Question

1. xe

i. write the full electron configuration, short - hand electron configuration, and fill in
the orbital diagrams, for the following elements.

1. nitrogen
2. chlorine
3. sodium
4. neon

Explanation:

Step1: Determine the atomic number

• Nitrogen (\(N\)) has an atomic number \(Z = 7\).
• Chlorine (\(Cl\)) has an atomic number \(Z=17\).
• Sodium (\(Na\)) has an atomic number \(Z = 11\).
• Neon (\(Ne\)) has an atomic number \(Z=10\).

Step2: Write the full electron configuration

• Nitrogen:
• Using the Aufbau principle (\(1s<2s < 2p\)), the full electron configuration is \(1s^{2}2s^{2}2p^{3}\).
• Chlorine:
• The order of filling is \(1s<2s < 2p<3s<3p\). So, the full electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\).
• Sodium:
• Filling order gives \(1s^{2}2s^{2}2p^{6}3s^{1}\).
• Neon:
• With \(Z = 10\), \(1s^{2}2s^{2}2p^{6}\).

Step3: Write the short - hand electron configuration

• Nitrogen:
• The nearest noble gas with lower atomic number is \(He\) (\(Z = 2\)). So, \([He]2s^{2}2p^{3}\).
• Chlorine:
• The nearest noble gas is \(Ne\) (\(Z = 10\)). So, \([Ne]3s^{2}3p^{5}\).
• Sodium:
• Using \(Ne\) (\(Z = 10\)), \([Ne]3s^{1}\).
• Neon:
• Since it is a noble gas, \([Ne]\) (or \([He]2s^{2}2p^{6}\)).

Step4: Orbital diagram rules

• For each orbital (represented by a box), electrons are filled following Pauli's exclusion principle (maximum 2 electrons per orbital with opposite spins) and Hund's rule (electrons fill degenerate orbitals singly first with parallel spins).
• Nitrogen (\(1s^{2}2s^{2}2p^{3}\)):
• \(1s\) orbital: 2 electrons (opposite spins).
• \(2s\) orbital: 2 electrons (opposite spins).
• \(2p\) orbitals: 3 electrons, each in separate \(2p\) orbitals with parallel spins.
• Chlorine (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)):
• \(1s\): 2 electrons. \(2s\): 2 electrons. \(2p\): 6 electrons (fully filled). \(3s\): 2 electrons. \(3p\): 5 electrons (2 orbitals have 2 electrons with opposite spins, 1 orbital has 1 electron).
• Sodium (\(1s^{2}2s^{2}2p^{6}3s^{1}\)):
• \(1s\): 2 electrons. \(2s\): 2 electrons. \(2p\): 6 electrons. \(3s\): 1 electron.
• Neon (\(1s^{2}2s^{2}2p^{6}\)):
• \(1s\): 2 electrons. \(2s\): 2 electrons. \(2p\): 6 electrons (fully filled).

Answer:

1. Nitrogen:

• Full: \(1s^{2}2s^{2}2p^{3}\)
• Short - hand: \([He]2s^{2}2p^{3}\)

1. Chlorine:

• Full: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
• Short - hand: \([Ne]3s^{2}3p^{5}\)

1. Sodium:

• Full: \(1s^{2}2s^{2}2p^{6}3s^{1}\)
• Short - hand: \([Ne]3s^{1}\)

1. Neon:

• Full: \(1s^{2}2s^{2}2p^{6}\)
• Short - hand: \([Ne]\) (or \([He]2s^{2}2p^{6}\))