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COORDINATION COMPOUNDS

274133 The hybridization and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ , respectively are :

1

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

and paramagnetic

2

{sp}^{3} d

and paramagnetic

3

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

and diamagnetic

4

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

and diamagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4}$
Let, O.N. of $Mn = x$
$∴ x + (- 1) \times 6 = - 4$
$x - 6 = - 4$
$x = + 2$
${Mn}^{2 +} : [Ar] 3 d^{5}$
Since, co-ordination no. is 6 . So it is an octahedral complex and ${CN}^{-}$ is a strong field ligand so hybridization will be $d^{2} {sp}^{3}$ ie., inner orbital complex and low spin.
one unpaired electron is present so, it is paramagnetic

JEE Main 25-02-2021

COORDINATION COMPOUNDS

274134 What is the hybridization present in the complex ${[Co {({NH}_{3})}_{6}]}^{3 +}$

1

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

2

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

3

{dsp}^{3}

4

{sp}^{3} d

Explanation:

(A) : Given complex : ${[Co {({NH}_{3})}_{6}]}^{3 +}$
Oxidation state of metal $= + 3$
Geometry $=$ octahedral
Electronic configuration of ${Co}^{3 +} = [Ar] 3 d^{6}$
original image
${[Co {({NH}_{3})}_{6}]}^{+ 3} = d^{2} {sp}^{3}$ Hybridized (Octahedral and diamagnetic]

TS- EAMCET 09.08.2021

COORDINATION COMPOUNDS

274135 The hybridisation and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ and ${[Fe (CN)_{6}]}^{3 -}$ , respectively are

1

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

and paramagnetic

2

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

and diamagnetic

3

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

and diamagnetic

4

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

and paramagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4 -}$
$x + 6 (- 1) = - 4$
$x = + 2$
${Mn}^{2 +} = [Ar] 3 d^{5}$ (Pairing of electron occur because $CN$ is SFL).
Ground state
Excited state
There is one unpaired electron.
six d'sp' hybrid orbital
Hence, ${[Mn (CN)_{6}]}^{4 -}$ is paramagnetic
And, ${[Fe (CN)_{6}]}^{3 -}$
$x + 6 (- 1) = - 3$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5}$
The same method also apply for ${[Fe (CN)_{6}]}^{3 -}$ because it contain $[Ar] 3 d^{5}$ electronic configuration which is similar to ${[Mn (CN)_{6}]}^{4 -}$
Thus, both are paramagnetic with $d^{2} {sp}^{3}$ hybridization.

JEE Main 2021

COORDINATION COMPOUNDS

274136 According to the valence bond theory the hybridisation of central metal atom is ${dsp}^{2}$ for which one of the following compound?

1

{NiCl}_{2} \cdot 6 H_{2} O

2

K_{2} [Ni (CN)_{4}]

3

[Ni (CO)_{4}]

4

{Na}_{2} [{NiCl}_{4}]

Explanation:

(B) : The hybridization of central atom is ${dsp}^{2}$ that means the complex is square planar in geometry.
$K_{2} [Ni (CN)_{4}]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals.
$K_{2} [Ni (CN)_{4}]$
$2 (+ 1) + x + 4 (- 1) = 0$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8} 4 s^{0} 4 p^{0}$
original image

JEE Main 2021

COORDINATION COMPOUNDS

274133 The hybridization and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ , respectively are :

1

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

and paramagnetic

2

{sp}^{3} d

and paramagnetic

3

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

and diamagnetic

4

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

and diamagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4}$
Let, O.N. of $Mn = x$
$∴ x + (- 1) \times 6 = - 4$
$x - 6 = - 4$
$x = + 2$
${Mn}^{2 +} : [Ar] 3 d^{5}$
Since, co-ordination no. is 6 . So it is an octahedral complex and ${CN}^{-}$ is a strong field ligand so hybridization will be $d^{2} {sp}^{3}$ ie., inner orbital complex and low spin.
one unpaired electron is present so, it is paramagnetic

JEE Main 25-02-2021

COORDINATION COMPOUNDS

274134 What is the hybridization present in the complex ${[Co {({NH}_{3})}_{6}]}^{3 +}$

1

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

2

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

3

{dsp}^{3}

4

{sp}^{3} d

Explanation:

(A) : Given complex : ${[Co {({NH}_{3})}_{6}]}^{3 +}$
Oxidation state of metal $= + 3$
Geometry $=$ octahedral
Electronic configuration of ${Co}^{3 +} = [Ar] 3 d^{6}$
original image
${[Co {({NH}_{3})}_{6}]}^{+ 3} = d^{2} {sp}^{3}$ Hybridized (Octahedral and diamagnetic]

