Question

Before Mendeleev published his periodic table, German scientist Johann Döbereiner grouped elements with similar properties into "triads," in which the unknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities: (a) The atomic mass of \(\mathrm{K}\) from the atomic masses of \(\mathrm{Na}\) and \(\mathrm{Rb}\) (b) The melting point of \(\mathrm{Br}_{2}\) from the melting points of \(\mathrm{Cl}_{2}\) \(\left(-101.0^{\circ} \mathrm{C}\right)\) and \(\mathrm{I}_{2}\left(113.6^{\circ} \mathrm{C}\right)\left(\right.\) actual value \(\left.=-7.2^{\circ} \mathrm{C}\right)\)

Short answer

Predicted atomic mass of \(\mathrm{K}\) is 54.23 u. Predicted melting point of \(\mathrm{Br}_{2}\) is 6.3 \degree C.

Step-by-step solution

Identify the Atomic Mass of \mathrm{K}

Given the atomic masses of \(\mathrm{Na}\) (22.99 u) and \(\mathrm{Rb}\) (85.47 u), use the triad method to predict the atomic mass of \(\mathrm{K}\).

Calculate the Average

To find the predicted atomic mass of \(\mathrm{K}\), take the average of the atomic masses of \(\mathrm{Na}\) and \(\mathrm{Rb}\): \[\ \frac{22.99 + 85.47}{2} \]

Solve for the Atomic Mass

Calculate the average: \[ \frac{22.99 + 85.47}{2} = 54.23 \] Thus, the predicted atomic mass of \(\mathrm{K}\) is 54.23 u.

Identify the Melting Points for \mathrm{Br}_{2}

Given the melting points of \(\mathrm{Cl}_{2}\) (-101.0 \degree C) and \(\mathrm{I}_{2}\) (113.6 \degree C), use the triad method to predict the melting point of \(\mathrm{Br}_{2}\).

Calculate the Average for Melting Point

To find the predicted melting point of \(\mathrm{Br}_{2}\), take the average of the melting points of \(\mathrm{Cl}_{2}\) and \(\mathrm{I}_{2}\): \[ \frac{-101.0 + 113.6}{2} \]

Solve for Melting Point

Calculate the average: \[ \frac{-101.0 + 113.6}{2} = 6.3 \degree C \] Thus, the predicted melting point of \(\mathrm{Br}_{2}\) is 6.3 \degree C.

Key concepts

Döbereiner's Triads

Before the periodic table as we know it was created, Johann Döbereiner, a German scientist, observed that certain groups of three elements (which he called 'triads') exhibited similar chemical properties. He found that the unknown properties of an element could often be predicted by averaging the known properties of the other two elements in the triad. This was an early attempt to bring order to the chemical elements and was a significant step towards the development of the modern periodic table.
For instance, if we take three elements like Na, K, and Rb, which are known to form a triad, we can use the known properties (such as atomic masses) of Na and Rb to predict the corresponding property of K.

Atomic Mass Prediction

Using Döbereiner's triad method, we can predict the atomic mass of an unknown element based on the known masses of two other elements in the same triad.
To predict the atomic mass of potassium (\(\mathrm{K}\)), we are given the atomic masses of sodium (\(\mathrm{Na}\)) as 22.99 u and rubidium (\(\mathrm{Rb}\)) as 85.47 u.
Here's how you can do it:

• First, take the sum of the known atomic masses: \(22.99 + 85.47\)
• Next, find the average by dividing by 2: \(\frac{22.99 + 85.47}{2} = 54.23\)

So, based on this calculation, the predicted atomic mass of potassium (\(\mathrm{K}\)) would be 54.23 u. This method aligns with the observed trends within the triad group and helps highlight the predictive power of Döbereiner's approach.

Melting Point Prediction

In addition to predicting atomic mass, Döbereiner's triads can also be used to predict other properties, such as melting points. For example, let's predict the melting point of bromine (\(\mathrm{Br_{2}}\)) using the known melting points of chlorine (\(\mathrm{Cl_{2}}\)) which is \(-101.0 ^\text{°C}\), and iodine (\(\mathrm{I_{2}}\)) which is \(113.6 ^\text{°C}\).
To calculate the predicted melting point of bromine (\(\mathrm{Br_{2}}\)), we follow a similar averaging method:

• First, sum the known melting points: \(-101.0 ^\text{°C} + 113.6 ^\text{°C}\)
• Then, divide by 2 to find the average: \(\frac{-101.0 + 113.6}{2} = 6.3 ^\text{°C}\)

According to this method, the predicted melting point of bromine (\(\mathrm{Br_{2}}\)) is approximately \(6.3 ^\text{°C}\). However, it's important to note that the actual melting point of bromine is \(-7.2 ^\text{°C}\), indicating some limitations in the predictive power of Döbereiner's triads as it is a simple and early method.