GATE Life Sciences s, p and d Block Elements Syllabus

List two major limitations of Valence Bond Theory.

It does not quantitatively explain the colour (absorption spectra) of complexes, nor does it explain why some complexes are high-spin and others low-spin (no concept of crystal-field stabilization). It also cannot account for the relative energies/strengths of different ligands.

What is the central idea of Crystal Field Theory (CFT)?

CFT treats metal–ligand bonding as purely electrostatic (ionic). Ligands are point charges/dipoles whose electric field splits the degenerate metal $d$ orbitals into sets of different energies, the pattern depending on the geometry of the ligand arrangement.

How do the five d orbitals split in an octahedral crystal field?

They split into a lower-energy triply-degenerate $t_{2g}$ set ($d_{xy}, d_{yz}, d_{zx}$) and a higher-energy doubly-degenerate $e_{g}$ set ($d_{x^{2}-y^{2}}, d_{z^{2}}$), separated by $\Delta_{o}$. The $e_{g}$ orbitals point directly at the ligands, so they are raised in energy.

State the barycentre (centre-of-gravity) energies of the $t_{2g}$ and $e_{g}$ sets in an octahedral field relative to the average.

The $t_{2g}$ set is stabilized by $-0.4\,\Delta_{o}$ each and the $e_{g}$ set is destabilized by $+0.6\,\Delta_{o}$ each, preserving the barycentre: $3(-0.4\Delta_{o}) + 2(+0.6\Delta_{o}) = 0$.

How does the d-orbital splitting in a tetrahedral field compare with an octahedral field?

The pattern is inverted: the lower set is $e$ ($d_{x^{2}-y^{2}}, d_{z^{2}}$) and the upper set is $t_{2}$ ($d_{xy}, d_{yz}, d_{zx}$). The splitting is smaller: $\Delta_{t} = \tfrac{4}{9}\,\Delta_{o}$ for the same metal and ligands.

Why are tetrahedral complexes almost always high-spin?

Because $\Delta_{t} = \tfrac{4}{9}\Delta_{o}$ is small, it is almost always less than the pairing energy $P$. Electrons therefore prefer to occupy higher orbitals singly rather than pair up, giving high-spin configurations.

Define Crystal Field Stabilization Energy (CFSE) and give the octahedral formula.

CFSE is the net lowering of energy of the $d$-electrons due to crystal-field splitting relative to the spherical (unsplit) field. For octahedral: $$\text{CFSE} = \left[-0.4\,n_{t_{2g}} + 0.6\,n_{e_{g}}\right]\Delta_{o} + mP$$ where $m$ is the number of extra electron pairs formed.

What determines whether an octahedral complex is high-spin or low-spin, in terms of $\Delta_{o}$ and pairing energy $P$?

If $\Delta_{o} > P$ (strong field), electrons pair in the lower $t_{2g}$ set giving a low-spin complex. If $\Delta_{o} < P$ (weak field), electrons occupy $e_{g}$ singly to avoid pairing, giving a high-spin complex.

Write the spectrochemical series for common ligands (weak to strong field).

$$\ce{I^- < Br^- < S^2- < Cl^- < F^- < OH^- < C2O4^2- < H2O < NCS^- < NH3 < en < NO2^- < CN^- < CO}$$ Ligands on the left give small $\Delta$ (weak field); those on the right give large $\Delta$ (strong field).

For a $d^{6}$ octahedral ion, give the $t_{2g}/e_{g}$ occupation and number of unpaired electrons in both high-spin and low-spin cases.

High-spin (weak field): $t_{2g}^{4}e_{g}^{2}$, $4$ unpaired electrons. Low-spin (strong field): $t_{2g}^{6}e_{g}^{0}$, $0$ unpaired electrons (diamagnetic).

How does CFT explain the colour of a transition-metal complex?

A $d$-electron absorbs visible light of energy $E = \Delta_{o}$ and is promoted from $t_{2g}$ to $e_{g}$. The wavelength absorbed is $\lambda = \dfrac{hc}{\Delta_{o}}$; the colour seen is the complement of the absorbed colour. Larger $\Delta_{o}$ means absorption at shorter wavelength.

Why is $\ce{[Ti(H2O)6]^3+}$ purple/violet, and what transition is responsible?

$\ce{Ti^3+}$ is $d^{1}$ ($t_{2g}^{1}$). It absorbs green-yellow light (around $500\text{–}510\ \text{nm}$) for the single $t_{2g} \to e_{g}$ transition; the transmitted/complementary colour is purple. The absorption corresponds to $\Delta_{o} \approx 20300\ \text{cm}^{-1}$.

How does the magnitude of $\Delta_{o}$ affect the colour of a complex? Compare $\ce{[Cu(H2O)4]^2+}$ and $\ce{[Cu(NH3)4]^2+}$.

A stronger-field ligand gives a larger $\Delta_{o}$, so light of higher energy (shorter wavelength) is absorbed. $\ce{[Cu(H2O)4]^2+}$ is pale blue (weaker field) while $\ce{[Cu(NH3)4]^2+}$ is deep blue/violet (stronger field, larger $\Delta_{o}$).

What is the relationship between the colour absorbed and the colour observed for a complex?

The observed colour is the complementary colour of the light absorbed. For example, a complex that absorbs in the red region appears green, and one absorbing in the yellow-green region appears violet/purple.

