The Valence Bond Theory (VBT) explains the formation, magnetic beha...

Question

The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields.

Answer the following questions :
(a) In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion ? Give reason in support of your answer.
(b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following :
    (i) $[CoF_6]^{3-}$
    (ii) $[Co(NH_3)_6]^{3+}$   [At. No. : Co = 27]
(c) $[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why ?   [Atomic No. : Ni = 28]

Accepted Answer

(a) WHICH d-ORBITALS ARE RAISED IN AN OCTAHEDRAL CRYSTAL FIELD ANSWER: The two $e_g$ orbitals, namely $d_{z^2}$ and $d_{x^2-y^2}$, are raised in energy. Step 1 - Set up the geometry. In an octahedral complex, the six ligands approach the central metal ion ALONG the three Cartesian axes - two along $+x$ and $-x$, two along $+y$ and $-y$, and two along $+z$ and $-z$. Step 2 - Look at the orientation of the five d-orbitals. $d_{z^2}$ and $d_{x^2-y^2}$ (the $e_g$ set) have their lobes pointing DIRECTLY ALONG the axes - that is, straight at the incoming ligands. $d_{xy}$, $d_{yz}$ and $d_{zx}$ (the $t_{2g}$ set) have their lobes pointing BETWEEN the axes - that is, in between the ligands. Step 3 - Apply the electrostatic argument (the reason). The ligands are treated as point negative charges (or as dipoles with the negative end towards the metal). Electrons in the metal d-orbitals are REPELLED by these ligand electrons. Since the $e_g$ orbitals point straight at the ligands, they experience a MUCH GREATER ELECTROSTATIC REPULSION. Their energy is therefore RAISED. The $t_{2g}$ orbitals point away from the ligands, experience LESS repulsion, and their energy is LOWERED. Step 4 - Quantify the splitting (barycentre rule). The degeneracy of the five d-orbitals is destroyed. Taking the average (barycentre) energy as the reference: $$e_g \ (d_{z^2},\, d_{x^2-y^2}) : ext{raised by } +0.6\,\Delta_0$$ $$t_{2g}\ (d_{xy},\, d_{yz},\, d_{zx}) : ext{lowered by } -0.4\,\Delta_0$$ where $\Delta_0$ is the crystal field splitting energy for an octahedral field. (Check: $2 imes (+0.6\Delta_0) + 3 imes (-0.4\Delta_0) = 1.2\Delta_0 - 1.2\Delta_0 = 0$.) (b) ELECTRONIC CONFIGURATIONS USING CRYSTAL FIELD THEORY The rule: if $\Delta_0

P$, the ligand is STRONG FIELD and electrons pair up in $t_{2g}$ first (LOW SPIN). (i) $[CoF_6]^{3-}$ Step 1 - Oxidation state: $x + 6(-1) = -3 \Rightarrow x = +3$, so $Co^{3+}$. Step 2 - Configuration: $Co(Z=27) = [Ar]3d^7 4s^2$; $Co^{3+} = [Ar]3d^6$ Step 3 - $F^-$ is a WEAK FIELD ligand (it lies low in the spectrochemical series), so $\Delta_0

P$. All six electrons pair up in the lower $t_{2g}$ set. Step 4 - Configuration: $$\boxed{t_{2g}^6\ e_g^0}$$ This is a LOW SPIN complex with 0 unpaired electrons; diamagnetic, $\mu = 0$ BM. It is an inner orbital ($d^2sp^3$) complex. (c) WHY $[NiCl_4]^{2-}$ IS PARAMAGNETIC BUT $[Ni(CO)_4]$ IS DIAMAGNETIC The key difference is the OXIDATION STATE of nickel and the FIELD STRENGTH of the ligand. CASE 1 : $[NiCl_4]^{2-}$ Step 1 - Oxidation state of Ni. $x + 4(-1) = -2 \Rightarrow x = +2$, so nickel is $Ni^{2+}$. Step 2 - Electronic configuration. $Ni\,(Z=28) = [Ar]3d^8 4s^2$ $Ni^{2+} = [Ar]3d^8 4s^0$ The $3d^8$ arrangement is: three orbitals fully paired and TWO ORBITALS SINGLY OCCUPIED, i.e. 2 UNPAIRED ELECTRONS. Step 3 - Nature of the ligand. $Cl^-$ is a WEAK FIELD ligand. It cannot supply enough energy to force the $3d$ electrons to pair up. Step 4 - Hybridisation and magnetic behaviour. Since no $3d$ orbital is vacated, hybridisation involves $4s$ and $4p$ orbitals: $sp^3$ hybridisation $\Rightarrow$ TETRAHEDRAL geometry. The 2 unpaired electrons remain $\Rightarrow$ the complex is PARAMAGNETIC. $$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.83\ ext{BM}$$ CASE 2 : $[Ni(CO)_4]$ Step 1 - Oxidation state of Ni. CO is a NEUTRAL ligand and the complex is neutral, so $x + 4(0) = 0 \Rightarrow x = 0$. Nickel is in the ZERO oxidation state. Step 2 - Electronic configuration of the neutral atom. $Ni\,(Z=28) = [Ar]3d^8 4s^2$ - which has 2 unpaired electrons in $3d$. Step 3 - Effect of the strong field ligand. CO is a very STRONG FIELD ligand (it is at the top of the spectrochemical series). In its presence the TWO $4s$ ELECTRONS ARE PUSHED INTO THE $3d$ ORBITALS, pairing up all the $3d$ electrons: $$3d^8 4s^2 \longrightarrow 3d^{10} 4s^0$$ Step 4 - Hybridisation and magnetic behaviour. Now $3d$ is completely full and $4s$ is empty, so hybridisation uses $4s$ and $4p$: $sp^3$ hybridisation $\Rightarrow$ TETRAHEDRAL geometry. There are NO UNPAIRED ELECTRONS $\Rightarrow$ the complex is DIAMAGNETIC, $\mu = 0$ BM. CONCLUSION Both complexes are tetrahedral and both are $sp^3$ hybridised, but $[NiCl_4]^{2-}$ has $Ni^{2+}$ with a weak field ligand $\Rightarrow$ 2 unpaired electrons $\Rightarrow$ PARAMAGNETIC; $[Ni(CO)_4]$ has $Ni(0)$ with a strong field ligand that causes complete pairing $\Rightarrow$ 0 unpaired electrons $\Rightarrow$ DIAMAGNETIC.