TS- EAMCET 09.08.2021

COORDINATION COMPOUNDS

274135 The hybridisation and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ and ${[Fe (CN)_{6}]}^{3 -}$ , respectively are

1

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

and paramagnetic

2

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

and diamagnetic

3

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

and diamagnetic

4

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

and paramagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4 -}$
$x + 6 (- 1) = - 4$
$x = + 2$
${Mn}^{2 +} = [Ar] 3 d^{5}$ (Pairing of electron occur because $CN$ is SFL).
Ground state
Excited state
There is one unpaired electron.
six d'sp' hybrid orbital
Hence, ${[Mn (CN)_{6}]}^{4 -}$ is paramagnetic
And, ${[Fe (CN)_{6}]}^{3 -}$
$x + 6 (- 1) = - 3$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5}$
The same method also apply for ${[Fe (CN)_{6}]}^{3 -}$ because it contain $[Ar] 3 d^{5}$ electronic configuration which is similar to ${[Mn (CN)_{6}]}^{4 -}$
Thus, both are paramagnetic with $d^{2} {sp}^{3}$ hybridization.

JEE Main 2021

COORDINATION COMPOUNDS

274136 According to the valence bond theory the hybridisation of central metal atom is ${dsp}^{2}$ for which one of the following compound?

1

{NiCl}_{2} \cdot 6 H_{2} O

2

K_{2} [Ni (CN)_{4}]

3

[Ni (CO)_{4}]

4

{Na}_{2} [{NiCl}_{4}]

Explanation:

(B) : The hybridization of central atom is ${dsp}^{2}$ that means the complex is square planar in geometry.
$K_{2} [Ni (CN)_{4}]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals.
$K_{2} [Ni (CN)_{4}]$
$2 (+ 1) + x + 4 (- 1) = 0$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8} 4 s^{0} 4 p^{0}$
original image

JEE Main 2021

COORDINATION COMPOUNDS

274133 The hybridization and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ , respectively are :

1

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

and paramagnetic

2

{sp}^{3} d

and paramagnetic

3

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

and diamagnetic

4

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

and diamagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4}$
Let, O.N. of $Mn = x$
$∴ x + (- 1) \times 6 = - 4$
$x - 6 = - 4$
$x = + 2$
${Mn}^{2 +} : [Ar] 3 d^{5}$
Since, co-ordination no. is 6 . So it is an octahedral complex and ${CN}^{-}$ is a strong field ligand so hybridization will be $d^{2} {sp}^{3}$ ie., inner orbital complex and low spin.
one unpaired electron is present so, it is paramagnetic

JEE Main 25-02-2021

COORDINATION COMPOUNDS

274134 What is the hybridization present in the complex ${[Co {({NH}_{3})}_{6}]}^{3 +}$

1

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

2

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

3

{dsp}^{3}

4

{sp}^{3} d

Explanation:

(A) : Given complex : ${[Co {({NH}_{3})}_{6}]}^{3 +}$
Oxidation state of metal $= + 3$
Geometry $=$ octahedral
Electronic configuration of ${Co}^{3 +} = [Ar] 3 d^{6}$
original image
${[Co {({NH}_{3})}_{6}]}^{+ 3} = d^{2} {sp}^{3}$ Hybridized (Octahedral and diamagnetic]

TS- EAMCET 09.08.2021

COORDINATION COMPOUNDS

274135 The hybridisation and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ and ${[Fe (CN)_{6}]}^{3 -}$ , respectively are

1

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

and paramagnetic

2

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

and diamagnetic

3

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

and diamagnetic

4

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

and paramagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4 -}$
$x + 6 (- 1) = - 4$
$x = + 2$
${Mn}^{2 +} = [Ar] 3 d^{5}$ (Pairing of electron occur because $CN$ is SFL).
Ground state
Excited state
There is one unpaired electron.
six d'sp' hybrid orbital
Hence, ${[Mn (CN)_{6}]}^{4 -}$ is paramagnetic
And, ${[Fe (CN)_{6}]}^{3 -}$
$x + 6 (- 1) = - 3$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5}$
The same method also apply for ${[Fe (CN)_{6}]}^{3 -}$ because it contain $[Ar] 3 d^{5}$ electronic configuration which is similar to ${[Mn (CN)_{6}]}^{4 -}$
Thus, both are paramagnetic with $d^{2} {sp}^{3}$ hybridization.

