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
1) 17	37	18	17	19	0	cl
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 = \(p^+ - e^-\).

Step2: Solve for row 2

• \(Z = p^+ = e^- = 71\) (since \(e^- = 71\) and charge is 0? Wait, check \(A = Z + n^0 = 71 + 109 = 180\), which matches mass #. So \(Z = 71\), \(p^+ = 71\), charge = \(71 - 71 = 0\), symbol: element with \(Z=71\) is Lu (Lutetium).

Step3: Solve for row 3

• \(p^+ = 40\), so \(Z = 40\) (element Zr, Zirconium).
• \(e^- = 38\), so charge = \(40 - 38 = +2\).
• \(A = 40 + 46 = 86\).

Step4: Solve for row 4

• \(Z = 92\) (element U, Uranium), so \(p^+ = 92\).
• \(e^- = 86\), charge = \(92 - 86 = +6\).
• \(n^0 = A - Z = 238 - 92 = 146\).

Step5: Solve for row 5

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

Step6: Solve for row 6

• \(p^+ = 34\) (element Se, Selenium), so \(Z = 34\).
• \(n^0 = 45\), so \(A = 34 + 45 = 79\).
• Charge = \(-2\), so \(e^- = p^+ + 2 = 34 + 2 = 36\).

Step7: Solve for row 7

• \(p^+ = 48\) (element Cd, Cadmium? Wait, \(Z=48\) is Cd? Wait, \(Z=48\) is Cadmium. Wait, \(e^- = 49\), so charge = \(48 - 49 = -1\).
• \(A = 113\), so \(n^0 = 113 - 48 = 65\).

Answer:

Let's summarize each row (filling blanks):

1. (Given: \(Z=17\), \(A=36\)? Wait, original row 1: \(Z=17\), \(A=17\)? Wait, no, original row 1: \(Z=17\), \(A=36\)? Wait, user's table: row 1: Atomic #17, Mass #36? Wait, original table row 1: Atomic #17, Mass #36? Wait, \(n^0=19\), so \(A=17+19=36\), correct. \(p^+=17\), \(e^-=17\), charge 0, symbol Cl.

1. \(Z=71\), \(p^+=71\), charge \(0\), \(A=180\), symbol Lu.

1. \(Z=40\), \(A=86\), charge \(+2\), symbol Zr.

1. \(p^+=92\), charge \(+6\), \(n^0=146\), symbol U.

1. \(Z=82\), \(A=206\), \(p^+=82\), \(e^-=78\), \(n^0=124\), charge \(+4\).

1. \(Z=34\), \(A=79\), \(e^-=36\), symbol Se\(^{2-}\).

1. \(Z=48\), charge \(-1\), \(n^0=65\), symbol Cd\(^-\) (or check \(Z=48\) is Cd, \(e^-=49\), charge \(-1\), \(A=113\)).

(Note: For each row, the key is applying \(Z = p^+ = e^- (\text{if charge}=0)\), \(A = Z + n^0\), \(\text{charge} = p^+ - e^-\).)