Q: \$[NiCl_4]^{2-}\$ is paramagnetic while \$Ni(CO)_4]\$ diamagnetic though both are tetrahedral. Why? (Atomic no. of \$Ni = 28\$)

In \$Ni{\left( CO \right)}_{4}\$, the oxidation state of \$Ni\$ is zero whereas in \${\left[ Ni{Cl}_{4} \right]}^{2-}\$, the oxidation state of \$Ni\$ is \$+2\$.The presence of \$CO\$ ligand, which is a strong ligand, can pair all electrons in \$Ni{\left( CO \right)}_{4}\$ and thus it is diamagnetic in nature but \${Cl}^{-}\$ is a weak ligand and is unable to pair up the unpaired electrons and thus \${\left[ Ni{Cl}_{4} \right]}^{2-}\$ is paramagnetic in nature.

Q: Crystal field stabilization energy for high spin \$d^4\$ octahedral complex is:

For high spin \$d^4\$ Octahedral complex the splitting is given by(Refer to Image)\$CFSE=[-0.4\times 3+0.6\times 1]\triangle_0\$\$=[-1.2+0.6]\triangle_0\$\$=-0.6\triangle_0\$\$\therefore\$ Answer is option D.

Q: Draw figure to show the splitting of d orbitals in an octahedral crystal field.

The splitting of \$d-\$ orbitals in an octahedral crystal field is as shown.

Q: The crystal field splitting energy for octahedral \$(\Delta_0)\$ and tetrahedral \$(\Delta_t)\$ complexes are related as:

Since splitting in tetrahedral complex is \$\cfrac{2}{3}\$rd of octahedral complex ,so for one legand splitting in \$OH =\cfrac { \Delta _0 }{ 6 } \$,then for one legend splitting in tetrahedral is \$\cfrac { 2 }{ 3 } \left( \cfrac { \Delta _0 }{ 6 } \right) \$, so for \$4\$ legand ,\$\Delta t=4\times \cfrac { 2 }{ 3 } \times \cfrac { \Delta _0 }{ 6 } \\ \quad \quad \quad =\cfrac { 4 }{ 9 } \Delta _0\$

Q: Explain the violet colour of \$[Ti(H_2O)_6]^{3+}\$ complex on basis of the crystal field theory?

In the complex \$\displaystyle [Ti(H_2O)_6]^{3+}\$, the Ti atom has oxidation state of \$\displaystyle+3\$. Its outer electronic configuration is \$\displaystyle 3d^1\$. It has one unpaired electron. This unpaired electron is excited from \$\displaystyle t_{2g}\$ level to \$\displaystyle e_g\$ level by absorbing yellow light and hence appears violet coloured.

Q: What is the correct electronic configuration of the central atom is \$K_4(Fe(CN)_6]\$ on crystal field theory?

\$K_4[Fe(CN)_6]\$Fe ground state \$[Ar]3d^64s^2\$\$Fe^{2+}:3d^64s^0\$So, B is the correct option.

Q: The hexaaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Mn(II) has an electronic configuration of \$3d^54s^0\$. In the case of hexaaquomanganese (II) ion the ligand attached to metal is water which is a weak field ligand and is not able to pair up the electrons of 3d subshell and the configuration is \$t_2g^3\$ and \$eg^2\$. whereas in the case of hexacyano ion the \$CN^-\$ is a strong field ligand and it causes pairing of electrons in the d orbitals and hence thee exist only one unpaired electron in \$t_2g^5\$\$eg^0\$.

Q: Explain the following:\$Co^{2+}\$ is easily oxidised to \$Co^{3+}\$ in the presence of a strong ligand.

With the electronic configuration \$3d^2 4s^0, Co^{2+}\$ has three unpaired electrons. \$H_2O\$ being a weak ligand, the unpaired electrons present in \$3d\$ orbital of \$Co\ (II)\$ do not pair up. In the presence of strong ligands, two unpaired electrons in \$3d\$ orbitals pair up and the third unpaired \$e^-\$ shifts to higher energy orbital from where it can be easily lost and hence show an oxidation state of \$III\$.

