H.C.F. of two consecutive even numbers is :
Consider two consecutive even numbers. They are of the form 2n,2n+2 for some integer n
2n=2×n
2n+2=2×(n+1)
The numbers n,n+1, being two consecutive integers, are relatively prime
They do not have any factor other than 1 common to them.
Hence HCF of two consecutive even numbers is 2.
So the correct answer is option D