QBManager

COORDINATION COMPOUNDS

28696 Which of the following is paramagnetic

1

[N i {(C O)}_{4}]

2

[C o (N H_{3})_{6}]^{3 +}

3

{[N i {(C N)}_{4}]}^{2 -}

4

{[N i C l_{4}]}^{2 -}

Explanation:

In $N i (C O)_{4},$ the oxidation state of $N i$ is zero whereas in ${[N i C l_{4}]}^{2 -}$ , the oxidation state of $N i$ is. The presence of $C O$ ligand, which is a strong ligand, can pair all electrons in $N i (C O)_{4}$ and thus it is diamagnetic in nature but $C l^{-}$ is a weak ligand and is unable to pair up the unpaired electrons and thus ${[N i C l_{4}]}^{2 -}$ is paramagnetic in nature.

COORDINATION COMPOUNDS

28697 Which one of the following shows maximum paramagnetic character

1

{[F e {(C N)}_{6}]}^{3 -}

2

{[F e {(C N)}_{6}]}^{4 -}

3

{[C r {(H_{2} O)}_{6}]}^{3 +}

4

{[C u {(H_{2} O)}_{6}]}^{2 +}

Explanation:

(i) ${[F e (C N)_{6}]}^{3 -}$
Here, oxidation state of $F e = + 3$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{5} 4 s^{0}$
Number of unpaired electrons $= 5$
(ii) ${[F e (C N)_{6}]}^{4 -}$
Here, oxidation state of $F e = + 2$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{6} 4 s^{0}$
Number of unpaired electrons $= 4$
(iii) ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
Here, oxidation state of $C r = + 3$
Electronic configuration of $C r^{3 +} = [A r] 3 d^{3} 4 s^{0}$
Number of unpaired electrons $= 3$
(iv) ${[C u {(H_{2} O)}_{6}]}^{2 +}$
Here, oxidation state of $C u = + 2$
Electronic configuration of $C u^{2 +} = [A r] 3 d^{9} 4 s^{0}$
Number of unpaired electrons $= 1$
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, ${[F e (C N)_{6}]}^{3 -}$ has maximum paramagnetic character.

COORDINATION COMPOUNDS

28698 The complexes $[C o (N H_{3})_{6}] [C r (C_{2} O_{4})_{3}]$ and $[C r (N H_{3})_{6}] [C o (C_{2} O_{4})_{3}]$

1 Linkage isomerism

2 Geometrical isomerism

3 Coordination isomerism

4 Ionisation isomerism

Explanation:

(c) Co-ordination isomerism is possible when both $+ v e$ and $- v e$ ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and the anion.

COORDINATION COMPOUNDS

28699 Which of the following exhibits highest molar conductivity

1

[C o {(N H_{3})}_{6}] C l_{3}

2

[C o {(N H_{3})}_{5} C l] C l_{2}

3

[C o {(N H_{3})}_{4} C l_{2}] C l

4

[C o {(N H_{3})}_{3} C l_{3}]

COORDINATION COMPOUNDS

28696 Which of the following is paramagnetic

1

[N i {(C O)}_{4}]

2

[C o (N H_{3})_{6}]^{3 +}

3

{[N i {(C N)}_{4}]}^{2 -}

4

{[N i C l_{4}]}^{2 -}

Explanation:

In $N i (C O)_{4},$ the oxidation state of $N i$ is zero whereas in ${[N i C l_{4}]}^{2 -}$ , the oxidation state of $N i$ is. The presence of $C O$ ligand, which is a strong ligand, can pair all electrons in $N i (C O)_{4}$ and thus it is diamagnetic in nature but $C l^{-}$ is a weak ligand and is unable to pair up the unpaired electrons and thus ${[N i C l_{4}]}^{2 -}$ is paramagnetic in nature.

