Correct: This statement accurately describes the fundamental process of ionic bond formation. The metal atom loses its valence electrons (low ionization energy) and the nonmetal atom gains them (high electron affinity), creating cations and anions which are then held together by electrostatic attraction.

A: Incorrect. Sharing of electrons describes covalent bonding, not ionic bonding.

B: Incorrect. This also describes covalent bonding, focusing on the attraction for shared electrons.

D: Incorrect. The delocalization of electrons into a 'sea' describes metallic bonding, not ionic bonding.

Correct: Elements in Group 2, the alkaline earth metals, have two valence electrons. They tend to lose both of these electrons to achieve a stable, noble-gas electron configuration, resulting in the formation of cations with a +2 charge.

A: Incorrect. Group 1 elements have one valence electron and form +1 ions.

C: Incorrect. Group 16 elements have six valence electrons and form -2 ions.

D: Incorrect. Group 17 elements have seven valence electrons and form -1 ions.

Correct: Elements in the same column of the periodic table tend to form analogous compounds. Strontium (Sr) is in the same group (Group 2) as calcium (Ca), so it forms a $$Sr^{2+}$$ ion. Iodine (I) is in the same group (Group 17) as fluorine (F), so it forms an $$I^-$$ ion. Therefore, the formula is $$SrI_2$$.

A: Incorrect. This formula would imply a +1 charge for strontium, which is not typical for a Group 2 element.

B: Incorrect. This formula would imply a -2 charge for iodine and a +1 charge for strontium, which are both incorrect.

D: Incorrect. This formula would imply a +3 charge for strontium, which is incorrect.

Correct: A nonmetal with 5 valence electrons (Group 15) will typically gain 3 electrons to form an anion with a -3 charge ($$Z^{3-}$$). Magnesium (Mg), from Group 2, forms a cation with a +2 charge ($$Mg^{2+}$$). To achieve neutrality, three $$Mg^{2+}$$ ions (total +6) are needed to balance two $$Z^{3-}$$ ions (total -6), resulting in the formula $$Mg_3Z_2$$.

A: Incorrect. This formula does not balance the charges of the respective ions.

B: Incorrect. This formula confuses the charges, as if Mg were +3 and Z were -2.

D: Incorrect. This formula would result if Z formed a -1 ion, which is not typical for an element with 5 valence electrons.

Correct: Ionic compounds are typically formed between a metal and a nonmetal, where valence electrons are transferred. Potassium (K) is an alkali metal that readily loses one electron, and Bromine (Br) is a halogen that readily gains one electron.

A: Incorrect. Carbon and Oxygen are both nonmetals and are likely to form a covalent compound by sharing electrons.

B: Incorrect. Nitrogen and Hydrogen are both nonmetals and form the covalent compound ammonia.

D: Incorrect. Silicon is a metalloid and Chlorine is a nonmetal; they form a polar covalent compound, $$SiCl_4$$.

Correct: A 1:1 ratio requires the cation and anion to have charges of equal magnitude. Barium (Ba) is in Group 2 and forms a $$Ba^{2+}$$ ion. Sulfur (S) is in Group 16 and forms an $$S^{2-}$$ ion. These combine in a 1:1 ratio to form BaS.

A: Incorrect. Lithium ($$Li^+$$, Group 1) and Oxygen ($$O^{2-}$$, Group 16) form $$Li_2O$$ (2:1 ratio).

B: Incorrect. Calcium ($$Ca^{2+}$$, Group 2) and Nitrogen ($$N^{3-}$$, Group 15) form $$Ca_3N_2$$ (3:2 ratio).

D: Incorrect. Potassium ($$K^+$$, Group 1) and Phosphorus ($$P^{3-}$$, Group 15) form $$K_3P$$ (3:1 ratio).

This question tests understanding of valence electrons and ionic compounds. Potassium (Z=19) is in Group 1 with electron configuration [Ar]4s¹, meaning it has 1 valence electron in the fourth shell. To achieve the stable argon configuration (18 electrons), potassium loses its single valence electron to form K⁺. The K²⁺ option (A) is incorrect because losing 2 electrons would remove an electron from the filled 3p orbital, requiring much more energy and creating an unstable configuration that doesn't match any noble gas. Group 1 elements always form +1 cations by losing their single valence electron.

This question assesses the skill of valence electrons and ionic compounds. Valence electrons influence ion formation by showing how many electrons are gained or lost for stability. Element P with atomic number 9 is fluorine in Group 17, with 7 valence electrons, and it gains 1 electron to form a -1 ion. This achieves the stable configuration of neon. A tempting distractor is +1, but it is incorrect because halogens are nonmetals that gain electrons, not lose them. Main-group ions form to achieve noble-gas configurations by gaining or losing the fewest electrons possible.

This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine ion formation by indicating whether an atom will gain or lose electrons for stability. Group 16 elements have 6 valence electrons and typically gain 2 electrons to complete an octet, forming anions with a -2 charge. This results in a noble-gas-like configuration, making $Q^{2-}$ the most stable monatomic ion for element Q. A tempting distractor is $Q^{2+}$, but it is incorrect because nonmetals in Group 16 gain electrons rather than lose them to form negative ions. Main-group ions form to achieve noble-gas configurations by gaining or losing the fewest electrons possible.

This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine the charges of ions, which in turn dictate the formulas of ionic compounds. Element T in Group 1 has 1 valence electron and forms a +1 ion, while W in Group 16 has 6 valence electrons and forms a -2 ion. To balance charges, two T ions combine with one W ion, resulting in the formula T_2W. A tempting distractor is TW, but it is incorrect because it does not account for the -2 charge of W requiring two +1 ions for neutrality. Main-group ions form to achieve noble-gas configurations, and compound formulas balance total positive and negative charges.