Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the electron capacity of the \(n\) th energy level? What is the capacity of the fourth energy level?

Short Answer

Expert verified
The electron capacity of the nth energy level is \(2n^2\). The fourth energy level can hold 32 electrons.

Step by step solution

01

Identify the Formula

The electron capacity of a given energy level can be found using the formula: \[ 2n^2 \] where \( n \) is the principal quantum number (the energy level).
02

Calculate the Capacity of the nth Energy Level

Substitute \( n \) into the formula to find the capacity. For the nth energy level, it is: \[ 2n^2 \].
03

Calculate the Capacity of the Fourth Energy Level

To find the capacity of the fourth energy level, substitute \( n = 4 \) into the formula: \[ 2(4)^2 = 2 \times 16 = 32 \]. Therefore, the fourth energy level can hold 32 electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

principal quantum number
The principal quantum number () is a key concept in quantum mechanics that describes the size and energy of an electron's orbit around the nucleus of an atom. Denoted usually as 'n', it determines the main energy level or shell of an electron.

The principal quantum number can take any positive integer value starting from 1 (n = 1, 2, 3,...). Lower values of 'n' correspond to energy levels closer to the nucleus, with lower energy. Higher values indicate energy levels that are further away from the nucleus and have higher energy.

In summary, the principal quantum number not only helps to identify the location of an electron but also helps to estimate the energy associated with that electron's position.
electron configuration
Electron configuration is the arrangement of electrons in an atom's orbitals. This configuration follows specific rules and principles that dictate how electrons fill available energy levels and sublevels.

The most notable rules include the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
  • The Aufbau principle states that electrons occupy the lowest energy orbital available.
  • Hund's rule emphasizes that electrons will fill up degenerate orbitals (orbitals of the same energy) singly first before pairing.
  • Lastly, the Pauli exclusion principle asserts that no two electrons can have identical quantum numbers in the same atom.
For example, the electron configuration of an oxygen atom, which has 8 electrons, is written as: 1s² 2s² 2p⁴. This notation indicates the distribution of electrons across different orbitals.
energy level formula
The energy level formula is used to determine the number of electrons that can be accommodated within a specific energy level (shell) of an atom. This is crucial for understanding atomic structure and is expressed by: \ \ The formula is: \[ 2n^2 \]
where \( n \) denotes the principal quantum number (energy level).

Using this formula, you can calculate the maximum number of electrons that can occupy any given energy level. For instance, if n = 3 (the third energy level), the electron capacity is calculated as \[ 2(3)^2 = 2 \times 9 = 18 \]. Therefore, the third energy level can hold a maximum of 18 electrons. This formula is fundamental in determining the electron configuration and the placement of electrons in an atom.
n-th energy level
The concept of the n-th energy level refers to the position of electrons in different shells around an atom's nucleus, each defined by their principal quantum number 'n'.

Each n-th energy level can be pictured as an orbit where electrons move. As the value of 'n' increases, these orbits are further away from the nucleus and have higher energy. It’s essential to know that each n-th energy level can hold a specific number of electrons, calculated by the formula \[ 2n^2 \]. For instance:
  • The 1st energy level (n=1) can hold up to 2 electrons.
  • The 2nd energy level (n=2) can accommodate up to 8 electrons.
  • The 3rd energy level (n=3) can contain up to 18 electrons.
  • The 4th energy level (n=4) can hold up to 32 electrons.
By understanding the n-th energy level, we can predict how electrons are distributed in an atom and how this affects chemical properties and reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain the relationship between atomic size and reducing strength in Group \(1 \mathrm{~A}(1)\). Explain the relationship between IE and oxidizing strength in Group \(7 \mathrm{~A}(17)\)

Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic: (a) \(\mathrm{V}^{3+}\) (b) \(\mathrm{Cd}^{2+}\) (c) \(\mathrm{Co}^{3+}\) (d) \(\mathrm{Ag}^{+}\)

Name the element described in each of the following: (a) Smallest atomic radius in Group \(6 \mathrm{~A}(16)\) (b) Largest atomic radius in Period 6 (c) Smallest metal in Period 3 (d) Highest IE \(_{1}\) in Group \(4 \mathrm{~A}(14)\) (e) Lowest IE \(_{1}\) in Period 5 (f) Most metallic in Group \(5 \mathrm{~A}(15)\) (g) Group \(3 \mathrm{~A}(13)\) element that forms the most basic oxide (h) Period 4 element with highest energy level filled (i) Condensed ground-state electron configuration of [Ne] \(3 s^{2} 3 p^{2}\) (j) Condensed ground-state electron configuration of \([\mathrm{Kr}] 5 s^{2} 4 d^{6}\) (k) Forms \(2+\) ion with electron configuration [Ar] \(3 d^{3}\) (1) Period 5 element that forms \(3+\) ion with pseudo-noble gas configuration (m) Period 4 transition element that forms \(3+\) diamagnetic ion (n) Period 4 transition element that forms \(2+\) ion with a half-filled \(d\) sublevel (o) Heaviest lanthanide (p) Period 3 element whose \(2-\) ion is isoelectronic with Ar (q) Alkaline earth metal whose cation is isoelectronic with \(\mathrm{Kr}\) (r) Group \(5 \mathrm{~A}(15)\) metalloid with the most acidic oxide

Before Mendeleev published his periodic table, German scientist Johann Döbereiner grouped elements with similar properties into "triads," in which the unknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities: (a) The atomic mass of \(\mathrm{K}\) from the atomic masses of \(\mathrm{Na}\) and \(\mathrm{Rb}\) (b) The melting point of \(\mathrm{Br}_{2}\) from the melting points of \(\mathrm{Cl}_{2}\) \(\left(-101.0^{\circ} \mathrm{C}\right)\) and \(\mathrm{I}_{2}\left(113.6^{\circ} \mathrm{C}\right)\left(\right.\) actual value \(\left.=-7.2^{\circ} \mathrm{C}\right)\)

Niobium \((\mathrm{Nb} ; Z=41)\) has an anomalous ground-state electron configuration for a Group \(5 \mathrm{~B}(5)\) element: \([\mathrm{Kr}] 5 s^{1} 4 d^{4} .\) What is the expected electron configuration for elements in this group? Draw partial orbital diagrams to show how paramagnetic measurements could support niobium's actual configuration.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free