Question

name_____________________________ period ______ date ____________
isotopes, ions, and atoms worksheet
atomic # = # of protons. mass # = atomic # + neutrons. protons = electrons when charge is zero.

atomic #	mass #	# p⁺	# e⁻	# n⁰	charge	symbol
2)	180		71	109
3)		40	38	46
4) 92	238		86
5)						$\ce{_{82}^{206}pb^{4+}}$
6)		34		45	-2
7)	113	48	49

Explanation:

Step1: Recall formulas

Atomic number (\(Z\)) = number of protons (\(p^+\)) = number of electrons (\(e^-\)) when charge is 0.
Mass number (\(A\)) = Atomic number (\(Z\)) + number of neutrons (\(n^0\)).
Charge = number of protons - number of electrons.

Step2: Solve for row 1:

• \(Z = p^+ = 17\) (since atomic # = \(p^+\)).
• \(A = Z + n^0 = 17 + 19 = 36\).
• \(e^- = p^+\) (charge = 0) → \(e^- = 17\).
• Symbol: Element with \(Z = 17\) is Cl. So \(^{36}_{17}\text{Cl}\).

Step3: Solve for row 2:

• \(Z = p^+ = e^- = 71\) (since charge = 0, \(p^+ = e^-\)).
• \(A = 180\) (given).
• \(n^0 = A - Z = 180 - 71 = 109\) (matches).
• Charge = \(p^+ - e^- = 71 - 71 = 0\).
• Symbol: \(Z = 71\) is Lu. So \(^{180}_{71}\text{Lu}\).

Step4: Solve for row 3:

• \(Z = p^+ = 40\).
• \(A = Z + n^0 = 40 + 46 = 86\).
• \(e^- = 38\) (given).
• Charge = \(40 - 38 = +2\).
• Symbol: \(Z = 40\) is Zr. So \(^{86}_{40}\text{Zr}^{2+}\).

Step5: Solve for row 4:

• \(Z = 92\) (atomic #), so \(p^+ = 92\).
• \(A = 238\) (given).
• \(n^0 = A - Z = 238 - 92 = 146\).
• \(e^- = 86\) (given).
• Charge = \(92 - 86 = +6\).
• Symbol: \(Z = 92\) is U. So \(^{238}_{92}\text{U}^{6+}\).

Step6: Solve for row 5:

• Symbol: \(^{206}_{82}\text{Pb}^{4+}\).
• \(Z = 82\) (subscript) → \(p^+ = 82\).
• \(A = 206\) (superscript).
• \(n^0 = A - Z = 206 - 82 = 124\).
• Charge = \(+4\) → \(e^- = p^+ - \text{charge} = 82 - 4 = 78\).

Step7: Solve for row 6:

• \(p^+ = 34\) → \(Z = 34\) (element Se).
• \(A = p^+ + n^0 = 34 + 45 = 79\).
• \(e^- = p^+ - \text{charge} = 34 - (-2) = 36\) (since charge = \(p^+ - e^-\) → \(e^- = p^+ - \text{charge}\); charge = -2 → \(e^- = 34 + 2 = 36\)).
• Symbol: \(^{79}_{34}\text{Se}^{2-}\).

Step8: Solve for row 7:

• \(p^+ = 48\) → \(Z = 48\) (element Cd).
• \(A = 113\) (given).
• \(n^0 = A - Z = 113 - 48 = 65\).
• \(e^- = 49\) (given).
• Charge = \(48 - 49 = -1\).
• Symbol: \(^{113}_{48}\text{Cd}^{-}\).

Answer:

(for row 1 as example, extend for others):
Row 1:
Atomic # = 17, Mass # = 36, \(p^+ = 17\), \(e^- = 17\), \(n^0 = 19\), charge = 0, Symbol = \(^{36}_{17}\text{Cl}\)

(Repeat similar steps for all rows as needed.)