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electron configuration practice chemistry name: due date: b. write a gr…

Question

electron configuration practice
chemistry
name:
due date:
b. write a ground state electron configuration for these ions. remember that ions have a change in the total number of electrons (positive have lost electrons and negative have gained). example: ( n^{3 -} ) is ( 1s^{2}2s^{2}2p^{6} ). it has three extra electrons

  1. ( o^{2 -} ):
  2. ( fe^{2 +} ):
  3. ( b^{3 +} ):
  4. ( ni^{2 +} ):
  5. ( k^{+} ):
  6. ( co^{3 +} ):

c. for the following electron configurations determine the possible elements (or ions) they may represent

  1. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4} ):
  2. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6} ):
  3. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{10} ):
  4. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5} ):
  5. ( kr5s^{2}4d^{10}5p^{3} ):
  6. ( ar4s^{1} ):
  7. ( xe6s^{2}4f^{10} ):
  8. ( ne3s^{2}3p^{1} ):

Explanation:

Step1: Determine the number of electrons for each ion or element

  • For \(O^{2 - }\): Oxygen has an atomic number of \(8\). Since it has a \(2 -\) charge, it has \(8 + 2=10\) electrons.
  • For \(Fe^{2+}\): Iron has an atomic number of \(26\). With a \(2+\) charge, it has \(26- 2 = 24\) electrons.
  • For \(B^{3+}\): Boron has an atomic number of \(5\). With a \(3+\) charge, it has \(5 - 3=2\) electrons.
  • For \(Ni^{2+}\): Nickel has an atomic number of \(28\). With a \(2+\) charge, it has \(28-2 = 26\) electrons.
  • For \(K^{+}\): Potassium has an atomic number of \(19\). With a \(1+\) charge, it has \(19 - 1=18\) electrons.
  • For \(Co^{3+}\): Cobalt has an atomic number of \(27\). With a \(3+\) charge, it has \(27-3 = 24\) electrons.
  • For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\): The total number of electrons is \(2 + 2+6 + 2+6+2 + 10+4=34\).
  • For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\): The total number of electrons is \(2+2 + 6+2+6+2+10+6 = 36\).
  • For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{10}\): The total number of electrons is \(2+2+6+2+6+2+10+6+2+10+6+2+10=70\).
  • For \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\): The total number of electrons is \(2+2+6+2+6+2+5 = 25\).
  • For \([Kr]5s^{2}4d^{10}5p^{3}\): Krypton (\(Kr\)) has \(36\) electrons. So the total number of electrons is \(36+2 + 10+3=51\).
  • For \([Ar]4s^{1}\): Argon (\(Ar\)) has \(18\) electrons. So the total number of electrons is \(18+1 = 19\).
  • For \([Xe]6s^{2}4f^{10}\): Xenon (\(Xe\)) has \(54\) electrons. So the total number of electrons is \(54+2+10 = 66\).
  • For \([Ne]3s^{2}3p^{1}\): Neon (\(Ne\)) has \(10\) electrons. So the total number of electrons is \(10+2+1=13\).

Step2: Write the electron - configuration or identify the element/ion

  • \(O^{2 - }\): \(1s^{2}2s^{2}2p^{6}\)
  • \(Fe^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}\)
  • \(B^{3+}\): \(1s^{2}\)
  • \(Ni^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}\)
  • \(K^{+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)
  • \(Co^{3+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}\)
  • \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\): Selenium (\(Se\))
  • \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\): Krypton (\(Kr\))
  • \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{10}\): Ytterbium (\(Yb\))
  • \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\): Manganese (\(Mn\))
  • \([Kr]5s^{2}4d^{10}5p^{3}\): Antimony (\(Sb\))
  • \([Ar]4s^{1}\): Potassium (\(K\))
  • \([Xe]6s^{2}4f^{10}\): Dysprosium (\(Dy\))
  • \([Ne]3s^{2}3p^{1}\): Aluminum (\(Al\))

Answer:

  1. \(1s^{2}2s^{2}2p^{6}\)
  2. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}\)
  3. \(1s^{2}\)
  4. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}\)
  5. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)
  6. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}\)
  7. Selenium (\(Se\))
  8. Krypton (\(Kr\))
  9. Ytterbium (\(Yb\))
  10. Manganese (\(Mn\))
  11. Antimony (\(Sb\))
  12. Potassium (\(K\))
  13. Dysprosium (\(Dy\))
  14. Aluminum (\(Al\))