The atomic numbers of elements A,B,C and D are z–1, z, z+1 and z+2, respectively. If 'B' is a noble gas, choose the correct answers from the following statements:
(a) 'A' has higher electron affinity.
(b) 'C' exists in +2 oxidation state.
(c) 'D' is an alkaline Earth metal.

We know that  ' B ' is a noble gas and the atomic number of ' B ' is Z. We know that they are the most stable elements in the periodic table and they neither accept electrons nor lose electrons. 

When one electron is removed from a noble gas, the atomic number is reduced to $\mathrm{Z}-1$. The element with $\mathrm{Z}-1$atomic number is ' A '.  It only takes one more electron to achieve noble gas electronic configuration, so there is a strong tendency to gain electrons. As a result, the element must have a high electron affinity. So statement (1) is correct.

Now, We know that 

The atomic number of the element C is  $\mathrm{Z}+1$. Its atomic number will be the same as that of a noble gas after losing one electron, so it will readily lose one electron and not more than one. The element C will exis +1 oxidation state and not +2. Hence, Statement (2) is incorrect.

Now,

Similarly, element D will lose two electrons because it will attain noble gas electronic configuration after losing two electrons. We know group number 2 elements lose 2 electrons and are known as alkaline earth metals, so the statement (3) is also correct.

Since Statement (1) and (2) are correct, Option(c) is the correct answer.