Question

Find the number of complexes in which configuration of central metal ion is $t_{2g}^6, e_g^0$.

(I) $[Co(NH_3)_5Cl]Cl_2$, (II) $(NH_4)_2[PtCl_4]$, (III) $K_4[Ni(CN)_6]$ (IV) $K_3[Co(NO_2)_6]$, (V) $K_3[Fe(CN)_6]$, (VI) $Na_2[Fe(CN)_5(NO)]$

Answer 1

3

Answer 2

To determine the number of complexes with a central metal ion configuration of $t_{2g}^6 e_g^0$, we need to analyze each complex:

The $t_{2g}^6 e_g^0$ configuration corresponds to a low-spin $d^6$ electronic configuration in an octahedral crystal field.

(I) $[Co(NH_3)_5Cl]Cl_2$

1. Central metal ion and oxidation state: Co. Let oxidation state be $x$. $x + 5(0) + (-1) = +2 \Rightarrow x = +3$. So, $Co^{3+}$.

2. d-electron count: Co is $[Ar]3d^74s^2$. $Co^{3+}$ is $[Ar]3d^6$. This is a $d^6$ ion.

3. Geometry: The coordination number is 6 (5 $NH_3$ and 1 $Cl^-$), so it is octahedral.

4. Ligands and spin state: $NH_3$ is a strong field ligand, and $Cl^-$ is a weak field ligand. For $Co^{3+}$ ($d^6$), strong field ligands cause low spin pairing. The presence of 5 strong field $NH_3$ ligands ensures a strong enough field.

   • For $d^6$ in an octahedral field with strong ligands: Low spin configuration is $t_{2g}^6 e_g^0$.

   • Therefore, (I) has the $t_{2g}^6 e_g^0$ configuration.

(II) $(NH_4)_2[PtCl_4]$

1. Central metal ion and oxidation state: Pt. Let oxidation state be $x$. $2(+1) + x + 4(-1) = 0 \Rightarrow x = +2$. So, $Pt^{2+}$.

2. d-electron count: Pt is $[Xe]4f^{14}5d^96s^1$. $Pt^{2+}$ is $[Xe]4f^{14}5d^8$. This is a $d^8$ ion.

3. Geometry: The coordination number is 4. $Pt^{2+}$ complexes with coordination number 4 are typically square planar.

4. Ligands and spin state: $Cl^-$ is a weak field ligand.

   • The notation $t_{2g}$ and $e_g$ is specific to octahedral complexes. Square planar complexes have a different orbital splitting pattern.

   • Therefore, (II) does not have the $t_{2g}^6 e_g^0$ configuration.

(III) $K_4[Ni(CN)_6]$

1. Central metal ion and oxidation state: Ni. Let oxidation state be $x$. $4(+1) + x + 6(-1) = 0 \Rightarrow x = +2$. So, $Ni^{2+}$.

2. d-electron count: Ni is $[Ar]3d^84s^2$. $Ni^{2+}$ is $[Ar]3d^8$. This is a $d^8$ ion.

3. Geometry: The coordination number is 6, so it is octahedral.

4. Ligands and spin state: $CN^-$ is a very strong field ligand.

   • For $d^8$ in an octahedral field, the configuration is always high spin, regardless of ligand field strength, because the last two electrons must occupy the $e_g$ orbitals.

   • Configuration for $d^8$ octahedral: $t_{2g}^6 e_g^2$.

   • Therefore, (III) does not have the $t_{2g}^6 e_g^0$ configuration.

(IV) $K_3[Co(NO_2)_6]$

1. Central metal ion and oxidation state: Co. Let oxidation state be $x$. $3(+1) + x + 6(-1) = 0 \Rightarrow x = +3$. So, $Co^{3+}$.

2. d-electron count: $Co^{3+}$ is $[Ar]3d^6$. This is a $d^6$ ion.

3. Geometry: The coordination number is 6, so it is octahedral.

4. Ligands and spin state: $NO_2^-$ is a strong field ligand.

   • For $d^6$ in an octahedral field with strong ligands: Low spin configuration is $t_{2g}^6 e_g^0$.

   • Therefore, (IV) has the $t_{2g}^6 e_g^0$ configuration.

(V) $K_3[Fe(CN)_6]$

1. Central metal ion and oxidation state: Fe. Let oxidation state be $x$. $3(+1) + x + 6(-1) = 0 \Rightarrow x = +3$. So, $Fe^{3+}$.

2. d-electron count: Fe is $[Ar]3d^64s^2$. $Fe^{3+}$ is $[Ar]3d^5$. This is a $d^5$ ion.

3. Geometry: The coordination number is 6, so it is octahedral.

4. Ligands and spin state: $CN^-$ is a very strong field ligand.

   • For $d^5$ in an octahedral field with strong ligands: Low spin configuration is $t_{2g}^5 e_g^0$.

   • Therefore, (V) does not have the $t_{2g}^6 e_g^0$ configuration.

(VI) $Na_2[Fe(CN)_5(NO)]$

1. Central metal ion and oxidation state: Fe. This is sodium nitroprusside. The NO ligand is typically considered as $NO^+$. Let oxidation state be $x$. $2(+1) + x + 5(-1) + (+1) = 0 \Rightarrow 2 + x - 5 + 1 = 0 \Rightarrow x = +2$. So, $Fe^{2+}$.

2. d-electron count: Fe is $[Ar]3d^64s^2$. $Fe^{2+}$ is $[Ar]3d^6$. This is a $d^6$ ion.

3. Geometry: The coordination number is 6, so it is octahedral.

4. Ligands and spin state: $CN^-$ is a very strong field ligand, and $NO^+$ is also a very strong field ligand.

   • For $d^6$ in an octahedral field with strong ligands: Low spin configuration is $t_{2g}^6 e_g^0$.

   • Therefore, (VI) has the $t_{2g}^6 e_g^0$ configuration.

The complexes with the $t_{2g}^6 e_g^0$ configuration are (I), (IV), and (VI). The number of such complexes is 3.