hybridization ${Mn}^{2 +} : d^{5}$ (low spin)
original image
Tetrahedral complexes of NI (II) mostly forms high spin but it can very rare to form low spin complex. Aqueous solution of Mn (II) ions is pink in colour. Hence, correct statement is (I), (II) and (III).

454. Complex(A) has a composition of

H_{12} O_{6} {Cl}_{3} Cr

. If the complex on treatment with conc.

H_{2} {SO}_{4}

loses

13.5 %

of its original mass

COORDINATION COMPOUNDS

274195 Which among the following is a neutral complex?

1

[Fe {(H_{2} O)}_{6}] {Cl}_{3}

2

[Ni {({NH}_{3})}_{6}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

4

K [Ag (CN)_{2}]

Explanation:

(C) : Complexes which does not contain any charge or the coordination sphere is at as neutral ligand.
(a) $[Fe {(H_{2} O)}_{6}] {Cl}_{3} ⟶ {[Fe {(H_{2} O)}_{6}]}^{3 +} + 3 {Cl}^{-}$
(b) $[Ni {({NH}_{3})}_{6}] {Cl}_{2} ⟶ {[Ni {({NH}_{3})}_{6}]}^{2 +} + 2 {Cl}^{-}$
(c) $pt [{({NH}_{3})}_{2} {Cl}_{2}] ⟶$ No charge or coordiantion sphere
$(d) K [Ag (CN)_{2}] ⟶ K^{+} + {[Ag (CN)_{2}]}^{-}$
Hence, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ complex is a neutral complex.

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274297 The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is

Ignore pairing energy

1

{[Co (phen)_{3}]}^{2 +}

2

{[Ni (phen)_{3}]}^{2 +}

3

{[Zn (phen)_{3}]}^{2 +}

4

{[Fe (phen)_{3}]}^{2 +}

Explanation:

(D) : Here, phenanthroline is a strong field ligand.
${[Fe (phen)_{3}]}^{2 +} \overset{- e^{-}}{⟶} {[Fe (phen)_{3}]}^{+ 3}$
Oxidation state $= + 2$ Oxidation state $= + 3$
E.C. $= [Ar] 3 d^{6}$

$d^{6} - <↑↓ e^{g}$
CFSE $= [- 0.4 \times 6 + 0.6 \times 0] Δ_{0}$ CFSE $= [- 0.4 \times 5 + 0.6 \times 0] Δ_{0}$ CFSE $= - 2.4 Δ_{0}$
CFSE $= - 2.0 Δ_{0}$
Hence, CFSE decreases after the oxidation.

JEE Main 2019

COORDINATION COMPOUNDS

274255 Ethylenediaminetetraacetate ion is a

1 unidentate ligand

2 bidentate ligand

3 pentadentate ligand

4 hexadentate ligand.

Explanation:

(D) : Ethylene diaamine tetra acetate ion $({EDTA}^{4})$ is a hexadentate ligand i.e, having six donor atoms.
original image

J and K CET-(2013)

COORDINATION COMPOUNDS

274249 $[EDTA]^{4 -}$ is a:

1 monodentate ligand

2 bidentate ligand

3 quadridentate ligand

4 hexadentate ligand

Explanation:

(D) : $[EDTA]^{4 -}$ is a hexadentate ligand having six donor sites.
original image

BITSAT-2009

COORDINATION COMPOUNDS

274187 For octahedral $Mn (II)$ and tetrahedral $Ni (II)$ complexes, consider the following statements,
(I) Both the complexes can be high spin.
(II) $Ni$
(II) complex can very rarely be low spin.
(III) With strong field ligands, Mn(II) complexes can be low spin.
(IV) Aqueous solution of $Mn$ (II) ions is yellow in colour.
The correct statements is

1 (I) and (II) only

2 (II), (III) and (IV) only

3 (I), (II) and (III) only

4 (I), (III) and (IV) only

Explanation:

(C) : Tetrahedral complexes mostly form high spin complexes because metal form bonds with weak field ligand. Hence, Ni (II) tetrahedral complexes can be high spin.
$Mn (II)$ Octahedral complex form high spin when it form bonds with weak field ligand due to which outer dorbital's are used.
$Mn (II) : d^{5}$

Mn (II) complexes with strong field ligands form low spin complex i.e., pairing of d-electron occurs, and inner d-orbitals are used for hybridization ${Mn}^{2 +} : d^{5}$ (low spin)
original image
Tetrahedral complexes of NI (II) mostly forms high spin but it can very rare to form low spin complex. Aqueous solution of Mn (II) ions is pink in colour. Hence, correct statement is (I), (II) and (III).

