Question

Question: Examine the ratio of atomic mass to atomic number for the elements with evenatomic number through calcium.

This ratio is approximately the ratio of the average mass number to the atomic number.

(a)Which two elements stand out as different in this set

of ten?

(b)What would be the “expected” atomic mass of argon, based on the correlation considered here?

(c)Show how the anomaly in the ordering of natural atomic masses of argon and potassium can be accounted for by the formation of “extra”${}^{\mathbf{40}}\mathbf{Ar}$ via

decay of data-custom-editor="chemistry" ${}^{\mathbf{40}}\mathbf{K}$ atoms.

Short answer

1. All the elements with even number of atomic numbers through calcium shows $\raisebox{1ex}{$\mathrm{M}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Z}$}\right.$ ratio of $2$. $\mathrm{Be}$and $\mathrm{Ar}$do not follow this trend.
2. The expected atomic mass of Argon should be $36.0\mathrm{u}$
3. Since ${}^{40}\mathrm{K}$decays to ${}^{40}\mathrm{Ar}$, the atomic mass of Argon increases.

Step-by-step solution

Ratio of atomic mass to atomic number

Let us make a table to find the ratio of atomic mass to atomic numbers for elements with even atomic number through calcium.

Element	Atomic number	Atomic mass	$\mathrm{M}:\mathrm{Z}$	Element	Atomic mass	Atomic number	$\mathrm{M}:\mathrm{Z}$
He	2	4.00	2	Mg	12	24.30	2.025
Be	4	9.01	2.25	Si	14	28.08	2.0
C	6	12.01	2	S	16	32.06	2.00
O	8	15.99	1.99	Ar	18	39.94	2.21
Ne	10	20.0	2	Ca	20	40.07	2.00

It is evident that the ratio of atomic mass to atomic number is approximately $2.$The anomalies are shown by $\mathrm{Be}$and $\mathrm{Ar}$

Expected Atomic mass of Argon

According to the correlation, the atomic mass of Argon should be two times the atomic number of argon. Therefore, the expected atomic mass of argon should be$36\mathrm{u}$.

Explanation for anomalous behavior of Argon

The $\raisebox{1ex}{$\mathrm{M}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Z}$}\right.$ ratio for potassium$=\frac{40}{19}=2.10$

Since the number of neutrons in ${}^{40}\mathrm{K}$ is higher than the number of protons, ${}^{40}\mathrm{K}$undergoes beta elimination to form.${}^{40}\mathrm{Ar}$.

. ${}_{19}{}^{40}\mathrm{K}\to {}_{18}{}^{40}\mathrm{Ar}+{{}_{-1}{}^0\mathrm{e}}^{-1}+\overline{\mathrm{V}}$

So as a result of presence of a isotope of Argon with higher atomic mass, the average atomic mass increase and the $\raisebox{1ex}{$\mathrm{M}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Z}$}\right.$ratio becomes greater than $2$.