JEE Main 2021

COORDINATION COMPOUNDS

274136 According to the valence bond theory the hybridisation of central metal atom is ${dsp}^{2}$ for which one of the following compound?

1

{NiCl}_{2} \cdot 6 H_{2} O

2

K_{2} [Ni (CN)_{4}]

3

[Ni (CO)_{4}]

4

{Na}_{2} [{NiCl}_{4}]

Explanation:

(B) : The hybridization of central atom is ${dsp}^{2}$ that means the complex is square planar in geometry.
$K_{2} [Ni (CN)_{4}]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals.
$K_{2} [Ni (CN)_{4}]$
$2 (+ 1) + x + 4 (- 1) = 0$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8} 4 s^{0} 4 p^{0}$
original image

JEE Main 2021

COORDINATION COMPOUNDS

274133 The hybridization and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ , respectively are :

1

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

and paramagnetic

2

{sp}^{3} d

and paramagnetic

3

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

and diamagnetic

4

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

and diamagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4}$
Let, O.N. of $Mn = x$
$∴ x + (- 1) \times 6 = - 4$
$x - 6 = - 4$
$x = + 2$
${Mn}^{2 +} : [Ar] 3 d^{5}$
Since, co-ordination no. is 6 . So it is an octahedral complex and ${CN}^{-}$ is a strong field ligand so hybridization will be $d^{2} {sp}^{3}$ ie., inner orbital complex and low spin.
one unpaired electron is present so, it is paramagnetic

JEE Main 25-02-2021

COORDINATION COMPOUNDS

274134 What is the hybridization present in the complex ${[Co {({NH}_{3})}_{6}]}^{3 +}$

1

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

2

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

3

{dsp}^{3}

4

{sp}^{3} d

Explanation:

(A) : Given complex : ${[Co {({NH}_{3})}_{6}]}^{3 +}$
Oxidation state of metal $= + 3$
Geometry $=$ octahedral
Electronic configuration of ${Co}^{3 +} = [Ar] 3 d^{6}$
original image
${[Co {({NH}_{3})}_{6}]}^{+ 3} = d^{2} {sp}^{3}$ Hybridized (Octahedral and diamagnetic]

TS- EAMCET 09.08.2021

COORDINATION COMPOUNDS

274135 The hybridisation and magnetic nature of ${[Mn (CN)_{6}]}^{4 -}$ and ${[Fe (CN)_{6}]}^{3 -}$ , respectively are

1

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

and paramagnetic

2

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

and diamagnetic

3

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

and diamagnetic

4

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

and paramagnetic

Explanation:

(A) : ${[Mn (CN)_{6}]}^{4 -}$
$x + 6 (- 1) = - 4$
$x = + 2$
${Mn}^{2 +} = [Ar] 3 d^{5}$ (Pairing of electron occur because $CN$ is SFL).
Ground state
Excited state
There is one unpaired electron.
six d'sp' hybrid orbital
Hence, ${[Mn (CN)_{6}]}^{4 -}$ is paramagnetic
And, ${[Fe (CN)_{6}]}^{3 -}$
$x + 6 (- 1) = - 3$
$x = + 3$
${Fe}^{3 +} = [Ar] 3 d^{5}$
The same method also apply for ${[Fe (CN)_{6}]}^{3 -}$ because it contain $[Ar] 3 d^{5}$ electronic configuration which is similar to ${[Mn (CN)_{6}]}^{4 -}$
Thus, both are paramagnetic with $d^{2} {sp}^{3}$ hybridization.

JEE Main 2021

COORDINATION COMPOUNDS

274136 According to the valence bond theory the hybridisation of central metal atom is ${dsp}^{2}$ for which one of the following compound?

1

{NiCl}_{2} \cdot 6 H_{2} O

2

K_{2} [Ni (CN)_{4}]

3

[Ni (CO)_{4}]

4

{Na}_{2} [{NiCl}_{4}]

Explanation:

(B) : The hybridization of central atom is ${dsp}^{2}$ that means the complex is square planar in geometry.
$K_{2} [Ni (CN)_{4}]$ is only complex which will form the square planar geometry because it has the strong field ligand due to which pairing of electron occurred in inner orbitals.
$K_{2} [Ni (CN)_{4}]$
$2 (+ 1) + x + 4 (- 1) = 0$
$x = + 2$
${Ni}^{2 +} = [Ar] 3 d^{8} 4 s^{0} 4 p^{0}$
original image

JEE Main 2021

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