Question

1. write the electron configuration for the following elements:

a. p
b. in
c. pt

Explanation:

Part a: Electron Configuration of P (Phosphorus)

Step 1: Determine the atomic number of P

Phosphorus (P) has an atomic number of 15, meaning it has 15 electrons.

Step 2: Apply the Aufbau principle, Pauli exclusion principle, and Hund's rule

Start filling orbitals in the order: \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), etc.

• \(1s\) orbital can hold 2 electrons: \(1s^2\)
• \(2s\) orbital can hold 2 electrons: \(2s^2\)
• \(2p\) orbital can hold 6 electrons: \(2p^6\)
• \(3s\) orbital can hold 2 electrons: \(3s^2\)
• Remaining electrons: \(15 - (2 + 2 + 6 + 2) = 3\), which go into the \(3p\) orbital: \(3p^3\)

Step 1: Determine the atomic number of In

Indium (In) has an atomic number of 49, so it has 49 electrons.

Step 2: Apply the electron filling order

The order of filling is \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), \(4p\), \(5s\), \(4d\), \(5p\), etc.

• \(1s^2\), \(2s^2\), \(2p^6\), \(3s^2\), \(3p^6\), \(4s^2\), \(3d^{10}\), \(4p^6\), \(5s^2\), \(4d^{10}\) (total electrons so far: \(2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 = 48\))
• Remaining 1 electron goes to \(5p\): \(5p^1\)

Step 1: Determine the atomic number of Pt

Platinum (Pt) has an atomic number of 78, so it has 78 electrons.

Step 2: Apply the electron filling order (considering exceptions for transition metals)

The noble gas before Pt is Xenon (Xe) with electron configuration \([Xe] = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (54 electrons).

• Remaining electrons: \(78 - 54 = 24\)
• The filling order for the remaining electrons: \(6s\), \(4f\) (but Pt has no \(4f\) electrons filled yet), \(5d\), \(6p\). However, due to electron configuration exceptions (stability from half - filled or filled d - orbitals), Pt has a configuration where we adjust the \(6s\) and \(5d\) electrons.
• The correct filling gives: \([Xe] 4f^{14}\) (wait, no, Pt is in the 5d series, before the lanthanides fully fill. Wait, correct approach:

The electron configuration of Pt is \([Xe] 4f^{14}\) is wrong. Let's do it step by step.
Wait, the correct order after Xe (\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)):
We fill \(6s\), then \(4f\) (but Pt is atomic number 78, Xe is 54. \(78 - 54=24\)). The \(4f\) orbitals start filling at atomic number 58 (Ce), so for Pt (78), \(4f\) is fully filled (\(4f^{14}\))? No, wait, the order of filling is \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), \(4p\), \(5s\), \(4d\), \(5p\), \(6s\), \(4f\), \(5d\), \(6p\).
So after Xe (\(5s^2 4d^{10} 5p^6\) is part of Xe? Wait, no, Xe electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (atomic number 54). Then for Pt (78), we have \(78 - 54 = 24\) electrons left.
We fill \(6s^2\) (2 electrons), then \(4f^{14}\) (14 electrons), then \(5d^8\) (8 electrons). But there is a special case for Pt, where one electron from \(6s\) moves to \(5d\) to make \(5d^9\) and \(6s^1\)? Wait, no, the correct electron configuration of Pt is \([Xe] 4f^{14} 5d^9 6s^1\)? No, actually, the correct electron configuration of platinum (Pt) is \([Xe] 4f^{14} 5d^9 6s^1\) is incorrect. Wait, let's calculate the number of electrons:
Xe has 54 electrons. \(4f^{14}\) is 14, \(5d^9\) is 9, \(6s^1\) is 1. \(54+14 + 9+1=78\). But actually, the correct electron configuration of Pt is \([Xe] 4f^{14} 5d^9 6s^1\) (due to the stability gained from having a half - filled or filled d - orbital and s - orbital adjustment). Wait, another way:
The electron configuration of Pt can also be written as \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1 4f^{14} 5d^9\)

Answer:

The electron configuration of P is \(1s^2 2s^2 2p^6 3s^2 3p^3\)

Part b: Electron Configuration of In (Indium)