Question

electron configuration practice
chemistry
name:
due date:
e. in the space below, write the full (unabbreviated) electron configurations of the
following elements:

1. sodium
2. iron
3. bromine
4. barium
5. neptunium

f. in the space below, write the noble gas (abbreviated) electron configurations of the
following elements:

1. cobalt
2. silver
3. tellurium
4. radium
5. lawrencium

g. determine what elements are denoted by the following electron configurations:

1. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4} )
2. ( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1} )
3. ( kr5s^{2}4d^{10}5p^{3} )
4. ( xe6s^{2}4f^{14}5d^{6} )
5. ( rn7s^{2}5f^{11} )

Explanation:

For electron configurations, we count the number of electrons in each orbital. The atomic number of an element is equal to the total number of electrons in its neutral atom. For noble - gas abbreviated configurations, we use the noble gas from the previous period as a shorthand.

For example, for cobalt (atomic number 27):

• The noble gas before cobalt is argon (atomic number 18).
• Then we add the remaining electrons: \(3d^74s^2\). So the noble - gas abbreviated configuration is \([Ar]3d^74s^2\).

For silver (atomic number 47):

• The noble gas before silver is krypton (atomic number 36).
• Then we add the remaining electrons: \(4d^{10}5s^1\). So the noble - gas abbreviated configuration is \([Kr]4d^{10}5s^1\).

For tellurium (atomic number 52):

• The noble gas before tellurium is krypton (atomic number 36).
• Then we add the remaining electrons: \(4d^{10}5s^25p^4\). So the noble - gas abbreviated configuration is \([Kr]4d^{10}5s^25p^4\).

For radium (atomic number 88):

• The noble gas before radium is radon (atomic number 86).
• Then we add the remaining electrons: \(7s^2\). So the noble - gas abbreviated configuration is \([Rn]7s^2\).

For lawrencium (atomic number 103):

• The noble gas before lawrencium is radon (atomic number 86).
• Then we add the remaining electrons: \(5f^{14}6d^17s^2\). So the noble - gas abbreviated configuration is \([Rn]5f^{14}6d^17s^2\).

For determining elements from electron configurations:

• For \(1s^22s^22p^63s^23p^4\), the total number of electrons is \(2 + 2+6 + 2+4=16\). The element with atomic number 16 is sulfur.
• For \(1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1\), the total number of electrons is \(2+2 + 6+2+6+2+10+6 + 1=37\). The element with atomic number 37 is rubidium.
• For \([Kr]5s^24d^{10}5p^3\), krypton has 36 electrons. Adding \(2 + 10+3=15\) more electrons. \(36+15 = 51\). The element with atomic number 51 is antimony.
• For \([Xe]6s^24f^{14}5d^6\), xenon has 54 electrons. Adding \(2+14 + 6=22\) more electrons. \(54+22=76\). The element with atomic number 76 is osmium.
• For \([Rn]7s^25f^{11}\), radon has 86 electrons. Adding \(2+11 = 13\) more electrons. \(86+13=99\). The element with atomic number 99 is einsteinium.

Answer:

1. \([Ar]3d^74s^2\)
2. \([Kr]4d^{10}5s^1\)
3. \([Kr]4d^{10}5s^25p^4\)
4. \([Rn]7s^2\)
5. \([Rn]5f^{14}6d^17s^2\)
6. Sulfur
7. Rubidium
8. Antimony
9. Osmium
10. Einsteinium