Question 1

$[NiCl_4]^{2-}$ is paramagnetic while $Ni(CO)_4]$ diamagnetic though both are tetrahedral. Why? (Atomic no. of $Ni = 28$)

Accepted Answer

In $Ni{\left( CO \right)}_{4}$, the oxidation state of $Ni$ is zero whereas in ${\left[ Ni{Cl}_{4} \right]}^{2-}$, the oxidation state of $Ni$ is $+2$.The presence of $CO$ ligand, which is a strong ligand, can pair all electrons in $Ni{\left( CO \right)}_{4}$ and thus it is diamagnetic in nature but ${Cl}^{-}$ is a weak ligand and is unable to pair up the unpaired electrons and thus ${\left[ Ni{Cl}_{4} \right]}^{2-}$ is paramagnetic in nature.

Question 2

Crystal field stabilization energy for high spin $d^4$ octahedral complex is:

Accepted Answer

For high spin $d^4$ Octahedral complex the splitting is given by(Refer to Image)$CFSE=[-0.4\times 3+0.6\times 1]\triangle_0$$=[-1.2+0.6]\triangle_0$$=-0.6\triangle_0$$\therefore$ Answer is option D.

Question 3

Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Accepted Answer

The splitting of $d-$ orbitals in an octahedral crystal field is as shown.

Question 4

The crystal field splitting energy for octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complexes are related as:

Accepted Answer

Since splitting in tetrahedral complex is $\cfrac{2}{3}$rd of octahedral complex ,so for one legand splitting in $OH =\cfrac { \Delta _0 }{ 6 } $,then for one legend splitting in tetrahedral is $\cfrac { 2 }{ 3 } \left( \cfrac { \Delta _0 }{ 6 }  \right) $, so for $4$ legand ,$\Delta t=4\times \cfrac { 2 }{ 3 } \times \cfrac { \Delta _0 }{ 6 } \ \quad \quad \quad =\cfrac { 4 }{ 9 } \Delta _0$

Question 5

Explain the violet colour of $[Ti(H_2O)_6]^{3+}$ complex on basis of the crystal field theory?

Accepted Answer

In the complex $\displaystyle [Ti(H_2O)_6]^{3+}$, the Ti atom has oxidation state of $\displaystyle+3$. Its outer electronic configuration is $\displaystyle 3d^1$. It has one unpaired electron. This unpaired electron is excited from $\displaystyle t_{2g}$ level to $\displaystyle e_g$ level by absorbing yellow light and hence appears violet coloured.

Question 6

What is crystal field splitting energy? How does the magnitude of $\Delta_{0}$ decide the actual configuration of d orbitals in a coordination entity?

Accepted Answer

The crystal field stabilisation energy (CFSE)  is the gain in the energy achieved by preferential filling up of orbitals by electrons. It is usually less than or equal to 0. When it is equal to 0, the complex is unstable. The magnitude of CFSE depends on the number and nature of ligands and the geometry of the complex.Consider octahedral $\displaystyle d^4$ system. Three electrons are in lower $\displaystyle t_{2g}$ level. Fourth electron will enter into higher $\displaystyle e_g$ level if $\displaystyle \Delta _o < P$. Fourth electron will enter into lower $\displaystyle t_{2g}$ level if $\displaystyle \Delta _o > P$. Here, P is the pairing energy. It is the energy required to pair two electrons against electron electron repulsion in the same orbital.

Question 7

What is the correct electronic configuration of the central atom is $K_4(Fe(CN)_6]$ on crystal field theory?

Accepted Answer

$K_4[Fe(CN)_6]$Fe ground state $[Ar]3d^64s^2$$Fe^{2+}:3d^64s^0$So, B is the correct option.

Question 8

The hexaaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Accepted Answer

Mn(II) has an electronic configuration of $3d^54s^0$. In the case of hexaaquomanganese (II) ion the ligand attached to metal is water which is a weak field ligand and is not able to pair up the electrons of 3d subshell and the configuration is $t_2g^3$ and $eg^2$. whereas in the case of hexacyano ion the $CN^-$ is a strong field ligand and it causes pairing of electrons in the d orbitals and hence thee exist only one unpaired electron in $t_2g^5$$eg^0$.

