When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have: a. different electron configuration from each other.
b. an unchanged electron configuration.
c. the same charge.
d. the same electron configuration (isoelectronic) as neon.
[Definition: The word isoelectronic means that when you write out the electron configuration they are the same. An exam would be He and Li whereby both of them have 2 electrons and therefore they are both are 1s2 in their electron configurations.]

d. the same electron configuration (isoelectronic) as neon.

Explanation:

Isoelectronic species are defined as the molecules which have the same number of electrons.

Atomic number of nitrogen is 7 and thus has 7 electrons. Nitrogen has electronic configuration of 2,5 and thus can gain 3 electrons and thus will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of oxygen is 8 and thus has 8 electrons. Oxygen has electronic configuration of 2,6 and thus can gain 2 electrons and thus will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of flourine is 9 and thus has 9 electrons. Flourine has electronic configuration of 2,7 and thus can gain 1 electron and thus will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of sodium is 11 and thus has 11 electrons. Sodium has electronic configuration of 2,8,1 and thus can lose 1 electron and thus will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of magnesium is 12 and thus has 12 electrons. Magnesium has electronic configuration of 2,8,2 and thus can lose 2 electrons and thus will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of aluminium is 13 and thus has 13 electrons. Aluminium has electronic configuration of 2,8,3 and thus can lose 3 electrons and thus will have electronic configuration of 2,8 ( same as that of neon)

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2