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practice: fill in the chart below with numbers of each subatomic partic…

Question

practice:
fill in the chart below with numbers of each subatomic particle. you may use the simulation
for elements up to neon. you may use a periodic table for any of the questions.

Explanation:

Step1: Recall atomic structure concepts

  • The mass number is written as a superscript before the element symbol.
  • The atomic number is the number of protons and for a neutral atom, the number of electrons equals the number of protons.
  • Number of neutrons = Mass number - Atomic number.
  • For ions, the number of electrons changes: for an anion (\(X^{n -}\)), number of electrons=atomic number + n; for a cation (\(X^{n+}\)), number of electrons = atomic number - n.

Step2: Analyze \(^{7}\text{Li}\)

  • Mass number (\(A\)) = 7 (from \(^{7}\text{Li}\)).
  • Atomic number (\(Z\)) of Li (from periodic table) = 3.
  • Number of electrons (neutral atom) = 3.
  • Number of protons = 3.
  • Number of neutrons=\(A - Z=7 - 3 = 4\).

Step3: Analyze \(^{4}\text{He}\)

  • Mass number (\(A\)) = 4 (from \(^{4}\text{He}\)).
  • Atomic number (\(Z\)) of He (from periodic table) = 2.
  • Number of electrons (neutral atom) = 2.
  • Number of protons = 2.
  • Number of neutrons=\(A - Z=4 - 2 = 2\).

Step4: Analyze the element with \(e^{-}=6\), \(p^{+}=6\), \(n = 6\)

  • Symbol: \(^{12}\text{C}\) (since \(A=Z + n=6 + 6 = 12\), atomic number \(Z = 6\) is for carbon).
  • Mass number (\(A\))=12.
  • Atomic number (\(Z\)) = 6.

Step5: Analyze the element with \(e^{-}=6\), \(p^{+}=6\), \(n = 7\)

  • Symbol: \(^{13}\text{C}\) (since \(A=Z + n=6+7 = 13\), atomic number \(Z = 6\) is for carbon).
  • Mass number (\(A\))=13.
  • Atomic number (\(Z\)) = 6.

Step6: Analyze \(^{19}\text{F}^{-1}\)

  • Mass number (\(A\)) = 19 (from \(^{19}\text{F}^{-1}\)).
  • Atomic number (\(Z\)) of F (from periodic table) = 9.
  • Number of electrons: for \(F^{-1}\), \(e^{-}=Z + 1=9 + 1 = 10\).
  • Number of protons = 9.
  • Number of neutrons=\(A - Z=19 - 9 = 10\).

Step7: Analyze the element with \(e^{-}=10\), \(p^{+}=8\), \(n = 8\)

  • Symbol: \(^{16}\text{O}^{2 -}\) (since \(A=Z + n=8 + 8 = 16\), atomic number \(Z = 8\) is for oxygen. For \(e^{-}=10\), \(n_{e^{-}}=Z + 2\) so it is \(O^{2-}\)).
  • Mass number (\(A\))=16.
  • Atomic number (\(Z\)) = 8.

Step8: Analyze the element with \(A = 9\), \(Z = 4\), \(e^{-}=2\)

  • Symbol: \(^{9}\text{Be}^{2+}\) (since \(n=A - Z=9 - 4 = 5\), and for \(e^{-}=Z - 2\) (cation), atomic number \(Z = 4\) is for beryllium).
  • Number of protons = 4.
  • Number of neutrons = 5.

Answer:

SymbolMass numberAtomic numberNumber of electronsNumber of protonsNumber of neutrons
\(^{4}\text{He}\)42222
\(^{12}\text{C}\)126666
\(^{13}\text{C}\)136667
\(^{19}\text{F}^{-1}\)19910910
\(^{16}\text{O}^{2 -}\)1681088
\(^{9}\text{Be}^{2+}\)94245