Question

bonding practice
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
calcium & nitrogen
ammonium & oxygen
carbon & hydrogen
tungsten & phosphate

Explanation:

(Note: Pauling electronegativity values used: S=2.58, F=3.98, Ca=1.00, N=3.04, C=2.55, H=2.20, W≈2.36, P=2.19, O=3.44, N=3.04, H=2.20)

Step1: Sulfur & Fluorine

Electronegativity difference: $3.98 - 2.58 = 1.40$ (polar covalent, since $0.5 < ΔEN < 1.7$). Formula: $SF_6$ (total atoms: $1+6=7$).

Step2: Calcium & Nitrogen

Electronegativity difference: $3.04 - 1.00 = 2.04$ (ionic). Formula: $Ca_3N_2$ (total atoms: $3+2=5$).

Step3: Carbon & Hydrogen

Electronegativity difference: $2.55 - 2.20 = 0.35$ (nonpolar covalent, $ΔEN < 0.5$). Formula: $CH_4$ (total atoms: $1+4=5$).

Step4: Tungsten & Phosphate

Tungsten (W, assume +6) and phosphate ($PO_4^{3-}$). Formula: $W_3(PO_4)_6$ → simplify to $W(PO_4)_2$ (total atoms: $1 + (1+4)×2 = 11$). Ionic (metal + polyatomic ion).

Step5: Ammonium & Oxygen

Ammonium ($NH_4^+$) and oxide ($O^{2-}$). Formula: $(NH_4)_2O$ (total atoms: $(1+4)×2 +1 = 11$). Ionic (polyatomic ion + ion).

Answer:

1. Sulfur & Fluorine: Polar Covalent, $SF_6$, 7 atoms
2. Calcium & Nitrogen: Ionic, $Ca_3N_2$, 5 atoms
3. Carbon & Hydrogen: Nonpolar Covalent, $CH_4$, 5 atoms
4. Tungsten & Phosphate: Ionic, $W(PO_4)_2$, 11 atoms
5. Ammonium & Oxygen: Ionic, $(NH_4)_2O$, 11 atoms