Potassium has a density of $\pu{0.86g/cm3}$ and sodium has a density of $\pu{0.97g/cm3}$,even though potassium is below sodium and one might expect the alkali metals to exhibit monotonously increasing densities down the group. Why is this discrepancy the case? All of the answers I found were either unrelated or vague.

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    $\begingroup$ Googling leads quickly to things like quora.com/… $\endgroup$
    – Jon Custer
    Commented Jun 6, 2021 at 2:36
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    $\begingroup$ I don't know why you were downvoted, although phrasing of the question could definitely be improved. I could not find an answer on the site (although perhaps should have looked a little longer, sometimes answers are hidden within posts that are actually asking something else). The answer (obviously) has to do with differences in electronic structure, and the presence of low energy 3d orbitals. It would be nice to see a more detailed answer to this, it is unlikely to be simple. To get to the heart of the question you need to look into the electronic structure of atoms. $\endgroup$
    – Buck Thorn
    Commented Jun 6, 2021 at 9:36
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    $\begingroup$ @JonCuster after some 20 minutes searching I could not find a particularly good answer (other than trivial ones, or something like "3d orbitals are available" which is next to trivial). Answers on other sites are often terrible. A good answer might however require some discussion of electronic structure beyond an introductory chem 'level. $\endgroup$
    – Buck Thorn
    Commented Jun 6, 2021 at 9:44
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    $\begingroup$ Related: How does density of alkali metals increase down the group? $\endgroup$
    – Loong
    Commented Jun 6, 2021 at 14:19
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    $\begingroup$ @BuckThorn I doubt it is electronic structure, more likely it's just that K is just lighter than you expect - after all it is one of the elements that's put in the wrong place if you order things by atomic weight, being lighter than Argon. And I agree "3d orbitals are available" is dreadful, I see that more of an explanation for K to be denser than expected, due to the 3d contraction $\endgroup$
    – Ian Bush
    Commented Jun 6, 2021 at 14:36

2 Answers 2


Although density may be a relatively easily measurable property of solid materials, it may not suggest the most fundamental relationship between mass and volume for the elements. Molar volume (the volume required to contain $6 \times 10^{23}$ atoms of each the elements) is pictured in the graph below (WebElements):

Trend of Molar volume of elements

It is amazing that ~100 elements can be arranged in a table with so many regularities - and even the irregularities can be explained if you try hard enough. There are some that are more unexpected and greater than the $\ce{Na-K}$ difference.

It is good that both $\ce{Na}$ and $\ce{K}$ have the same crystal structure (body-centered cubic). To explain the $\ce{Na-K}$ difference, some other differences should also be taken into account. Both have one electron outside a completed shell system - but sodium's $\mathrm{3s}$ electron (2 nodes) is dealing with a different kind of shielding compared to potassium's $\mathrm{4s}$ electron (3 nodes). In fact, the outer $\mathrm{3s}$ electron of sodium is ionized at $\pu{5.139 eV} \ (\pu{496 kJ/mol})$, while potassium is ionized more easily, at $\pu{4.34 eV} \ (\pu{418.8 kJ/mol})$. Since potassium holds onto its last electron so weakly (relative to sodium), the "sea of electrons" that helps hold the crystalline metal together isn't attracting the potassium ions as well as the sea does for sodium - not binding the crystal together as tightly. Result: lower density. Complete explanation? Probably not, but it's a start.

Although the enthalpy of fusion for both $\ce{Na}$ and $\ce{K}$ is given as $\pu{20.5 kJ/mol}$, the melting points are $\pu{63.38 ^\circ C}$ of $\ce{K}$ and $\pu{97.7 ^\circ C}$ for $\ce{Na}$, suggesting that sodium is slightly more tightly bound in the crystal. Even the boiling points $\pu{759 ^\circ C}$ of $\ce{K}$ vs $\pu{883 ^\circ C}$ of $\ce{Na}$ suggest that sodium atoms are bound more tightly than potassium in the liquid.

Interestingly, if you eliminate the $\mathrm{3s}$ electron of sodium and the $\mathrm{4s}$ electron of potassium, you drop to neon, which has a solid density of 1.444 and argon, with a solid density of 1.616. These two densities are about what you would expect; i.e., increasing as you go down the column. Note that the corresponding molar volumes are $\pu{13.23 cm^3}$ for $\ce{Ne}$ and $\pu{22.56 cm^3}$ for $\ce{Ar}$.

The same trend is found in the molar volumes of $\ce{Na}$ $(\pu{23.78 cm^3})$ and $\ce{K}$ $(\pu{45.94 cm^3})$, and the "discrepancy" is obscured.


