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The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265).  (Unfortunately, I haven't been able to find an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

Added: This, does, however, depend on the fact that the sampling is from the normal distribution. The authors classify the distribution of extreme spacings as ES short, if $S_{n:n}$ converges in probability to $0$ as $n \to \infty$; ES medium, if $S_{n:n}$ is bounded but non-zero in probability; and ES long, if $S_{n:n}$ diverges in probability. While the $N(0,1)$ distribution has ES short right tails, the authors show that the gamma family has ES medium right tails (see Shai Covo's answer for the special case of the exponential) and the Pareto family has ES long right tails.

The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Here's an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

Added: This, does, however, depend on the fact that the sampling is from the normal distribution. The authors classify the distribution of extreme spacings as ES short, if $S_{n:n}$ converges in probability to $0$ as $n \to \infty$; ES medium, if $S_{n:n}$ is bounded but non-zero in probability; and ES long, if $S_{n:n}$ diverges in probability. While the $N(0,1)$ distribution has ES short right tails, the authors show that the gamma family has ES medium right tails (see Shai Covo's answer for the special case of the exponential) and the Pareto family has ES long right tails.

The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Unfortunately, I haven't been able to find an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

Added: This, does, however, depend on the fact that the sampling is from the normal distribution. The authors classify the distribution of extreme spacings as ES short, if $S_{n:n}$ converges in probability to $0$ as $n \to \infty$; ES medium, if $S_{n:n}$ is bounded but non-zero in probability; and ES long, if $S_{n:n}$ diverges in probability. While the $N(0,1)$ distribution has ES short right tails, the authors show that the gamma family has ES medium right tails (see Shai Covo's answer for the special case of the exponential) and the Pareto family has ES long right tails.

The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Unfortunately, I haven't been able to find an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

Added: This, does, however, depend on the fact that the sampling is from the normal distribution. The authors classify the distribution of extreme spacings as ES short, if $S_{n:n}$ converges in probability to $0$ as $n \to \infty$; ES medium, if $S_{n:n}$ is bounded but non-zero in probability; and ES long, if $S_{n:n}$ diverges in probability. While the $N(0,1)$ distribution has ES short right tails, the authors show that the gamma family has ES medium right tails (see Shai Covo's answer for the special case of the exponential) and the Pareto family has ES long right tails.

The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Unfortunately, I haven't been able to find an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Unfortunately, I haven't been able to find an online copy.)

Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.

The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.

Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)

In other words, (2) is more surprising.

Added: This, does, however, depend on the fact that the sampling is from the normal distribution. The authors classify the distribution of extreme spacings as ES short, if $S_{n:n}$ converges in probability to $0$ as $n \to \infty$; ES medium, if $S_{n:n}$ is bounded but non-zero in probability; and ES long, if $S_{n:n}$ diverges in probability. While the $N(0,1)$ distribution has ES short right tails, the authors show that the gamma family has ES medium right tails (see Shai Covo's answer for the special case of the exponential) and the Pareto family has ES long right tails.