Problem 13
Question
Using the Group 18 element shorthand, give the electronic configurations for the following elements or ions: (a) arsenic; (b) cobalt; (c) holmium; (d) bromide ion. How many unpaired electrons are there in each case? (Section 3.6)
Step-by-Step Solution
Verified Answer
(a) As: 3 unpaired electrons; (b) Co: 3 unpaired electrons; (c) Ho: 5 unpaired electrons; (d) Br⁻: 0 unpaired electrons.
1Step 1: Understand Group 18 (Noble Gas) Shorthand
Group 18 elements, also known as noble gases, include Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn). In the electronic configuration, these elements can be used as shorthand to simplify the representation by indicating the configuration of the previous noble gas in brackets.
2Step 2: Determine Electronic Configuration for Arsenic (As)
Arsenic is the 33rd element. The nearest noble gas before arsenic is Argon (Ar), which has an atomic number of 18. Using noble gas shorthand, the electronic configuration is: \[ [Ar] ext{ 3d}^{10} ext{ 4s}^2 ext{ 4p}^3 \] The outermost p subshell (4p) has 3 electrons, so arsenic has 3 unpaired electrons.
3Step 3: Determine Electronic Configuration for Cobalt (Co)
Cobalt is the 27th element. The nearest noble gas before cobalt is Argon (Ar), atomic number 18. Using noble gas shorthand, the electronic configuration is: \[ [Ar] ext{ 3d}^7 ext{ 4s}^2 \] The 3d subshell with 7 electrons has 3 unpaired electrons.
4Step 4: Determine Electronic Configuration for Holmium (Ho)
Holmium is the 67th element. The nearest noble gas before holmium is Xenon (Xe), atomic number 54. Using noble gas shorthand, the electronic configuration is: \[ [Xe] ext{ 4f}^{11} ext{ 6s}^2 \] In the 4f subshell with 11 electrons, there are 5 unpaired electrons.
5Step 5: Determine Electronic Configuration for Bromide Ion (Br⁻)
Bromine ion (Br⁻) is the same as an elemental bromine with one extra electron, making it 36 instead of 35. The nearest noble gas is Krypton (Kr), atomic number 36. So, its configuration is simply: \[ [Kr] \] As it is stable like a noble gas, the bromide ion has 0 unpaired electrons.
Key Concepts
Noble Gas ShorthandUnpaired ElectronsElectron Configuration of Transition Metals
Noble Gas Shorthand
Using the noble gas shorthand is a clever way to write electronic configurations. It significantly shortens the process. Instead of writing out every single electron, we use the nearest noble gas to cover part of the configuration. For example, if you're working with cobalt, you use argon in brackets:
- Since argon has 18 electrons, writing `[Ar]` stands in for those 18.
- The electronic configuration of cobalt is then `\([Ar] \text{ 3d}^7 \text{ 4s}^2\)`.
Unpaired Electrons
Unpaired electrons play a crucial role in an element's magnetic and chemical properties. When we talk about unpaired electrons, we're focusing on those electrons that don't have a "partner" in their orbital. Here’s why they matter:
- Magnetism: Elements with unpaired electrons, like iron, have magnetic properties. The unpaired electrons cause spin, which aligns in the presence of a magnetic field.
- Chemical Reactivity: Unpaired electrons are more likely to participate in bonds. They seek out partnerships to complete their orbitals, influencing how an element reacts with others.
Electron Configuration of Transition Metals
Transition metals have electron configurations that might seem a bit tricky at first. However, understanding these helps in grasping their unique characteristics. Here are some key points:
- Partially Filled d Subshell: Transition metals like cobalt have an incomplete d subshell, which gives them unique properties like color and various oxidation states.
- Variable Oxidation States: Cobalt with its `3d^7` configuration can lose different numbers of electrons, leading to multiple oxidation states.
- Strong Catalysts: Many transition metals serve as catalysts, owing to their ability to lend and take electrons easily.
Other exercises in this chapter
Problem 7
How many orbitals are possible for \(n=5 ?\) Identify the orbital types giving the number of each. (Section 3.5 )
View solution Problem 9
For the following atomic orbitals, give the values of the quantum numbers \(n\) and \(L\) In each case indicate what values for \(m\), are allowed. (Section 3.5
View solution Problem 17
For each of the following pairs of elements, which has the larger first ionization energy: (a) sodium and magnesium; (b) magnesium and aluminium; (c) magnesium
View solution Problem 18
Account for the following trends in ionization energy. (Section \(3.7)\) (a) The increase from oxygen to fluorine. (b) The decrease from nitrogen to phosphorus.
View solution