Although I liked James Gaidis's answer, I do not agree with some of arguments because they are all parts of one or more continuous trends. For instance, look at the melting points and boiling points trends of alkali metals and other trends as illustrated in the following table:

$$ \begin{array}{l|ccc} \hline \bf{\text{Physical property}} & \ce{Li} && \ce{Na} && \ce{K} && \ce{Rb} && \ce{Cs} \\ \hline \text{Boiling point, }\pu{^\circ C} & 180.54 & \gt & 97.72 & \gt & 63.38 & \gt & 39.31 & \gt & 28.44 \\ \text{Melting point, }\pu{^\circ C} & 1342 & \gt & 883 & \gt & 759 & \gt & 688 & \gt & 671 \\ \text{Heat of fusion, }\pu{kJ mol-1} & 3.00 & \gt & 2.60 & \gt & 2.32 & \gt & 2.19 & \gt & 2.09 \\ \text{Heat of vaporization, }\pu{kJ mol-1} & 136 & \gt & 97.42 & \gt & 79.1 & \gt & 69 & \gt & 66.1 \\ \text{First ionization energy, }\pu{kJ mol-1} & 520.2 & \gt & 495.8 & \gt & 418.8 & \gt & 403.0 & \gt & 375.7 \\ \text{Density, }\pu{g cm-3} & 0.534 & \lt & 0.968 & \gt & 0.89 & \lt & 1.532 & \lt & 1.93 \\ \hline \end{array}\\ \text{Source of the data: https://en.wikipedia.org/wiki/Alkali_metal} $$

As shown in the above table, only density has broken the trend. Therefore the reasoning for this density inconsistance using other physical measures in the table is somewhat valid but not solid (see Buck Thorn's comments elsewhere). I'd like to give a different angle on this uneven trend.

Density is generally defind as the mass of unit volume $(\rho = \frac{m}{v})$. When you consider the density of an element, it is better to consider the molar mass and molar volume in place of mass and volume of other solids, respectively. Since the question is about the density difference of sodium $(\ce{Na})$ and potassium $(\ce{K})$, let's concentrate on alkali metal group. As usual and easity understand, the mass of a metal increases going down on the group. It is also clear that the principle quantam number $(n)$ increases as going down the group and hence, volume should have increased as well (adding more shells). The question is, the rate of increment of mass is comparable with the rate of increment of volume in each element? Let's make a table for this trend:

$$ \begin{array}{l|ccc} \hline \bf{\text{Property}} & \ce{_{3}Li} & \ce{_{11}Na} & \ce{_{19}K} & \ce{_{37}Rb} & \ce{_{55}Cs} \\ \hline \text{Molar mass, }\pu{g mol-1} & 6.94 & 22.99 & 39.10 & 85.47 & 132.91 \\ \text{Percent mass increment} & - & \approx 231.3\% & \approx 70.1\% & \approx 118.6\% & \approx 55.5\% \\ \text{Proton/nutron} & 3/4 & 11/12 & 19/20 & 37/48 & 55/78 \\ \text{Calculated atomic radii, }\pu{pm} & 167 & 190 & 243 & 265 & 298 \\ \text{measured covalent radii, }\pu{pm} & 145 & 180 & 220 & 235 & 260 \\ a\text{ (Side of the unit cell), }\pu{pm} & 351 & 429.6 & 532.8 & 598.5 & 614.1 \\ \text{Atomic radii from $a$, }\pu{pm} & 152 & 186 & 231 & 242 & 266 \\ \text{Percent volume increment (exp)} & - & \approx 22.4\% & \approx 24.2\% & \approx 4.8\% & \approx 9.9\% \\ \hline \end{array}\\ \text{Source of the data: https://en.wikipedia.org/wiki/Atomic_radius} $$

In the table, the atomic radii computed from theoretical models (calculated atomic radii) are from published work of Enrico Clementi and others (Ref.1) while the measured covalent radii for the metals are from published work of J. C. Slater (Ref.2). The experimental mattalic radii are from published crystal data.

As illustrated in the above table, when going from $\ce{Li}$ to $\ce{Na}$, the mass increased by $231\%$ but volume increased by only $22\%$. Thus, without a doubt, $\ce{Na}$ should have higher density than that of $\ce{Li}$. Similarly, we can argue that same principle would apply when going from $\ce{K}$ to $\ce{Rb}$ and going from $\ce{Ru}$ to $\ce{Cs}$ to explain the trend. Howevr, to explain the broken trend when going from $\ce{Na}$ to $\ce{K}$, we shoul look at the rates.

When going from $\ce{Li}$ to $\ce{Na}$ and going from $\ce{Na}$ to $\ce{K}$, the percent volume change is basically the same (22% vs 24%). Yet, with that increment togethher with $+131\%$ mass increment (more than 2.3 time increment) has given only $0.964 - 0.534 = 0.43$ density change. Therefore, one can expect much less density increment (if there is any) when going from $\ce{Na}$ to $\ce{K}$ because the mass increment is minimal (less than 100%) when compared to the percent volume increment.

