Comprehensive Study Guide: Bond Types, Electronegativity, and Chemical Calculations

Catalyst Review and Fundamental Stoichiometry

This section covers foundational chemistry concepts, calculations involving molarity and volume, and the application of stoichiometry to balanced chemical reactions.

Molarity and Solution Calculations

• Problem 1: How many liters ($dm^3$) of $HBr$ are present if $20$ moles of Hydrogen bromide are present within a $4$ molar ($M$) $HBr$ solution?
  • Formula: $M = \frac{n}{V}$
  • Rearranged: $V = \frac{n}{M}$
  • Calculation: $V = \frac{20}{4} = 5\,dm^3$
• Problem 2: How many moles are present in a $3$ molar ($M$) solution of $SrCl_2$ with a volume of $3\,dm^3$?
  • Formula: $n = M \times V$
  • Calculation: $n = 3\,mol\,dm^{-3} \times 3\,dm^3 = 9\,moles$

Definitions

• Cation: A positively charged ion that has lost one or more electrons.
• Zwitterion: A molecule or ion having separate positively and negatively charged groups, resulting in an overall neutral charge.
• Valence Electron: An electron in the outer shell of an atom which can participate in the formation of chemical bonds.

Lewis Structure Drawings

Drawings are required for the following compounds:

• $I_2$
• $HBr$
• $SrCl_2$
• $K_3As$

Stoichiometry and Mass Calculations

Balanced Equation:$15Ca + 2P_5 \rightarrow 5Ca_3P_2$

• Stoichiometry Question: How many grams of calcium phosphide ($Ca_3P_2$) are needed if $10$ moles of $Ca_3P_2$ will be made?
• Molar Conversion Question: How many moles of $Ca$ are needed for $160$ grams of Calcium?
  • (Note: Atomic mass of $Ca \approx 40.08\,g\,mol^{-1}$. Calculation: $\frac{160}{40.08} \approx 4\,moles$).
• Yield Question: How many moles of calcium phosphide are made from $888$ grams of $Ca_3P_2$?

Objectives and Bond Classification

Content Objectives

• Learn the different types of chemical bonds: Ionic, Covalent, and Polar Covalent.
• Learn how to classify these bonds using electronegativity ($\chi$) calculations.

Language Objectives

• Practice determining electronegativity differences by reading and interpreting the periodic table chart.

Electronegativity Values of Elements

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The following values are provided for classification calculations:

• Group 1: $H = 2.1$, $Li = 1.0$, $Na = 0.9$, $K = 0.8$, $Rb = 0.8$, $Cs = 0.7$, $Fr = 0.7$
• Group 2: $Be = 1.6$, $Mg = 1.2$, $Ca = 1.0$, $Sr = 1.0$, $Ba = 0.9$, $Ra = 0.9$
• Transition Metals (Select):
  • $Sc = 1.3$, $Ti = 1.5$, $V = 1.6$, $Cr = 1.6$, $Mn = 1.5$, $Fe = 1.8$, $Co = 1.8$, $Ni = 1.8$, $Cu = 1.9$, $Zn = 1.6$
  • $Y = 1.2$, $Zr = 1.4$, $Nb = 1.6$, $Mo = 1.8$, $Tc = 1.9$, $Ru = 2.2$, $Rh = 2.2$, $Pd = 2.2$, $Ag = 1.9$, $Cd = 1.7$
  • $Hf = 1.3$, $Ta = 1.5$, $W = 1.7$, $Re = 1.9$, $Os = 2.2$, $Ir = 2.2$, $Pt = 2.2$, $Au = 2.4$, $Hg = 1.9$
• Group 13: $B = 2.0$, $Al = 1.5$, $Ga = 1.6$, $In = 1.7$, $Tl = 1.8$
• Group 14: $C = 2.5$, $Si = 1.8$, $Ge = 1.8$, $Sn = 1.8$, $Pb = 1.9$
• Group 15: $N = 3.0$, $P = 2.1$, $As = 2.0$, $Sb = 1.9$, $Bi = 1.9$
• Group 16: $O = 3.5$, $S = 2.5$, $Se = 2.4$, $Te = 2.1$, $Po = 2.0$
• Group 17 (Halogens): $F = 4.0$, $Cl = 3.0$, $Br = 2.8$, $I = 2.5$, $At = 2.2$

Bond Types and Classification Rules

Chemical bonds are categorized based on the difference in electronegativity ($\Delta\chi$) between the two bonding atoms.

Ionic Bonds

• Definition: The bond formed when one element gives electrons to another element.
• Characteristics:
  • Usually occurs between a metal and a non-metal.
  • Example: In water, $NaCl$ dissociates into $Na^+$ and $Cl^-$.
• Calculation Rule: $\Delta\chi \ge 1.7$
• Example calculation for $NaCl$:
  • $Na = 0.9$, $Cl = 3.0$
  • $|3.0 - 0.9| = 2.1$
  • Result: Since $2.1 \ge 1.7$, the bond is Ionic.

Covalent Bonds (Non-polar)

• Definition: The bond formed when atoms share electrons equally.
• Example: Methane ($CH_4$).
• Calculation Rule: $\Delta\chi \le 0.4$
• Example calculation for $CH_4$:
  • $C = 2.5$, $H = 2.1$
  • $|2.5 - 2.1| = 0.4$
  • Result: Since $0.4 \le 0.4$, the bond is Covalent.

Polar Covalent Bonds

• Definition: The bond formed when atoms share electrons, but the electrons are more strongly drawn toward one element than the other.
• Examples: Water ($HOH$), Acetic Acid ($CH_3COOH$/vinegar), and Fluoromethane ($CH_3F$).
• Calculation Rule: $0.4 < \Delta\chi < 1.7$
• Example calculation for $CH_3F$ (specifically the $C-F$ bond):
  • $C = 2.5$, $F = 4.0$
  • $|4.0 - 2.5| = 1.5$
  • Result: Since $1.5$ falls between $0.4$ and $1.7$, the bond is Polar Covalent.

Practice Problems

Classify the following substances as Ionic, Covalent, or Polar Covalent based on the electronegativity table provided:

1. $MgCl_2$
2. $S_2O_3$
3. $C_3H_8$
4. $MnS_2$