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umka2103 [35]
3 years ago
6

Lewis structure XeH4? The hydrogen atom is not actually electronegative enough to form bonds to xenon. Were the xenon-hydrogen b

ond to exist, what would be the structure of XeH4?

Chemistry
2 answers:
nalin [4]3 years ago
8 0

The Lewis structure of {\mathbf{Xe}}{{\mathbf{H}}_{\mathbf{4}}} is attached in the image. The xenon hydrogen bonds will exist.

Further explanation:

Chemical bond is formed by the attraction of atoms or ions. Due to electronegativity, atoms are able to attract electrons in a chemical bond. The chemical bond is also formed with the help of valence electrons in an atom. The bond forms should obey the octet rule. The covalent bond is the bond formed by mutual sharing of electrons between the bonded atoms.

The bonding between the different atoms in covalent molecules is shown by structures known as the Lewis structures. These structures reveal the presence of lone pairs in the molecule. Also known as Lewis dot diagrams, electron dot diagrams, Lewis dot structures.

Hydrogen is not an electronegative atom. Due to the presence of valence electrons in hydrogen, it forms covalent bond with other atoms.

Hydrogen has one valence electron. Xe has eight electrons present in the outermost shell. So Xe has 8 valence electrons.

In {\text{Xe}}{{\text{H}}_{\text{4}}} , four hydrogen atom has 4 valence electrons as each hydrogen atom has one valence electron. Also, Xe has eight valence electrons. The 4 valence electrons of hydrogen will form four sigma covalent bonds with 4 electrons of Xe, and the remaining 4 electrons of Xe will form 2 lone pairs of electrons. Therefore, xenon hydrogen bonds will exist.

Lewis structure of{\mathbf{Xe}}{{\mathbf{H}}_{\mathbf{4}}} (Refer to the structure in the attached image):

The total number of valence electrons of {\text{Xe}}{{\text{H}}_{\text{4}}}is calculated as follows:

Total valence electrons = [(1) (Valence electrons of Xe) + (4) (Valence electrons of H)]

\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}}\right)&=\left[{\left({\text{1}}\right)\left(8\right)+\left({\text{4}}\right)\left(1\right)}\right]\\&=12\\\end{aligned}  

Learn more:

1. Molecular shape around each central atoms in the amino acid glycine: brainly.com/question/4341225

2. How many molecules will be present on completion of reaction?:brainly.com/question/4414828

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, valence electrons, XeH4, Xe, hydrogen, chemical bond, bonding electrons, total valence electrons.

neonofarm [45]3 years ago
4 0

 <span>Xenon has 8 valence electrons, each hydrogen atom has 1, therefore you have 8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule (180 degrees from each other). The resulting molecular structure will be square planar.</span><span>

































8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule . The resulting  structure is square planar.</span>
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Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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You have a sample of 75.6 g of C3H8. How many moles of C3H8 are in the sample?
dsp73

The number of moles that are contained in the given mass of propane (C_3H_8 is 1.7143 moles.

<u>Given the following data:</u>

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<u>Scientific data:</u>

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To calculate the number of moles that are contained in the given mass of propane (C_3H_8):

<h3>How to calculate the moles of a compound.</h3>

In this exercise, you're required to determine the number of moles of propane that are contained in the given sample:

Mathematically, the number of moles contained in a chemical compound is given by this formula:

Number\;of\;moles = \frac{mass}{molar\;mass}

Substituting the given parameters into the formula, we have;

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Thus, right answer is:

A)  positive; added

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