How is the magnetic moment of a transition-metal complex calculated using the spin-only formula?

$$\mu_{\text{s}} = \sqrt{n(n+2)}\ \text{BM}$$ where $n$ is the number of unpaired electrons and BM is the Bohr magneton. Example: for $n = 4$, $\mu = \sqrt{4\times 6} = \sqrt{24} \approx 4.90\ \text{BM}$.

Distinguish paramagnetic and diamagnetic substances.

Paramagnetic substances have one or more unpaired electrons and are attracted by a magnetic field. Diamagnetic substances have all electrons paired and are weakly repelled by a magnetic field.

Calculate the spin-only magnetic moment of $\ce{Fe^3+}$ (high-spin, $d^{5}$).

High-spin $d^{5}$ has $n = 5$ unpaired electrons. $$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}$$

Calculate the spin-only magnetic moment of $\ce{[Ni(CN)4]^2-}$ and state its geometry.

$\ce{Ni^2+}$ is $d^{8}$; $\ce{CN^-}$ is strong-field, giving square planar $dsp^{2}$ with all electrons paired. Thus $n = 0$ and $$\mu = \sqrt{0(0+2)} = 0\ \text{BM}$$ (diamagnetic).

How many unpaired electrons and what magnetic moment does the high-spin $d^{4}$ ion $\ce{[Cr(H2O)6]^2+}$ have?

High-spin $d^{4}$ has configuration $t_{2g}^{3}e_{g}^{1}$, giving $n = 4$ unpaired electrons. $$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM}$$

What are the common geometries for coordination numbers 2, 4 and 6?

CN $2$: linear (e.g. $\ce{[Ag(NH3)2]^+}$). CN $4$: tetrahedral or square planar. CN $6$: octahedral (the most common geometry for transition-metal complexes).

What is the difference between structural isomerism and stereoisomerism in coordination compounds?

Structural (constitutional) isomers differ in the connectivity/arrangement of atoms or bonds (different chemical bonds). Stereoisomers have the same bonds/connectivity but differ in the spatial arrangement of the ligands around the metal.

Define ionization isomerism and give an example.

Ionization isomers give different ions in solution because a ligand inside the coordination sphere and a counter-ion outside are interchanged. Example: $\ce{[Co(NH3)5Br]SO4}$ (gives $\ce{SO4^2-}$) and $\ce{[Co(NH3)5SO4]Br}$ (gives $\ce{Br^-}$).

Define linkage isomerism and give an example.

Linkage isomerism arises when an ambidentate ligand coordinates through different donor atoms. Example: $\ce{[Co(NH3)5(NO2)]^2+}$ (nitro, bound via N) and $\ce{[Co(NH3)5(ONO)]^2+}$ (nitrito, bound via O).

Define coordination isomerism and give an example.

Coordination isomerism occurs in salts where both cation and anion are complex ions, and the ligands are distributed differently between the two metal centres. Example: $\ce{[Co(NH3)6][Cr(CN)6]}$ and $\ce{[Cr(NH3)6][Co(CN)6]}$.

Define hydrate (solvate) isomerism and give the classic example.

Hydrate isomers differ in whether water molecules are coordinated to the metal (inside the sphere) or present as water of crystallization (outside). Example: $\ce{[Cr(H2O)6]Cl3}$ (violet), $\ce{[Cr(H2O)5Cl]Cl2.H2O}$ (blue-green), $\ce{[Cr(H2O)4Cl2]Cl.2H2O}$ (dark green).

What are geometrical (cis–trans) isomers, and in which complexes do they occur?

Geometrical isomers differ in the relative positions of ligands: cis (adjacent, $90^{\circ}$) versus trans (opposite, $180^{\circ}$). They occur in square planar $\ce{MA2B2}$ complexes and octahedral $\ce{MA4B2}$ and $\ce{MA2B2C2}$ type complexes, but not in tetrahedral complexes.

What is fac–mer isomerism, and in which type of octahedral complex does it occur?

It occurs in octahedral $\ce{MA3B3}$ complexes. In the facial (fac) isomer the three identical ligands occupy one triangular face (mutually cis); in the meridional (mer) isomer they lie in a plane around the meridian (two trans, one cis).

What is optical isomerism in coordination compounds, and when does it arise?

Optical isomers are non-superimposable mirror images (enantiomers, chiral) that rotate plane-polarized light in opposite directions. It commonly arises in octahedral complexes with bidentate (chelate) ligands, e.g. $cis$-$\ce{[CoCl2(en)2]^+}$ and $\ce{[Co(en)3]^3+}$.

Why does the cis isomer of $\ce{[CoCl2(en)2]^+}$ show optical activity while the trans isomer does not?

The cis isomer lacks a plane of symmetry, so it is chiral and exists as two non-superimposable enantiomers (optically active). The trans isomer possesses a plane/centre of symmetry, making it superimposable on its mirror image (optically inactive, achiral).

Why are tetrahedral complexes of the type $\ce{MA2B2}$ unable to show geometrical (cis–trans) isomerism?

In a tetrahedron every ligand position is adjacent (equivalent) to every other position; there are no distinct cis or trans arrangements. Hence $\ce{MA2B2}$ tetrahedral complexes show only one form and no geometrical isomerism.