Question

60)the ratio of chlorine to zinc ions in the compound zinc chloride would be
a) 1:2
b) 3:1
c) 1:1
d) 2:1
e) 4:1

1. of the following metal ions, which would be most likely to have a +4 charge?

a) cu
b) pb
c) li
d) zn
e) cd
62)how many total carbonate ions would be present in the formula for aluminum carbonate
a) 3
b) 2
c) 1
d) 4
e) 5
63)which of the following compounds would be most ionic?
a) al₂o₃
b) bes
c) na₂s
d) mgcl₂
e) csf

Explanation:

Question 60

Step1: Determine the formula of zinc chloride

Zinc has a charge of \(+2\) (\(Zn^{2+}\)) and chlorine has a charge of \(-1\) (\(Cl^{-}\)). To form a neutral compound, the formula is \(ZnCl_{2}\).

Step2: Calculate the ratio of ions

In \(ZnCl_{2}\), for every \(1\) \(Zn^{2+}\) ion, there are \(2\) \(Cl^{-}\) ions. So the ratio of chlorine to zinc ions is \(2:1\).

Question 61

Step1: Analyze the common charges of each metal

• \(Cu\) commonly has \(+1\) or \(+2\) charges.
• \(Pb\) can have \(+2\) or \(+4\) charges (due to its position in the periodic table and electron - configuration).
• \(Li\) has a \(+1\) charge (as an alkali metal).
• \(Zn\) has a \(+2\) charge (a common oxidation state).
• \(Cd\) has a \(+2\) charge (similar to \(Zn\) in the periodic table).

Question 62

Step1: Write the formula of aluminum carbonate

Aluminum has a charge of \(+3\) (\(Al^{3+}\)) and carbonate has a charge of \(-2\) (\(CO_{3}^{2 -}\)). Using the criss - cross method, the formula is \(Al_{2}(CO_{3})_{3}\).

Step2: Count the carbonate ions

In \(Al_{2}(CO_{3})_{3}\), the number of carbonate (\(CO_{3}^{2 -}\)) ions is \(3\).

Question 63

Step1: Recall the concept of ionic character

Ionic character is related to the difference in electronegativity between the two elements in a compound. The greater the difference, the more ionic the compound.

• For \(Al_{2}O_{3}\): Electronegativity of \(Al = 1.61\), \(O=3.44\), difference \(= 3.44 - 1.61=1.83\)
• For \(BeS\): Electronegativity of \(Be = 1.57\), \(S = 2.58\), difference \(=2.58 - 1.57 = 1.01\)
• For \(Na_{2}S\): Electronegativity of \(Na = 0.93\), \(S = 2.58\), difference \(=2.58 - 0.93=1.65\)
• For \(MgCl_{2}\): Electronegativity of \(Mg = 1.31\), \(Cl = 3.16\), difference \(=3.16 - 1.31 = 1.85\)
• For \(CsF\): Electronegativity of \(Cs = 0.79\), \(F = 3.98\), difference \(=3.98 - 0.79=3.19\)

Answer:

1. D. \(2:1\)
2. B. \(Pb\)
3. A. \(3\)
4. E. \(CsF\)