COORDINATION COMPOUNDS

28697 Which one of the following shows maximum paramagnetic character

1

{[F e {(C N)}_{6}]}^{3 -}

2

{[F e {(C N)}_{6}]}^{4 -}

3

{[C r {(H_{2} O)}_{6}]}^{3 +}

4

{[C u {(H_{2} O)}_{6}]}^{2 +}

Explanation:

(i) ${[F e (C N)_{6}]}^{3 -}$
Here, oxidation state of $F e = + 3$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{5} 4 s^{0}$
Number of unpaired electrons $= 5$
(ii) ${[F e (C N)_{6}]}^{4 -}$
Here, oxidation state of $F e = + 2$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{6} 4 s^{0}$
Number of unpaired electrons $= 4$
(iii) ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
Here, oxidation state of $C r = + 3$
Electronic configuration of $C r^{3 +} = [A r] 3 d^{3} 4 s^{0}$
Number of unpaired electrons $= 3$
(iv) ${[C u {(H_{2} O)}_{6}]}^{2 +}$
Here, oxidation state of $C u = + 2$
Electronic configuration of $C u^{2 +} = [A r] 3 d^{9} 4 s^{0}$
Number of unpaired electrons $= 1$
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, ${[F e (C N)_{6}]}^{3 -}$ has maximum paramagnetic character.

COORDINATION COMPOUNDS

28698 The complexes $[C o (N H_{3})_{6}] [C r (C_{2} O_{4})_{3}]$ and $[C r (N H_{3})_{6}] [C o (C_{2} O_{4})_{3}]$

1 Linkage isomerism

2 Geometrical isomerism

3 Coordination isomerism

4 Ionisation isomerism

Explanation:

(c) Co-ordination isomerism is possible when both $+ v e$ and $- v e$ ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and the anion.

COORDINATION COMPOUNDS

28699 Which of the following exhibits highest molar conductivity

1

[C o {(N H_{3})}_{6}] C l_{3}

2

[C o {(N H_{3})}_{5} C l] C l_{2}

3

[C o {(N H_{3})}_{4} C l_{2}] C l

4

[C o {(N H_{3})}_{3} C l_{3}]

COORDINATION COMPOUNDS

28696 Which of the following is paramagnetic

1

[N i {(C O)}_{4}]

2

[C o (N H_{3})_{6}]^{3 +}

3

{[N i {(C N)}_{4}]}^{2 -}

4

{[N i C l_{4}]}^{2 -}

Explanation:

In $N i (C O)_{4},$ the oxidation state of $N i$ is zero whereas in ${[N i C l_{4}]}^{2 -}$ , the oxidation state of $N i$ is. The presence of $C O$ ligand, which is a strong ligand, can pair all electrons in $N i (C O)_{4}$ and thus it is diamagnetic in nature but $C l^{-}$ is a weak ligand and is unable to pair up the unpaired electrons and thus ${[N i C l_{4}]}^{2 -}$ is paramagnetic in nature.

COORDINATION COMPOUNDS

28697 Which one of the following shows maximum paramagnetic character

1

{[F e {(C N)}_{6}]}^{3 -}

2

{[F e {(C N)}_{6}]}^{4 -}

3

{[C r {(H_{2} O)}_{6}]}^{3 +}

4

{[C u {(H_{2} O)}_{6}]}^{2 +}

Explanation:

(i) ${[F e (C N)_{6}]}^{3 -}$
Here, oxidation state of $F e = + 3$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{5} 4 s^{0}$
Number of unpaired electrons $= 5$
(ii) ${[F e (C N)_{6}]}^{4 -}$
Here, oxidation state of $F e = + 2$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{6} 4 s^{0}$
Number of unpaired electrons $= 4$
(iii) ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
Here, oxidation state of $C r = + 3$
Electronic configuration of $C r^{3 +} = [A r] 3 d^{3} 4 s^{0}$
Number of unpaired electrons $= 3$
(iv) ${[C u {(H_{2} O)}_{6}]}^{2 +}$
Here, oxidation state of $C u = + 2$
Electronic configuration of $C u^{2 +} = [A r] 3 d^{9} 4 s^{0}$
Number of unpaired electrons $= 1$
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, ${[F e (C N)_{6}]}^{3 -}$ has maximum paramagnetic character.

