Question

1. write condensed electron configurations and orbital diagrams for each element: a. n b. as c. as³⁺ d. sn⁴⁺

Answer:

To solve this, we'll find the condensed electron configurations and orbital diagrams for each species. The condensed electron configuration uses the noble gas core (the electron configuration of the nearest noble gas with fewer electrons) followed by the valence electrons. Orbital diagrams show the distribution of electrons in orbitals (s, p, d, f) with arrows representing electrons (up for spin +1/2, down for spin -1/2), following the Aufbau principle (fill lower energy orbitals first), Pauli exclusion principle (max 2 electrons per orbital with opposite spins), and Hund's rule (electrons fill degenerate orbitals singly first with parallel spins).

Part (a): Nitrogen (N)

Nitrogen has an atomic number of 7. The nearest noble gas with fewer electrons is Helium (He, atomic number 2).

• Condensed Electron Configuration: The electron configuration of He is \(1s^2\). Nitrogen has \(7 - 2 = 5\) valence electrons. The valence orbitals for N (in period 2, group 15) are \(2s\) and \(2p\). So the condensed configuration is \([\text{He}]2s^22p^3\).
• Orbital Diagram:
• \(1s\) orbital (core, from He): \(\boldsymbol{\uparrow \downarrow}\)
• \(2s\) orbital (valence): \(\boldsymbol{\uparrow \downarrow}\)
• \(2p\) orbitals (valence, 3 degenerate orbitals):
• \(2p_x\): \(\boldsymbol{\uparrow}\)
• \(2p_y\): \(\boldsymbol{\uparrow}\)
• \(2p_z\): \(\boldsymbol{\uparrow}\) (follows Hund’s rule: single electrons in each \(p\) orbital with parallel spins)

Part (b): Arsenic (As)

Arsenic has an atomic number of 33. The nearest noble gas with fewer electrons is Argon (Ar, atomic number 18).

• Condensed Electron Configuration: The electron configuration of Ar is \(1s^22s^22p^63s^23p^6\). Arsenic has \(33 - 18 = 15\) valence electrons. Arsenic is in period 4, group 15, so its valence orbitals are \(4s\), \(3d\) (filled, since it’s a main - group element; \(3d\) is filled before \(4p\) for period 4 elements), and \(4p\). The \(3d\) orbitals hold 10 electrons (from \(3d^{10}\)), \(4s\) holds 2, and \(4p\) holds \(15-(10 + 2)=3\). So the condensed configuration is \([\text{Ar}]3d^{10}4s^24p^3\).
• Orbital Diagram:
• Core orbitals (from Ar): \(1s^22s^22p^63s^23p^6\) (all filled with \(\uparrow \downarrow\) in each orbital).
• \(3d\) orbitals (10 electrons, 5 orbitals): Each \(3d\) orbital has \(\uparrow \downarrow\) (since \(3d^{10}\) means all 5 \(d\) orbitals are filled with 2 electrons each).
• \(4s\) orbital: \(\boldsymbol{\uparrow \downarrow}\)
• \(4p\) orbitals (3 degenerate orbitals):
• \(4p_x\): \(\boldsymbol{\uparrow}\)
• \(4p_y\): \(\boldsymbol{\uparrow}\)
• \(4p_z\): \(\boldsymbol{\uparrow}\) (follows Hund’s rule)

Part (c): Arsenic(III) ion (\(\text{As}^{3+}\))

Arsenic has 33 electrons. When it forms \(\text{As}^{3+}\), it loses 3 electrons (since it’s a cation, losing electrons from the outermost shell). The outermost shell for As is the \(4p\) subshell (valence electrons are in \(4s^24p^3\); we lose electrons from the highest energy orbital first, which is \(4p\) here).

• Electrons in \(\text{As}^{3+}\): \(33 - 3 = 30\) electrons. The nearest noble gas with fewer electrons is Argon (Ar, 18 electrons).
• Condensed Electron Configuration: The electron configuration of Ar is \(1s^22s^22p^63s^23p^6\). After Ar, we have \(30 - 18 = 12\) electrons. These fill the \(3d\) (10 electrons) and \(4s\) (2 electrons) orbitals (since we removed electrons from \(4p\), the \(4s\) and \(3d\) are still filled). So the condensed configuration is \([\text{Ar}]3d^{10}4s^2\).
• Orbital Diagram:
• Core orbitals (from Ar): \(1s^22s^22p^63s^23p^6\) (all filled with \(\uparrow \downarrow\)).
• \(3d\) orbitals (10 electrons, 5 orbitals): Each \(3d\) orbital has \(\uparrow \downarrow\) (filled).
• \(4s\) orbital: \(\boldsymbol{\uparrow \downarrow}\)
• \(4p\) orbitals: All \(4p\) orbitals are empty (since we lost the 3 \(4p\) electrons).

