mgx₂\nvery low density\nhighly reactive\ndiatomic\nusing the periodic table, determine which element these…

Brief Explanations:

1. Analyze the formula ( MgX_2 ): Magnesium has a +2 oxidation state, so ( X ) should have a -1 oxidation state (since ( 2\times(-1)+2 = 0 )). Halogens (like fluorine, iodine) and oxygen (usually -2, but in some cases -1, but diatomic oxygen is ( O_2 ), and for ( MgO_2 ) it's peroxide, but oxygen's common diatomic is ( O_2 ) and its density is not very low compared to halogens in gaseous state? Wait, no: Fluorine (( F_2 )) is a diatomic gas, has very low density (gaseous), is highly reactive (halogens are reactive, fluorine is most reactive halogen), and forms ( MgF_2 ) (since F is -1, Mg is +2: ( Mg^{2+} + 2F^- \rightarrow MgF_2 )).
2. Iodine (( I_2 )) is a solid at room temp, so density is higher (not very low). Oxygen (( O_2 )): Magnesium oxide is ( MgO ) (O is -2, so ( MgO ), not ( MgO_2 ) usually, unless peroxide, but ( O_2^{2-} ) is peroxide, but oxygen's diatomic is ( O_2 ), and its reactivity: while reactive, the formula here is ( MgX_2 ), so X should be -1. Oxygen is -2 (or -1 in peroxide), so ( MgO_2 ) is magnesium peroxide, but oxygen's diatomic is ( O_2 ), but density of ( O_2 ) gas is higher than ( F_2 )? Wait, molar mass: ( F_2 ) is 38 g/mol, ( O_2 ) is 32 g/mol, ( I_2 ) is 254 g/mol (solid). Wait, density of gas: at same conditions, density is proportional to molar mass. So ( F_2 ) (38) vs ( O_2 ) (32): ( F_2 ) is denser than ( O_2 )? Wait, no, wait: density = mass/volume. At STP, 1 mole of gas is 22.4 L. So molar mass of ( F_2 ) is 38 g/mol, so density is 38/22.4 ≈ 1.696 g/L. ( O_2 ) is 32/22.4 ≈ 1.429 g/L. Wait, so ( O_2 ) is less dense? But the problem says "very low density". Wait, maybe I messed up. Wait, fluorine is a gas, very low density (gaseous), highly reactive, diatomic, and forms ( MgF_2 ). Iodine is solid, so density is high (as solid). Sodium is a metal, monatomic, not diatomic, so D is out. So between A (fluorine) and C (oxygen). Wait, the formula: ( MgX_2 ), so X has oxidation state -1. Fluorine is -1, so ( MgF_2 ) is correct. Oxygen is -2 (so ( MgO )) or -1 (peroxide, ( MgO_2 )). But the problem says "diatomic": ( F_2 ), ( O_2 ), ( I_2 ) are diatomic. Now, "very low density": gaseous diatomics. Fluorine and oxygen are gases, iodine is solid. So between F and O. Now, reactivity: fluorine is more reactive than oxygen (fluorine is the most reactive nonmetal). Also, the formula ( MgX_2 ): if X is O, then ( MgO_2 ) is magnesium peroxide, but is that common? ( MgF_2 ) is common (magnesium fluoride). Also, fluorine is a halogen, diatomic gas, very low density (gaseous), highly reactive. So the characteristics match fluorine. So option A (fluorine) is correct.

Answer:

A. fluorine