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CHXII09:COORDINATION COMPOUNDS

322294 Among the following ions which one has the highest paramagnetism?

1

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

2

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

3

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

4

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

Explanation:

Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of
${Cr}^{3 +} (3 d^{3}), {Fe}^{2 +} (3 d^{6}), {Cu}^{2 +} (3 d^{9})$ and
${Zn}^{2 +} (3 d^{10})$ are outer orbital complex ions.
Hence ${Fe}^{2 +}$ has maximum unpaired electrons.

CHXII09:COORDINATION COMPOUNDS

322295 The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ and ${[Fe (CN)_{6}]}^{3 -}$ complexes respectively is

1 5.92 B.M. and 1.732 B.M.

2 3.87 B.M. and 1.732 B.M.

3 4.89 B.M. and 6.92 B.M.

4 6.92 B.M. in both.

Explanation:

In both the complexes, Fe is in $+ 3$ oxidation state. ${CN}^{-}$ is a strong field ligand, but $H_{2} O$ is a weak field ligand.
supporting img

There are 5 unpaired electrons. Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)} = \sqrt{35} = 5.92$ B.M.

There is only one unpaired electron.
Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, option (1) is correct.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322296 The $d$ -electronic configuration of ${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +} {Ni}^{2 +}$ are $3 d^{4}, 3 d^{5}, 3 d^{6}$ and $3 d^{8}$ respectively.
Which of the following complex will show minimum paramagnetic behaviour?

1

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

2

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

3

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

4

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

Explanation:

${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +}$ and ${Ni}^{2 +}$ have $4, 5, 4$ and 2 unpaired electrons. No of unpaired electrons $α$ paramagnetism.

CHXII09:COORDINATION COMPOUNDS

322297 Match Column I with Column II and choose the correct combination from the options given.
supporting img

1

A - Q, B - R, C - S, D - T, E - P

2

A - T, B - S, C - R, D - P, E - Q

3

A - U, B - S, C - P, D - Q, E - R

4

A - P, B - Q, C - R, D - S, E - T

Explanation:

(A) ${[Ni (CN)_{4}]}^{- 2} - {dsp}^{2}$ hybridization square planar – Diamagnetic.
(B) Chlorophyll has ${Mg}^{+ 2}$ .
(C) Ziegler – Natta catalyst contains ${Ti}^{+ 4}$ .
(D) ${[{NiCl}_{4}]}^{- 2} - {sp}^{3}$ hybridization – tetrahedral – paramagnetic.
(E) Deoxy haemoglobin – non planar.

CHXII09:COORDINATION COMPOUNDS

322294 Among the following ions which one has the highest paramagnetism?

1

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

2

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

3

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

4

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

Explanation:

Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of
${Cr}^{3 +} (3 d^{3}), {Fe}^{2 +} (3 d^{6}), {Cu}^{2 +} (3 d^{9})$ and
${Zn}^{2 +} (3 d^{10})$ are outer orbital complex ions.
Hence ${Fe}^{2 +}$ has maximum unpaired electrons.

CHXII09:COORDINATION COMPOUNDS

322295 The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ and ${[Fe (CN)_{6}]}^{3 -}$ complexes respectively is

1 5.92 B.M. and 1.732 B.M.

2 3.87 B.M. and 1.732 B.M.

3 4.89 B.M. and 6.92 B.M.

4 6.92 B.M. in both.

Explanation:

In both the complexes, Fe is in $+ 3$ oxidation state. ${CN}^{-}$ is a strong field ligand, but $H_{2} O$ is a weak field ligand.
supporting img

There are 5 unpaired electrons. Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)} = \sqrt{35} = 5.92$ B.M.

There is only one unpaired electron.
Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, option (1) is correct.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322296 The $d$ -electronic configuration of ${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +} {Ni}^{2 +}$ are $3 d^{4}, 3 d^{5}, 3 d^{6}$ and $3 d^{8}$ respectively.
Which of the following complex will show minimum paramagnetic behaviour?

1

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

2

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

3

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

4

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

Explanation:

${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +}$ and ${Ni}^{2 +}$ have $4, 5, 4$ and 2 unpaired electrons. No of unpaired electrons $α$ paramagnetism.

CHXII09:COORDINATION COMPOUNDS

322297 Match Column I with Column II and choose the correct combination from the options given.
supporting img

1

A - Q, B - R, C - S, D - T, E - P

2

A - T, B - S, C - R, D - P, E - Q

3

A - U, B - S, C - P, D - Q, E - R

4

A - P, B - Q, C - R, D - S, E - T

Explanation:

(A) ${[Ni (CN)_{4}]}^{- 2} - {dsp}^{2}$ hybridization square planar – Diamagnetic.
(B) Chlorophyll has ${Mg}^{+ 2}$ .
(C) Ziegler – Natta catalyst contains ${Ti}^{+ 4}$ .
(D) ${[{NiCl}_{4}]}^{- 2} - {sp}^{3}$ hybridization – tetrahedral – paramagnetic.
(E) Deoxy haemoglobin – non planar.