On the other hand, when going down through a group, each period of the periodic table would add one shell. However, when going down from $\ce{Li}$ to $\ce{Na}$ and from $\ce{Na}$ to $\ce{K}$, the relavent nucleous gets only 7 protons. After that the addition of protons increases significantly (from $\ce{K}$ to $\ce{Rb}$, gains 18 protons and $\ce{Rb}$ to $\ce{Cs}$, gain another 18 protons). Near equal percent volume increment in either case suggests that the effects of opposite charge attractions and electron repulsions ate minimal in both cases. Bottom line is there is a clear discripancy of percent mass increment and percent volume increment at transition from $\ce{Na}$ to $\ce{K}$ stage.

I have calculated the percent volume increment using experimental crystal data as follows:

All of alkali metals crystallize with same crystal packing called body-centered cubic (bcc), which consists of one atom in the center of the cube and eight other atoms at the eight corners of the cube surrounding the cetral atom, each of them touching the central atom. Thus, the dioganal of the cube is equal to $4r$ where $r$ is the radius of the atom. If the length of each side of the cube is $a$, then diagonal is equal to $\sqrt{3}a$. Thus,

$$r = \frac{\sqrt{3}a}{4} \tag1$$

Keep in mind that $a$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For instance, it has been reported that the closest $\ce{Na-Na}$ separation is $\pu{372 pm}$ implying a sodium metallic radius of $\pu{186 pm}$ while the closest $\ce{K-K}$ separation is $\pu{461 pm}$ implying a potassium metallic radius of $\pu{231 pm}$ (as both calculated by using the equation $(1)$). The molar volume of the metal is not just the calculated volume of an atom considering it is a sphere times Avegadro number. The void volume between the atoms must be put into concern, packing is a big factor in volume calculations, thus density of the metal. The following calculations would show you the effect of the packing on the density:

If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each center atom shares with only one unit cell. Thus, total atoms per unit cell is $8 \times \frac{1}{8} + 1 = 2$. The volume of the unit cell is $a^3$, which is considered to be the volume of $\pu{2 atoms}$. Now, you can calculate the molar volume of each metal ($V_\ce{M}$):

$$V_\ce{M} = a_\ce{M}^3 \times \frac{1}{\pu{2 atoms}} \times N_A \tag2$$

The units of the volume would be depend on the units you have used for $a$. For instance, for $\ce{Na}$:

$$V_\ce{Na} = (\pu{429.6 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{2 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} \\ = (429.6)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{23.87 cm3 mol-1}$$

Similarly, for $\ce{K}$ and other alkali metals:

$$V_\ce{K} = (532.8)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{45.54 cm3 mol-1}$$ $$V_\ce{Li} = (351)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{13.02 cm3 mol-1}$$ $$V_\ce{Rb} = (558.5)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{52.45 cm3 mol-1}$$ $$V_\ce{Cs} = (614.1)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{69.73 cm3 mol-1}$$

Once we know the molar volume of each metal, we can calculate the density ($\rho$) by using $\rho=\frac{\text {molar mass}}{\text {molar volume}}$:

$$\rho_\ce{Li} = \frac{\pu{6.94 g mol-1}}{\pu{13.02 cm3 mol-1}} = \pu{0.533 g cm-3}$$

$$\rho_\ce{Na} = \frac{\pu{22.99 g mol-1}}{\pu{23.87 cm3 mol-1}} = \pu{0.963 g cm-3}$$

$$\rho_\ce{K} = \frac{\pu{39.10 g mol-1}}{\pu{45.54 cm3 mol-1}} = \pu{0.859 g cm-3}$$

$$\rho_\ce{Rb} = \frac{\pu{85.47 g mol-1}}{\pu{52.45 cm3 mol-1}} = \pu{1.630 g cm-3}$$

$$\rho_\ce{Cs} = \frac{\pu{132.91 g mol-1}}{\pu{69.73 cm3 mol-1}} = \pu{1.906 g cm-3}$$

These calculated values are consistence with the experimental values. Therefore, it is safe to say that the increasing trend of alkali metals are broken at $\ce{Na}$ and $\ce{K}$. After that, it again continues as follows:

$$\rho_\ce{Li} \lt \rho_\ce{Na} \gt \rho_\ce{K} \lt \rho_\ce{Rb} \lt \rho_\ce{Cs}$$


  1. E. Clementi, D. L. Raimondi, and W. P. Reinhardt, "Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons," J. Chem. Phys. 1967, 47(4), 1300–1307 (DOI: https://doi.org/10.1063/1.1712084).
  2. J. C. Slater, "Atomic Radii in Crystals," J. Chem. Phys. 1964, 41(4), 3199–3205 (DOI: https://doi.org/10.1063/1.1725697).
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    $\begingroup$ BTW, thanks for the edit on my answer! I totally flubbed the bcc. $\endgroup$ Commented Jun 9, 2021 at 13:49

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