An important reason for finding the sample standard deviation $S$ is that $S$ is an estimate of the population standard deviation $\sigma.$ For larger $n$ the estimate is more precise. Confidence intervals can help to give an idea of the precision.
Suppose you are sampling from a normal distribution with standard deviation $\sigma = 5.$
I will show you the method for finding a confidence interval for $\sigma$ based on a sample (because it's no secret), but for purposes of this answer the point is the length of the resulting confidence interval. (You can skip the 'Method' section if it gives more technical detail than you want.)
Method: One can show that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1),$ the chi-squared distribution with $n-1$ degrees of freedom. From that distribution I can find boundaries $L$ and $U$ with $P\left(L < \frac{(n-1)S^2}{\sigma^2} < U\right) = 0.95.$ By manipulating inequalities, this becomes
$P\left(\frac{(n-1)S^2}{U} < \sigma^2 < \frac{(n-1)S^2}{L}\right) = 0.95.$ and
$P\left(S\sqrt{\frac{n-1}{U}} < \sigma < S\sqrt{\frac{n-1}{L}}\right) = 0.95.$
Finally, we say that $\left(S\sqrt{\frac{n-1}{U}}. S\sqrt{\frac{n-1}{L}}\right)$
is a 95% confidence interval for $\sigma.$ Roughly speaking, this means that for 05% of samples this interval includes $\sigma.$
Computation: Suppose I have sample standard deviations $S$ for two samples
from a normal population with mean $\sigma = 7.$ The first sample is of size $n=10$ and the second is of size $n = 100.$ Below are simulated samples and their respective confidence intervals, using R statistical software.
Size 10: $S = 7.25.$ The 95% confidence interval is $(4.99, 13.24)$ of length $9.25.$
set.seed(330)
s = sd(rnorm(10, 50, 7)); s
[1] 7.252305
CI = sqrt(9*s^2/qchisq(c(.975,.025),9))
CI; diff(CI)
[1] 4.988391 13.239882 # CI
[1] 8.251491 # length of CI
Size 100: $S = 7.20.$ The 95% confidence interval is $(6.32, 8.36)$ of length $2.04.$
set.seed(331)
s = sd(rnorm(100, 50, 7)); s
[1] 7.200381
CI = sqrt(99*s^2/qchisq(c(.975,.025),99))
CI; diff(CI)
[1] 6.321984 8.364505
[1] 2.042521
And just for illustration, without showing the computations, with $n = 2,$ I got $S = 3.34,$ giving a confidence interval $(1.49, 106.60)$ of length $105.11.$
The confidence interval does include $\sigma = 7,$ but it is too long to be of
any practical use.
I wouldn't say that your readers would be 'irked' if you
give a sample standard deviation for a sample of size $n = 2,$ but they might
wonder why you bothered to report it. (Maybe it's better just to give the two values.)
Finally, there is another difficulty with sample standard deviations based on
samples of only two observations. they are seriously biased downward as estimates of $\sigma.$
On average, $S$ based on two normal observations will have the value
$0.798\sigma.$ That is $E(S) \approx 0.798\sigma.$ [It is always true, for all $n,$ that $E(S) < \sigma,$ but for $n$ as large as 10, the downward
bias is seldom of practical importance, and the bias becomes truly negligible for large $n.]$
