The electronic configuration of an element is ALS2, 3s2 3p2. What is the element? Is it a metal or non-metal? Explain.

The number of electron shells occupied by electrons determines the period.
The configuration 2, 8, 8, 2 indicates 4 electron shells.
Therefore, element X is in Period 4.

For main group elements, the number of valence electrons (electrons in the outermost shell) determines the group number.
The outermost shell has 2 electrons.
Therefore, element X is in Group 2.

b) Is the element X a metal or non-metal? Explain.

Element X has 2 valence electrons. Elements in Group 2 are known as alkaline earth metals.

Metals tend to lose their valence electrons to achieve a stable electron configuration. Element X will readily lose its 2 valence electrons.

c) Which ion will be formed by element X to achieve a stable electron configuration?

To achieve a stable octet, element X will lose its 2 valence electrons.

Losing 2 electrons results in a net charge of +2.
The ion formed will be $\boxed{\text{X}^{2+}}$.

Question 2:

a) If the element V is in period 4 and group 5, write the electronic configuration of V.

Period 4 means 4 electron shells. Group 5 (main group, also known as Group 15) means 5 valence electrons.
The electron configuration for an element in Period 4, Group 15 (like Arsenic, As) is:
Shell 1: 2 electrons
Shell 2: 8 electrons
Shell 3: 18 electrons (including d-block electrons)
Shell 4: 5 electrons (valence electrons)
Total electrons = $2 + 8 + 18 + 5 = 33$.

The electronic configuration of V is $\boxed{2, 8, 18, 5}$.

b) What is the family name (class name) for the group to which V belongs?

Element V belongs to Group 5 (or Group 15 in IUPAC notation).

The family name for elements in Group 15 is $\boxed{\text{Pnictogens}}$.