Part (d): Tin(IV) ion (\(\text{Sn}^{4+}\))

Tin has an atomic number of 50. When it forms \(\text{Sn}^{4+}\), it loses 4 electrons (tin is in group 14, and losing 4 electrons gives a stable cation). Tin’s electron configuration (neutral) is \([\text{Kr}]4d^{10}5s^25p^2\) (Kr is the noble gas core, atomic number 36). When forming \(\text{Sn}^{4+}\), it loses electrons from the outermost shell. The outermost shell for Sn is the \(5p\) and \(5s\) subshells (and we also lose from the \(4d\)? No, wait: for transition metals and post - transition metals, the order of electron loss is from the highest energy orbitals. The electron configuration of neutral Sn is \(1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^2\) (or \([\text{Kr}]4d^{10}5s^25p^2\) with Kr: \(1s^22s^22p^63s^23p^63d^{10}4s^24p^6\)). When losing 4 electrons, Sn loses the 2 \(5p\) electrons and 2 \(5s\) electrons (since \(5p\) is higher in energy than \(5s\), and \(5s\) is higher than \(4d\); for post - transition metals like Sn, electron loss is from \(n s\) and \(n p\) first, then \( (n - 1)d\) in some cases, but for \( \text{Sn}^{4+}\), it loses \(5s^25p^2\), leaving the \(4d^{10}\) filled).

• Condensed Electron Configuration: The noble gas core is Krypton (Kr, atomic number 36). After losing 4 electrons, Sn⁴⁺ has \(50 - 4 = 46\) electrons. The electron configuration of Kr is \(1s^22s^22p^63s^23p^63d^{10}4s^24p^6\). After Kr, we have \(46 - 36 = 10\) electrons, which fill the \(4d\) orbital (since \(4d\) can hold 10 electrons). So the condensed configuration is \([\text{Kr}]4d^{10}\).
• Orbital Diagram:
• Core orbitals (from Kr): \(1s^22s^22p^63s^23p^63d^{10}4s^24p^6\) (all filled with \(\uparrow \downarrow\) in each orbital).
• \(4d\) orbitals (10 electrons, 5 orbitals): Each \(4d\) orbital has \(\uparrow \downarrow\) (filled, since \(4d^{10}\)).
• \(5s\) and \(5p\) orbitals: All \(5s\) and \(5p\) orbitals are empty (since we lost the \(5s^25p^2\) electrons).

Final Answers

a. Condensed: \(\boldsymbol{[\text{He}]2s^22p^3}\); Orbital Diagram: \(1s:\uparrow \downarrow\), \(2s:\uparrow \downarrow\), \(2p_x:\uparrow\), \(2p_y:\uparrow\), \(2p_z:\uparrow\)
b. Condensed: \(\boldsymbol{[\text{Ar}]3d^{10}4s^24p^3}\); Orbital Diagram: \(1s - 3p\) filled (\(\uparrow \downarrow\) in each), \(3d\) filled (\(\uparrow \downarrow\) in each \(3d\) orbital), \(4s:\uparrow \downarrow\), \(4p_x:\uparrow\), \(4p_y:\uparrow\), \(4p_z:\uparrow\)
c. Condensed: \(\boldsymbol{[\text{Ar}]3d^{10}4s^2}\); Orbital Diagram: \(1s - 3p\) filled (\(\uparrow \downarrow\) in each), \(3d\) filled (\(\uparrow \downarrow\) in each \(3d\) orbital), \(4s:\uparrow \downarrow\), \(4p\) orbitals empty
d. Condensed: \(\boldsymbol{[\text{Kr}]4d^{10}}\); Orbital Diagram: \(1s - 4p\) filled (\(\uparrow \downarrow\) in each), \(4d\) filled (\(\uparrow \downarrow\) in each \(4d\) orbital), \(5s\) and \(5p\) orbitals empty