Question

1. write the unabbreviated (full) electron configuration for a neutral atom of sodium (na).
2. what element has the unabbreviated electron configuration 1s²2s²2p⁶3s²3p⁴?
3. write the unabbreviated (full) electron configuration for the chloride ion (cl⁻).
4. a neutral atom has the unabbreviated electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁷. identify the element.
5. write the unabbreviated (full) electron configuration for a neutral atom of argon (ar). what is special about this configuration?
6. write the unabbreviated (full) electron configuration for the iron (iii) ion (fe³⁺).
7. identify the element with the following unabbreviated electron configuration: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹. what is notable about this configuration?
8. write the unabbreviated (full) electron configuration for a neutral atom of vanadium (v)

Explanation:

Step1: Recall electron - filling rules

Electrons fill orbitals in the order of increasing energy: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s etc. The s - orbital can hold 2 electrons, the p - orbital can hold 6 electrons, and the d - orbital can hold 10 electrons.

Step2: Determine atomic number for Na

Sodium (Na) has an atomic number of 11. So the electron configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.

Step3: Calculate electrons for $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$

Count the electrons: 2 + 2+6 + 2+4 = 16. The element with atomic number 16 is sulfur (S).

Step4: Determine electrons for $Cl^{-}$

Chlorine (Cl) has an atomic number of 17. The $Cl^{-}$ ion has 18 electrons. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$.

Step5: Identify element for $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}$

Count the electrons: 2+2 + 6+2+6+2 + 7=27. The element with atomic number 27 is cobalt (Co).

Step6: Electron - configuration of Ar

Argon (Ar) has an atomic number of 18. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. It is a noble - gas configuration, which is stable with a full outer shell.

Step7: Electron - configuration of $Fe^{3 + }$

Iron (Fe) has an atomic number of 26. The $Fe^{3+}$ ion has 23 electrons. First, the neutral Fe has $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$. When it forms $Fe^{3+}$, 2 electrons are removed from the 4s and 1 from the 3d. So the configuration of $Fe^{3+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$.

Step8: Identify element for $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}$

Count the electrons: 2+2+6 + 2+6+2+10+6 + 1=39. The element with atomic number 39 is yttrium (Y). It has a single electron in the 5s orbital, which is characteristic of group 1 elements in the 5th period.

Step9: Electron - configuration of V

Vanadium (V) has an atomic number of 23. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}$.

Answer:

1. $1s^{2}2s^{2}2p^{6}3s^{1}$
2. Sulfur (S)
3. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$
4. Cobalt (Co)
5. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$, it is a noble - gas configuration (full outer shell)
6. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$
7. Yttrium (Y), has a single 5s electron
8. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}$