454. Complex(A) has a composition of

H_{12} O_{6} {Cl}_{3} Cr

. If the complex on treatment with conc.

H_{2} {SO}_{4}

loses

13.5 %

of its original mass

COORDINATION COMPOUNDS

274195 Which among the following is a neutral complex?

1

[Fe {(H_{2} O)}_{6}] {Cl}_{3}

2

[Ni {({NH}_{3})}_{6}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

4

K [Ag (CN)_{2}]

Explanation:

(C) : Complexes which does not contain any charge or the coordination sphere is at as neutral ligand.
(a) $[Fe {(H_{2} O)}_{6}] {Cl}_{3} ⟶ {[Fe {(H_{2} O)}_{6}]}^{3 +} + 3 {Cl}^{-}$
(b) $[Ni {({NH}_{3})}_{6}] {Cl}_{2} ⟶ {[Ni {({NH}_{3})}_{6}]}^{2 +} + 2 {Cl}^{-}$
(c) $pt [{({NH}_{3})}_{2} {Cl}_{2}] ⟶$ No charge or coordiantion sphere
$(d) K [Ag (CN)_{2}] ⟶ K^{+} + {[Ag (CN)_{2}]}^{-}$
Hence, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ complex is a neutral complex.

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274297 The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is

Ignore pairing energy

1

{[Co (phen)_{3}]}^{2 +}

2

{[Ni (phen)_{3}]}^{2 +}

3

{[Zn (phen)_{3}]}^{2 +}

4

{[Fe (phen)_{3}]}^{2 +}

Explanation:

(D) : Here, phenanthroline is a strong field ligand.
${[Fe (phen)_{3}]}^{2 +} \overset{- e^{-}}{⟶} {[Fe (phen)_{3}]}^{+ 3}$
Oxidation state $= + 2$ Oxidation state $= + 3$
E.C. $= [Ar] 3 d^{6}$

$d^{6} - <↑↓ e^{g}$
CFSE $= [- 0.4 \times 6 + 0.6 \times 0] Δ_{0}$ CFSE $= [- 0.4 \times 5 + 0.6 \times 0] Δ_{0}$ CFSE $= - 2.4 Δ_{0}$
CFSE $= - 2.0 Δ_{0}$
Hence, CFSE decreases after the oxidation.

JEE Main 2019

COORDINATION COMPOUNDS

274255 Ethylenediaminetetraacetate ion is a

1 unidentate ligand

2 bidentate ligand

3 pentadentate ligand

4 hexadentate ligand.

Explanation:

(D) : Ethylene diaamine tetra acetate ion $({EDTA}^{4})$ is a hexadentate ligand i.e, having six donor atoms.
original image

J and K CET-(2013)

COORDINATION COMPOUNDS

274249 $[EDTA]^{4 -}$ is a:

1 monodentate ligand

2 bidentate ligand

3 quadridentate ligand

4 hexadentate ligand

Explanation:

(D) : $[EDTA]^{4 -}$ is a hexadentate ligand having six donor sites.
original image

BITSAT-2009

COORDINATION COMPOUNDS

274187 For octahedral $Mn (II)$ and tetrahedral $Ni (II)$ complexes, consider the following statements,
(I) Both the complexes can be high spin.
(II) $Ni$
(II) complex can very rarely be low spin.
(III) With strong field ligands, Mn(II) complexes can be low spin.
(IV) Aqueous solution of $Mn$ (II) ions is yellow in colour.
The correct statements is

1 (I) and (II) only

2 (II), (III) and (IV) only

3 (I), (II) and (III) only

4 (I), (III) and (IV) only

Explanation:

(C) : Tetrahedral complexes mostly form high spin complexes because metal form bonds with weak field ligand. Hence, Ni (II) tetrahedral complexes can be high spin.
$Mn (II)$ Octahedral complex form high spin when it form bonds with weak field ligand due to which outer dorbital's are used.
$Mn (II) : d^{5}$

Mn (II) complexes with strong field ligands form low spin complex i.e., pairing of d-electron occurs, and inner d-orbitals are used for hybridization ${Mn}^{2 +} : d^{5}$ (low spin)
original image
Tetrahedral complexes of NI (II) mostly forms high spin but it can very rare to form low spin complex. Aqueous solution of Mn (II) ions is pink in colour. Hence, correct statement is (I), (II) and (III).

454. Complex(A) has a composition of

H_{12} O_{6} {Cl}_{3} Cr

. If the complex on treatment with conc.

H_{2} {SO}_{4}

loses

13.5 %

of its original mass

COORDINATION COMPOUNDS

274195 Which among the following is a neutral complex?

1

[Fe {(H_{2} O)}_{6}] {Cl}_{3}

2

[Ni {({NH}_{3})}_{6}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

4

K [Ag (CN)_{2}]

Explanation:

(C) : Complexes which does not contain any charge or the coordination sphere is at as neutral ligand.
(a) $[Fe {(H_{2} O)}_{6}] {Cl}_{3} ⟶ {[Fe {(H_{2} O)}_{6}]}^{3 +} + 3 {Cl}^{-}$
(b) $[Ni {({NH}_{3})}_{6}] {Cl}_{2} ⟶ {[Ni {({NH}_{3})}_{6}]}^{2 +} + 2 {Cl}^{-}$
(c) $pt [{({NH}_{3})}_{2} {Cl}_{2}] ⟶$ No charge or coordiantion sphere
$(d) K [Ag (CN)_{2}] ⟶ K^{+} + {[Ag (CN)_{2}]}^{-}$
Hence, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ complex is a neutral complex.

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274297 The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is

Ignore pairing energy

1

{[Co (phen)_{3}]}^{2 +}

2

{[Ni (phen)_{3}]}^{2 +}

3

{[Zn (phen)_{3}]}^{2 +}

4

{[Fe (phen)_{3}]}^{2 +}

Explanation:

(D) : Here, phenanthroline is a strong field ligand.
${[Fe (phen)_{3}]}^{2 +} \overset{- e^{-}}{⟶} {[Fe (phen)_{3}]}^{+ 3}$
Oxidation state $= + 2$ Oxidation state $= + 3$
E.C. $= [Ar] 3 d^{6}$

$d^{6} - <↑↓ e^{g}$
CFSE $= [- 0.4 \times 6 + 0.6 \times 0] Δ_{0}$ CFSE $= [- 0.4 \times 5 + 0.6 \times 0] Δ_{0}$ CFSE $= - 2.4 Δ_{0}$
CFSE $= - 2.0 Δ_{0}$
Hence, CFSE decreases after the oxidation.

JEE Main 2019

COORDINATION COMPOUNDS

274255 Ethylenediaminetetraacetate ion is a

1 unidentate ligand

2 bidentate ligand

3 pentadentate ligand

4 hexadentate ligand.

Explanation:

(D) : Ethylene diaamine tetra acetate ion $({EDTA}^{4})$ is a hexadentate ligand i.e, having six donor atoms.
original image

J and K CET-(2013)

COORDINATION COMPOUNDS

274249 $[EDTA]^{4 -}$ is a:

1 monodentate ligand

2 bidentate ligand

3 quadridentate ligand

4 hexadentate ligand

Explanation:

(D) : $[EDTA]^{4 -}$ is a hexadentate ligand having six donor sites.
original image

BITSAT-2009

COORDINATION COMPOUNDS

274187 For octahedral $Mn (II)$ and tetrahedral $Ni (II)$ complexes, consider the following statements,
(I) Both the complexes can be high spin.
(II) $Ni$
(II) complex can very rarely be low spin.
(III) With strong field ligands, Mn(II) complexes can be low spin.
(IV) Aqueous solution of $Mn$ (II) ions is yellow in colour.
The correct statements is