Question 9

Explain the following:$Co^{2+}$ is easily oxidised to $Co^{3+}$ in the presence of a strong ligand.

Accepted Answer

With the electronic configuration $3d^2 4s^0, Co^{2+}$ has three unpaired electrons. $H_2O$ being a weak ligand, the unpaired electrons present in $3d$ orbital of $Co\ (II)$ do not pair up. In the presence of strong ligands, two unpaired electrons in $3d$ orbitals pair up and the third unpaired $e^-$ shifts to higher energy orbital from where it can be easily lost and hence show an oxidation state of $III$.

Question 10

On the basis of crystal field theory write the electronic configuration for $d^{4}$ ion, if $\Delta_{0} < P$.

Accepted Answer

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Crystal Field Theory

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Colour In Coordination Compounds

$[N i C l_{4}]^{2 -}$ is paramagnetic while $N i (C O)_{4}]$ diamagnetic though both are tetrahedral. Why? (Atomic no. of $N i = 28$ )

Crystal field stabilization energy for high spin $d^{4}$ octahedral complex is:

Draw figure to show the splitting of d orbitals in an octahedral crystal field.

The crystal field splitting energy for octahedral $(Δ_{0})$ and tetrahedral $(Δ_{t})$ complexes are related as:

Explain the violet colour of $[T i (H_{2} O)_{6}]^{3 +}$ complex on basis of the crystal field theory?

What is crystal field splitting energy? How does the magnitude of $Δ_{0}$ decide the actual configuration of d orbitals in a coordination entity?

What is the correct electronic configuration of the central atom is $K_{4} (F e (C N)_{6}]$ on crystal field theory?

The hexaaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Explain the following:
$C o^{2 +}$ is easily oxidised to $C o^{3 +}$ in the presence of a strong ligand.

On the basis of crystal field theory write the electronic configuration for $d^{4}$ ion, if $Δ_{0} < P$ .

Revise with Concepts

Crystal Field Theory

Quick Summary With Stories

Colour In Coordination Compounds

[NiCl4​]2− is paramagnetic while Ni(CO)4​] diamagnetic though both are tetrahedral. Why? (Atomic no. of Ni=28)

Crystal field stabilization energy for high spin d4 octahedral complex is:

Draw figure to show the splitting of d orbitals in an octahedral crystal field.

The crystal field splitting energy for octahedral (Δ0​) and tetrahedral (Δt​) complexes are related as:

Explain the violet colour of [Ti(H2​O)6​]3+ complex on basis of the crystal field theory?

What is crystal field splitting energy? How does the magnitude of Δ0​ decide the actual configuration of d orbitals in a coordination entity?

What is the correct electronic configuration of the central atom is K4​(Fe(CN)6​] on crystal field theory?

The hexaaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Explain the following: Co2+ is easily oxidised to Co3+ in the presence of a strong ligand.

On the basis of crystal field theory write the electronic configuration for d4 ion, if Δ0​<P.

$[N i C l_{4}]^{2 -}$ is paramagnetic while $N i (C O)_{4}]$ diamagnetic though both are tetrahedral. Why? (Atomic no. of $N i = 28$ )

Crystal field stabilization energy for high spin $d^{4}$ octahedral complex is:

The crystal field splitting energy for octahedral $(Δ_{0})$ and tetrahedral $(Δ_{t})$ complexes are related as:

Explain the violet colour of $[T i (H_{2} O)_{6}]^{3 +}$ complex on basis of the crystal field theory?

What is crystal field splitting energy? How does the magnitude of $Δ_{0}$ decide the actual configuration of d orbitals in a coordination entity?

What is the correct electronic configuration of the central atom is $K_{4} (F e (C N)_{6}]$ on crystal field theory?

Explain the following:
$C o^{2 +}$ is easily oxidised to $C o^{3 +}$ in the presence of a strong ligand.

On the basis of crystal field theory write the electronic configuration for $d^{4}$ ion, if $Δ_{0} < P$ .