Can IQR ever be larger than standard deviation? As I understand it the IQR specifies the dispersion of the data by taking the difference between Q3 and Q1 (i.e., the range that the middle 50% of the data lies), while std specifies dispersion using all the data, to where 68% of the data belong. Does this imply that the std is always larger than (or equal to?) iqr? 
 A: Your argument about $68\%$ of the density contained within $1$ standard deviation (of the mean) is true for the Normal distribution, but not in general. There are examples where the standard deviation exceeds the IQR and examples in the other direction as well. 
Let $X$ have a t-distribution with $v$ degrees of freedom. 
The standard deviation of $X$ is
$$\sigma = \sqrt{\text{Var}(X)} = \sqrt{\frac{v}{v-2}}$$
The IQR can be found using the quantile function of the t-distribution. In R, we have
qt(.75, v) - qt(.25, v)

We have that $\sigma = IQR$ for $v\approx 3.61$. When $v < 3.61$, we have $\sigma > IQR$ and we have $\sigma < IQR$ otherwise. Since a normal distribution can be achieved in the limit as $v\rightarrow \infty$, this demonstrates that $\sigma < IQR$ for a normal distribution.


Sal considers the interesting question, can IQR be larger than $2\sigma$?. 
Indeed, this will occur for the symmetric Beta distribution when the shape parameter is small. 
Let $X \sim Beta(\alpha, \alpha)$. Then $IQR = 2\sigma$ when $\alpha = 0.5$, $IQR > 2\sigma$ when $\alpha < 0.5$ and $IQR < 2\sigma$ otherwise. 


We have established examples, where $IQR > \sigma$ and examples where $IQR > 2\sigma$. It is not possible, however, for $IQR$ to exceed $4\sigma$ when the distribution of $X$ is symmetric about its mean $\mu$.  Applying Chebyshev's Inequality, we have 
$$P(|X-\mu| \leq 2\sigma) \geq \frac{3}{4}.$$
If the density function of $X$ is symmetric about $\mu$, we have
$$P(|X-\mu| \leq \text{IQR}/2) = \frac{3}{4}.$$
This implies that $2\sigma \geq \text{IQR}/2$, or equivalently
$$\text{IQR} \leq 4\sigma$$
A: Trying some toy distributions, it appears that IQR is often less than standard deviation (sd), but that IQR will be larger than sd for bimodal distributions.
If the question is supposed to be asking Is IQR ever larger than 2 times sd ?, then I have been unsuccessful finding distributions that satisfies this question.  But for distributions meeting this criterion, see the answer by @knrumsey-ReinstateMonica .
A = c(1,2,99,100)

sd(A)

   ### 56.58327

IQR(A)

   ### 97.5

set.seed(12345)

B = c(rnorm(25,10,1 ), rnorm(25,15,1))

hist(B)

sd(B)

   ### 2.915855

IQR(B)

   ### 5.586538


There's also the case of some binomial distributions, such as mentioned by whuber in the comments to the original post.
set.seed(12345)

C = rbinom (100, 30, 0.26)

hist(C)

sd(C)

   ### 2.569125

IQR(C)

   ### 3


