Question

for each of the following element pairs determine if it is ionic, polar covalent, nonpolar covalent, or metallic (use the pauling scale). for each ionic pair write the formula using the swap and drop method and count the total number of atoms in the ionic compounds.
sulfur & fluorine - sf(6), 7 atoms
calcium & nitrogen- ca(3)n(2), 5 atoms
ammonium & oxygen- ionic, (nh(4))(2) o
carbon & hydrogen- nonpolar covalent, ch(4), 5 atoms
tungsten & phosphate - ionic, w(po(4) (2), 11 atoms
copper & zinc
fluorine & hydrogen
titanium (iv) & selenium
iron (iii) & nickel (ii)
niobium (vi) & phosphate

Explanation:

Step1: Identify bond type for Copper & Zinc

Both are metals → Metallic bond.

Step2: Identify bond type for Fluorine & Hydrogen

Pauling electronegativity: F=4.0, H=2.1. Difference=1.9 (0.5-1.9 → Polar Covalent).

Step3: Determine bond type for Titanium (IV) & Selenium

Ti(IV) is metal cation, Se is nonmetal anion → Ionic. Swap charges: Ti⁴⁺, Se²⁻ → Ti₂Se₄ → Simplify to TiSe₂. Total atoms: 1+2=3.

Step4: Identify bond type for Iron (III) & Nickel (II)

Both are metal cations → Metallic bond.

Step5: Determine bond type for Niobium (V) & Phosphate

Nb(V) is metal cation, PO₄³⁻ is polyatomic anion → Ionic. Swap charges: Nb⁵⁺, PO₄³⁻ → Nb₃(PO₄)₅. Total atoms: 3 + 5×(1+4) = 3+25=28.

Answer:

Copper & Zinc: Metallic
Fluorine & Hydrogen: Polar Covalent
Titanium (IV) & Selenium: Ionic, TiSe₂, 3 atoms
Iron (III) & Nickel (II): Metallic
Niobium (V) & Phosphate: Ionic, Nb₃(PO₄)₅, 28 atoms