CHXII09:COORDINATION COMPOUNDS

322294 Among the following ions which one has the highest paramagnetism?

1

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

2

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

3

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

4

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

Explanation:

Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of
${Cr}^{3 +} (3 d^{3}), {Fe}^{2 +} (3 d^{6}), {Cu}^{2 +} (3 d^{9})$ and
${Zn}^{2 +} (3 d^{10})$ are outer orbital complex ions.
Hence ${Fe}^{2 +}$ has maximum unpaired electrons.

CHXII09:COORDINATION COMPOUNDS

322295 The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ and ${[Fe (CN)_{6}]}^{3 -}$ complexes respectively is

1 5.92 B.M. and 1.732 B.M.

2 3.87 B.M. and 1.732 B.M.

3 4.89 B.M. and 6.92 B.M.

4 6.92 B.M. in both.

Explanation:

In both the complexes, Fe is in $+ 3$ oxidation state. ${CN}^{-}$ is a strong field ligand, but $H_{2} O$ is a weak field ligand.
supporting img

There are 5 unpaired electrons. Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)} = \sqrt{35} = 5.92$ B.M.

There is only one unpaired electron.
Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, option (1) is correct.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322296 The $d$ -electronic configuration of ${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +} {Ni}^{2 +}$ are $3 d^{4}, 3 d^{5}, 3 d^{6}$ and $3 d^{8}$ respectively.
Which of the following complex will show minimum paramagnetic behaviour?

1

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

2

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

3

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

4

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

Explanation:

${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +}$ and ${Ni}^{2 +}$ have $4, 5, 4$ and 2 unpaired electrons. No of unpaired electrons $α$ paramagnetism.

CHXII09:COORDINATION COMPOUNDS

322297 Match Column I with Column II and choose the correct combination from the options given.
supporting img

1

A - Q, B - R, C - S, D - T, E - P

2

A - T, B - S, C - R, D - P, E - Q

3

A - U, B - S, C - P, D - Q, E - R

4

A - P, B - Q, C - R, D - S, E - T

Explanation:

(A) ${[Ni (CN)_{4}]}^{- 2} - {dsp}^{2}$ hybridization square planar – Diamagnetic.
(B) Chlorophyll has ${Mg}^{+ 2}$ .
(C) Ziegler – Natta catalyst contains ${Ti}^{+ 4}$ .
(D) ${[{NiCl}_{4}]}^{- 2} - {sp}^{3}$ hybridization – tetrahedral – paramagnetic.
(E) Deoxy haemoglobin – non planar.

CHXII09:COORDINATION COMPOUNDS

322294 Among the following ions which one has the highest paramagnetism?

1

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

2

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

3

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

4

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

Explanation:

Larger the number of unpaired electrons, the more it has paramagnetism. The configuration of
${Cr}^{3 +} (3 d^{3}), {Fe}^{2 +} (3 d^{6}), {Cu}^{2 +} (3 d^{9})$ and
${Zn}^{2 +} (3 d^{10})$ are outer orbital complex ions.
Hence ${Fe}^{2 +}$ has maximum unpaired electrons.

CHXII09:COORDINATION COMPOUNDS

322295 The magnetic moment is measured in Bohr Magneton (B.M.). Spin only magnetic moment of Fe in ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ and ${[Fe (CN)_{6}]}^{3 -}$ complexes respectively is

1 5.92 B.M. and 1.732 B.M.

2 3.87 B.M. and 1.732 B.M.

3 4.89 B.M. and 6.92 B.M.

4 6.92 B.M. in both.

Explanation:

In both the complexes, Fe is in $+ 3$ oxidation state. ${CN}^{-}$ is a strong field ligand, but $H_{2} O$ is a weak field ligand.
supporting img

There are 5 unpaired electrons. Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{5 (5 + 2)} = \sqrt{35} = 5.92$ B.M.

There is only one unpaired electron.
Spin only magnetic moment value $= \sqrt{n (n + 2)} = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.732$ B.M.
Thus, option (1) is correct.

JEE - 2023

CHXII09:COORDINATION COMPOUNDS

322296 The $d$ -electronic configuration of ${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +} {Ni}^{2 +}$ are $3 d^{4}, 3 d^{5}, 3 d^{6}$ and $3 d^{8}$ respectively.
Which of the following complex will show minimum paramagnetic behaviour?

1

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

2

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

3

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

4

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

Explanation:

${Cr}^{2 +}, {Mn}^{2 +}, {Fe}^{2 +}$ and ${Ni}^{2 +}$ have $4, 5, 4$ and 2 unpaired electrons. No of unpaired electrons $α$ paramagnetism.

CHXII09:COORDINATION COMPOUNDS

322297 Match Column I with Column II and choose the correct combination from the options given.
supporting img

1

A - Q, B - R, C - S, D - T, E - P

2

A - T, B - S, C - R, D - P, E - Q

3

A - U, B - S, C - P, D - Q, E - R

4

A - P, B - Q, C - R, D - S, E - T

Explanation:

(A) ${[Ni (CN)_{4}]}^{- 2} - {dsp}^{2}$ hybridization square planar – Diamagnetic.
(B) Chlorophyll has ${Mg}^{+ 2}$ .
(C) Ziegler – Natta catalyst contains ${Ti}^{+ 4}$ .
(D) ${[{NiCl}_{4}]}^{- 2} - {sp}^{3}$ hybridization – tetrahedral – paramagnetic.
(E) Deoxy haemoglobin – non planar.

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