Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b

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Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b

onding electrons followed by the number of nonbonding electrons, separated by a comma, in the dot structure of this molecule (e.g., 1,2. 2,9

Chemistry

2 answers:

KiRa [2.8K] · Answer 1 · 2026-03-27T00:17:33+03:00

Result: The count of bonding electrons and non-bonding electrons amounts to (4, 18).

Explanation:

The Lewis-dot structure reveals the number of bonding and non-bonding electrons in $XeF_2$ .

Lewis-dot representation: It illustrates the valence electron count for atoms in a molecule and shows how they bond, as well as any lone pairs of electrons.

In this structure, 'Xe' is the central atom while 'F' is the terminal atom.

Xenon comprises 8 valence electrons, whereas fluorine contains 7.

The total number of valence electrons in $XeF_2$ is calculated as 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we can determine

The count of electrons involved in bonding = 4

The count of electrons involved in non-bonding (lone-pairs) = 22 - 4 = 18

Thus, the bonding and non-bonding electron counts are (4, 18).

Below is the Lewis-dot structure for $XeF_2$ .

VMariaS [2.8K] · Answer 2 · 2026-03-27T00:19:02+03:00

The counts for bonding electrons and non-bonding electrons are $\boxed{(4,18)}$

Further説明:

The Lewis structure represents the molecular configuration of an element along with any non-bonding pairs. In covalent compounds, both the bonding electrons and the non-bonding electron pairs can be illustrated through Lewis structures. These structures, along with the formal charges, assist in predicting the geometry, polarity, and reactivity of the molecules.

Typically, the least electronegative atom (excluding hydrogen) serves as the central atom. Therefore, out of xenon and fluorine, xenon will be the central atom. Then, arrange the fluorine atoms around the xenon atom. Next, assess the total number of valence electrons.

The total valence electrons in ${\text{Xe}}{{\text{F}}_2}$ can be calculated as follows:

${\text{Total valence electrons}}=\left[{\left({\text{1}}\right)\left({{\text{Valence electrons of Xenon}}}\right){+}}\left(2\right)\left({{\text{Valence electrons of Fluorine}}}\right)}\right]$

Xenon has 8 valence electrons while fluorine has 7. The valence electrons for ${\text{Xe}}{{\text{F}}_2}$ are as follows:

$\begin{aligned}{\text{Total valence electrons}}\left( {{\text{TVE}}}\right)&=\left[{\left(1\right)\left(8\right)+\left(2\right)\left(7\right)}\right]\\&=22\\\end{aligned}$

The formal charge for an atom can be computed as follows:

${\mathbf{Formal charge=}}\left[\begin{gathered}\left[\begin{gathered}{\mathbf{total number of valence electrons }}\hfill\\{\mathbf{in the free atom}}\hfill\\\end{gathered}\right]{\mathbf{}}-\\\left[{{\mathbf{total number of non - bonding electrons}}}\right]-\\\frac{{\left[{{\mathbf{total number of bonding electrons}}}\right]}}{{\mathbf{2}}}\\\end{gathered}\right]$

Each fluorine forms a single bond with the xenon atom. Each fluorine atom has 3 lone pairs on it.

In a free fluorine atom, there are 7 total valence electrons.

The total number of nonbonding electrons for each fluorine is 6.

The total bonding electrons in each fluorine is 2.

Use these numbers in equation (1) to find the formal charge on F.

$\begin{aligned}{\text{Formal charge on F}&}=\left[{7-6-\frac{2}{2}}\right]\\=0\\\end{aligned}$

Each fluorine has a formal charge of

Xenon creates two single bonds and has three lone pairs.

The total valence electrons in free Xenon is 7.

The lone pairs in Xe total 6.

The total number of bonding electrons in Xe is 4.

Plug these into equation (1) to find the formal charge on Xe.

$\begin{aligned}{\text{Formal charge on Xe}}&=\left[{8-6-\frac{4}{2}}\right]\\&=2-2\\&=0\\\end{aligned}$

Of the 22 electrons, four are used in the two Xe-F bonds. Thus, 18 electrons remain. These account for the non-bonding (lone pairs). Each fluorine has 3 lone pairs and xenon has three as well. (See the attached image for the Lewis structure).

Consequently, the counts of bonding and non-bonding electrons are (4, 18).

To learn more:

1. Molecular shape around the central atoms in glycine:

2. Rules for Lewis structures:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, number of valence electrons, XeF2, formal charge, single bond, bonding electrons, non-bonding electrons, total valence electrons.