1 (I) and (II) only

2 (II), (III) and (IV) only

3 (I), (II) and (III) only

4 (I), (III) and (IV) only

Explanation:

(C) : Tetrahedral complexes mostly form high spin complexes because metal form bonds with weak field ligand. Hence, Ni (II) tetrahedral complexes can be high spin.
$Mn (II)$ Octahedral complex form high spin when it form bonds with weak field ligand due to which outer dorbital's are used.
$Mn (II) : d^{5}$

Mn (II) complexes with strong field ligands form low spin complex i.e., pairing of d-electron occurs, and inner d-orbitals are used for hybridization ${Mn}^{2 +} : d^{5}$ (low spin)
original image
Tetrahedral complexes of NI (II) mostly forms high spin but it can very rare to form low spin complex. Aqueous solution of Mn (II) ions is pink in colour. Hence, correct statement is (I), (II) and (III).

454. Complex(A) has a composition of

H_{12} O_{6} {Cl}_{3} Cr

. If the complex on treatment with conc.

H_{2} {SO}_{4}

loses

13.5 %

of its original mass

COORDINATION COMPOUNDS

274195 Which among the following is a neutral complex?

1

[Fe {(H_{2} O)}_{6}] {Cl}_{3}

2

[Ni {({NH}_{3})}_{6}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

4

K [Ag (CN)_{2}]

Explanation:

(C) : Complexes which does not contain any charge or the coordination sphere is at as neutral ligand.
(a) $[Fe {(H_{2} O)}_{6}] {Cl}_{3} ⟶ {[Fe {(H_{2} O)}_{6}]}^{3 +} + 3 {Cl}^{-}$
(b) $[Ni {({NH}_{3})}_{6}] {Cl}_{2} ⟶ {[Ni {({NH}_{3})}_{6}]}^{2 +} + 2 {Cl}^{-}$
(c) $pt [{({NH}_{3})}_{2} {Cl}_{2}] ⟶$ No charge or coordiantion sphere
$(d) K [Ag (CN)_{2}] ⟶ K^{+} + {[Ag (CN)_{2}]}^{-}$
Hence, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ complex is a neutral complex.

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274297 The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is

Ignore pairing energy

1

{[Co (phen)_{3}]}^{2 +}

2

{[Ni (phen)_{3}]}^{2 +}

3

{[Zn (phen)_{3}]}^{2 +}

4

{[Fe (phen)_{3}]}^{2 +}

Explanation:

(D) : Here, phenanthroline is a strong field ligand.
${[Fe (phen)_{3}]}^{2 +} \overset{- e^{-}}{⟶} {[Fe (phen)_{3}]}^{+ 3}$
Oxidation state $= + 2$ Oxidation state $= + 3$
E.C. $= [Ar] 3 d^{6}$

$d^{6} - <↑↓ e^{g}$
CFSE $= [- 0.4 \times 6 + 0.6 \times 0] Δ_{0}$ CFSE $= [- 0.4 \times 5 + 0.6 \times 0] Δ_{0}$ CFSE $= - 2.4 Δ_{0}$
CFSE $= - 2.0 Δ_{0}$
Hence, CFSE decreases after the oxidation.

JEE Main 2019

COORDINATION COMPOUNDS

274255 Ethylenediaminetetraacetate ion is a

1 unidentate ligand

2 bidentate ligand

3 pentadentate ligand

4 hexadentate ligand.

Explanation:

(D) : Ethylene diaamine tetra acetate ion $({EDTA}^{4})$ is a hexadentate ligand i.e, having six donor atoms.
original image

J and K CET-(2013)

COORDINATION COMPOUNDS

274249 $[EDTA]^{4 -}$ is a:

1 monodentate ligand

2 bidentate ligand

3 quadridentate ligand

4 hexadentate ligand

Explanation:

(D) : $[EDTA]^{4 -}$ is a hexadentate ligand having six donor sites.
original image

BITSAT-2009

COORDINATION COMPOUNDS

274187 For octahedral $Mn (II)$ and tetrahedral $Ni (II)$ complexes, consider the following statements,
(I) Both the complexes can be high spin.
(II) $Ni$
(II) complex can very rarely be low spin.
(III) With strong field ligands, Mn(II) complexes can be low spin.
(IV) Aqueous solution of $Mn$ (II) ions is yellow in colour.
The correct statements is

