# Why does higher volatility for ATM Call Option lead to a lower risk-neutral probability of expiring ITM?

This is a follow-up question on the discussion in the thread here, from which I borrow the graph below depicting $$N(d_2)$$ (i.e. the risk neutral probability of a Call option expiring in the money) against Volatility:

I have the following two questions:

Question (i): As we can see in the graph above, increasing volatility eventually leads to a decreasing risk-neutral probability of an option expiring in the money (irrespective of whether the option is struck OTM, ITM, or ATM): in fact, the claim in the comments in the original thread is that "The probability of ending ITM under the risk-neutral measure should tend to 0".

Why would intuitively a large volatility lead to a reduction in the risk-neutral probability of an option expiring in the money, with a zero limit? Is it because the "large" (i.e. "infinite") volatility guarantees that the underlying price hits zero, effectively defaulting?

Question (ii): In the graph, we see that the ATM option "dislikes" increasing volatility with regard to the probability of ending up ITM, across the entire Volatility domain (not just for large values of vol).

Intuitively, this doesn't make much sense to me. I can see why the ITM option "dislikes" increasing vol, and I can see why the OTM option (initially) "likes" increasing vol.

But why does (intuitively) the ATM option dislike vol with regard to the (risk-neutral) probability of expiring in the money?