Tuning NTE: how the metal ion and crystal field stabilisation set bond strength

A stiffer link wobbles less, so it shrinks less when heated

Intuition

A skipping rope held tight at both ends barely bows. A loose rope bows a lot. A cyanide link tightly held by its metal ions swings only a little, so the framework shrinks only a little. A loosely held link swings widely and the framework shrinks a lot. So to tune the shrinking, you tune how tight the metal holds the link.

The previous topic said that the phonon energies, and so the sideways amplitudes, depend on the strength of the $M(1)\text{-}C{\equiv}N\text{-}M(2)$ bonding. Tightly bound cyanides give small amplitudes and small NTE. Loosely bound ones give large amplitudes and large NTE.

Lecture 37 asks what happens when you swap the metal $M$ in the Prussian-blue type family $M^{II}Pt^{IV}(CN)_6$. The result is

$\alpha_{Mn} > \alpha_{Fe} > \alpha_{Co} > \alpha_{Ni} < \alpha_{Cu} < \alpha_{Zn} \ll \alpha_{Cd}$

Read it as a valley. The NTE is largest at Mn, falls to its lowest at Ni, then rises again through Cu and Zn, and Cd is much larger still. The slide only gives this ordering and a plot, not numbers for each metal, so no values are quoted here.

Smaller ions bind harder, and crystal field stabilisation digs the valley at nickel

Two things change the bond strength as you go from Mn to Zn.

Ionic radius. Across the period the effective nuclear charge rises, so the ionic radius decreases. Smaller metal ions form stronger metal-ligand bonds because the electrostatic attraction is greater. Stronger bonds give smaller NTE. This explains the downward trend from Mn to Zn.

Crystal field stabilisation energy (CFSE). In an octahedral complex the d orbitals split. The $t_{2g}$ set sits at $-0.4\,\Delta_o$ and the $e_g$ set at $+0.6\,\Delta_o$, where $\Delta_o$ is the octahedral splitting. Electrons in $t_{2g}$ lower the energy and electrons in $e_g$ raise it. The net total is the CFSE. The lecture lists:

Ion	CFSE	Bond label on slide
Mn(II), d⁵ high spin	0	weak
Fe(II)	$-0.4\,\Delta_o$	partially weak
Co(II)	$-0.8\,\Delta_o$	partially strong
Ni(II)	$-1.2\,\Delta_o$	strong
Cu(II)	$-0.6\,\Delta_o$	intermediate
Zn(II)	0	weak

The CFSE contribution creates the trough shape on the $\alpha$ against $M$ plot. Ni(II) has the largest CFSE, hence the strongest bonds, and so the smallest NTE.

Try it: worked example

Check the Ni(II) CFSE from the level energies on the slide. Ni(II) is d^8 in an octahedral field, with t_2g at -0.4 Delta_o and e_g at +0.6 Delta_o. (A check using the lecture's levels.)

Show answer and steps

AnswerCFSE = -1.2 Delta_o, matching the slide.

1. Ni(II) has 8 d electrons. In the octahedral field they fill all six $t_{2g}$ places first, then put two in $e_g$.
2. $t_{2g}$ contribution: $6 \times (-0.4\,\Delta_o) = -2.4\,\Delta_o$.
3. $e_g$ contribution: $2 \times (+0.6\,\Delta_o) = +1.2\,\Delta_o$.
4. Total: $-2.4\,\Delta_o + 1.2\,\Delta_o = -1.2\,\Delta_o$.

Exam detail

Learn the six CFSE values, the trend equation, and the two-step explanation: radius gives the downward trend from Mn to Zn, CFSE gives the trough at Ni(II). The learning outcomes also ask for ways to vary thermal expansion, which here means changing the metal $M$ and the guests in the pores.

CFSE also shows up in lattice enthalpies and hydration enthalpies

Beyond the Born-Mayer equation, when there are unpaired electrons, the CFSE also adds to the lattice enthalpy. The lecture plots $-$lattice energy of $M^{II}Cl_2$ against the number of d electrons from Ca(II) (d⁰) to Zn(II) (d¹⁰). The curve rises from about 2260 kJ mol⁻¹ at d⁰ to a maximum near 2810 kJ mol⁻¹ at Cu(II) (d⁹), and sits above the Born-Mayer-type trend line across the Mn to Zn region. Hydration enthalpies show the same CFSE-related deviation. The slide gives no equation for this, and only refers to the Born-Mayer equation.

Pitfall

Do not read the $\alpha$ trend as a list of actual coefficients. The slide gives the order and a plot only. The lecture’s link between CFSE and thermal expansion is “stronger bonds, smaller NTE”.

Recall

Why is Ni(II) the best at resisting NTE in this family?

It has the biggest crystal field stabilisation energy, $-1.2\,\Delta_o$. That makes its metal-cyanide bonds the strongest. Strong bonds keep the sideways wobble small, so the framework shrinks the least.