1 (I) and (II) only

2 (II), (III) and (IV) only

3 (I), (II) and (III) only

4 (I), (III) and (IV) only

Explanation:

(C) : Tetrahedral complexes mostly form high spin complexes because metal form bonds with weak field ligand. Hence, Ni (II) tetrahedral complexes can be high spin.
$Mn (II)$ Octahedral complex form high spin when it form bonds with weak field ligand due to which outer dorbital's are used.
$Mn (II) : d^{5}$

Mn (II) complexes with strong field ligands form low spin complex i.e., pairing of d-electron occurs, and inner d-orbitals are used for hybridization ${Mn}^{2 +} : d^{5}$ (low spin)
original image
Tetrahedral complexes of NI (II) mostly forms high spin but it can very rare to form low spin complex. Aqueous solution of Mn (II) ions is pink in colour. Hence, correct statement is (I), (II) and (III).

454. Complex(A) has a composition of

H_{12} O_{6} {Cl}_{3} Cr

. If the complex on treatment with conc.

H_{2} {SO}_{4}

loses

13.5 %

of its original mass

COORDINATION COMPOUNDS

274195 Which among the following is a neutral complex?

1

[Fe {(H_{2} O)}_{6}] {Cl}_{3}

2

[Ni {({NH}_{3})}_{6}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

4

K [Ag (CN)_{2}]

Explanation:

(C) : Complexes which does not contain any charge or the coordination sphere is at as neutral ligand.
(a) $[Fe {(H_{2} O)}_{6}] {Cl}_{3} ⟶ {[Fe {(H_{2} O)}_{6}]}^{3 +} + 3 {Cl}^{-}$
(b) $[Ni {({NH}_{3})}_{6}] {Cl}_{2} ⟶ {[Ni {({NH}_{3})}_{6}]}^{2 +} + 2 {Cl}^{-}$
(c) $pt [{({NH}_{3})}_{2} {Cl}_{2}] ⟶$ No charge or coordiantion sphere
$(d) K [Ag (CN)_{2}] ⟶ K^{+} + {[Ag (CN)_{2}]}^{-}$
Hence, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ complex is a neutral complex.

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274297 The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is

Ignore pairing energy

1

{[Co (phen)_{3}]}^{2 +}

2

{[Ni (phen)_{3}]}^{2 +}

3

{[Zn (phen)_{3}]}^{2 +}

4

{[Fe (phen)_{3}]}^{2 +}

Explanation:

(D) : Here, phenanthroline is a strong field ligand.
${[Fe (phen)_{3}]}^{2 +} \overset{- e^{-}}{⟶} {[Fe (phen)_{3}]}^{+ 3}$
Oxidation state $= + 2$ Oxidation state $= + 3$
E.C. $= [Ar] 3 d^{6}$

$d^{6} - <↑↓ e^{g}$
CFSE $= [- 0.4 \times 6 + 0.6 \times 0] Δ_{0}$ CFSE $= [- 0.4 \times 5 + 0.6 \times 0] Δ_{0}$ CFSE $= - 2.4 Δ_{0}$
CFSE $= - 2.0 Δ_{0}$
Hence, CFSE decreases after the oxidation.

JEE Main 2019

COORDINATION COMPOUNDS

274255 Ethylenediaminetetraacetate ion is a

1 unidentate ligand

2 bidentate ligand

3 pentadentate ligand

4 hexadentate ligand.

Explanation:

(D) : Ethylene diaamine tetra acetate ion $({EDTA}^{4})$ is a hexadentate ligand i.e, having six donor atoms.
original image

J and K CET-(2013)

COORDINATION COMPOUNDS

274249 $[EDTA]^{4 -}$ is a:

1 monodentate ligand

2 bidentate ligand

3 quadridentate ligand

4 hexadentate ligand

Explanation:

(D) : $[EDTA]^{4 -}$ is a hexadentate ligand having six donor sites.
original image

BITSAT-2009

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