COORDINATION COMPOUNDS

28698 The complexes $[C o (N H_{3})_{6}] [C r (C_{2} O_{4})_{3}]$ and $[C r (N H_{3})_{6}] [C o (C_{2} O_{4})_{3}]$

1 Linkage isomerism

2 Geometrical isomerism

3 Coordination isomerism

4 Ionisation isomerism

Explanation:

(c) Co-ordination isomerism is possible when both $+ v e$ and $- v e$ ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and the anion.

COORDINATION COMPOUNDS

28699 Which of the following exhibits highest molar conductivity

1

[C o {(N H_{3})}_{6}] C l_{3}

2

[C o {(N H_{3})}_{5} C l] C l_{2}

3

[C o {(N H_{3})}_{4} C l_{2}] C l

4

[C o {(N H_{3})}_{3} C l_{3}]

COORDINATION COMPOUNDS

28696 Which of the following is paramagnetic

1

[N i {(C O)}_{4}]

2

[C o (N H_{3})_{6}]^{3 +}

3

{[N i {(C N)}_{4}]}^{2 -}

4

{[N i C l_{4}]}^{2 -}

Explanation:

In $N i (C O)_{4},$ the oxidation state of $N i$ is zero whereas in ${[N i C l_{4}]}^{2 -}$ , the oxidation state of $N i$ is. The presence of $C O$ ligand, which is a strong ligand, can pair all electrons in $N i (C O)_{4}$ and thus it is diamagnetic in nature but $C l^{-}$ is a weak ligand and is unable to pair up the unpaired electrons and thus ${[N i C l_{4}]}^{2 -}$ is paramagnetic in nature.

COORDINATION COMPOUNDS

28697 Which one of the following shows maximum paramagnetic character

1

{[F e {(C N)}_{6}]}^{3 -}

2

{[F e {(C N)}_{6}]}^{4 -}

3

{[C r {(H_{2} O)}_{6}]}^{3 +}

4

{[C u {(H_{2} O)}_{6}]}^{2 +}

Explanation:

(i) ${[F e (C N)_{6}]}^{3 -}$
Here, oxidation state of $F e = + 3$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{5} 4 s^{0}$
Number of unpaired electrons $= 5$
(ii) ${[F e (C N)_{6}]}^{4 -}$
Here, oxidation state of $F e = + 2$
Electronic configuration of $F e$ (ground state) $= [A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $= [A r] 3 d^{6} 4 s^{0}$
Number of unpaired electrons $= 4$
(iii) ${[Cr {(H_{2} O)}_{6}]}^{3 +}$
Here, oxidation state of $C r = + 3$
Electronic configuration of $C r^{3 +} = [A r] 3 d^{3} 4 s^{0}$
Number of unpaired electrons $= 3$
(iv) ${[C u {(H_{2} O)}_{6}]}^{2 +}$
Here, oxidation state of $C u = + 2$
Electronic configuration of $C u^{2 +} = [A r] 3 d^{9} 4 s^{0}$
Number of unpaired electrons $= 1$
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, ${[F e (C N)_{6}]}^{3 -}$ has maximum paramagnetic character.

COORDINATION COMPOUNDS

28698 The complexes $[C o (N H_{3})_{6}] [C r (C_{2} O_{4})_{3}]$ and $[C r (N H_{3})_{6}] [C o (C_{2} O_{4})_{3}]$

1 Linkage isomerism

2 Geometrical isomerism

3 Coordination isomerism

4 Ionisation isomerism

Explanation:

(c) Co-ordination isomerism is possible when both $+ v e$ and $- v e$ ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and the anion.

COORDINATION COMPOUNDS

28699 Which of the following exhibits highest molar conductivity

1

[C o {(N H_{3})}_{6}] C l_{3}

2

[C o {(N H_{3})}_{5} C l] C l_{2}

3

[C o {(N H_{3})}_{4} C l_{2}] C l

4

[C o {(N H_{3})}_{3} C l_{3}]

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