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3 years ago

In which of these diatomic molecules would you NOT find an octet of electrons

Chemistry

1 answer:

Semmy [17]3 years ago

8 0

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³ → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so each atom has an octet of electrons.

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹ → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, we can not find an octet.

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴ → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, each oxygen atom has an octet.

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵ → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, each fluorine atom has an octet of electrons around.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!

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La energía de ionización o potencial de ionización, es la energía necesaria para extraer un electrón de un átomo neutro. Por ejemplo, cuando un átomo de cloro neutro en forma gaseosa recoge un electrón para formar un ion Cl-, libera una energía de 349 kJ/mol o 3,6 eV/átomo.

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3 years ago

Earth’s mantle is a _____.

liubo4ka [24]

A layer of hot liquid rock

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3 years ago

Read 2 more answers

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

Match each chemical reaction with the type of reaction that best describes it.

r-ruslan [8.4K]

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂ (single displacement)

Ca + Br₂ → CaBr₂ (synthesis)

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O (combustion)

2 NaCl → 2 Na + Cl₂ (decomposition)

FeS + 2 HCl → FeCl₂ + H₂S (double displacement)

single displacement - is a chemical reaction of the following type: A + BC → AC + B

double displacement - is a chemical reaction of the following type: AB + CD → AC + BD

synthesis - the chemical product is obtained by combining in a synthesis the constituent elements

combustion - usually a exothermic reaction of a particular compound with oxygen

decomposition - degradation of a compound in simpler elements

8 0

3 years ago

How many moles of KBr will be produced from 10.51 moles of BaBr2?

gregori [183]

Answer:

21.02moles of KBr

Explanation:

Parameters given:

Number of moles BaBr₂ = 10.51moles

Complete reaction equation:

BaBr₂ + K₂SO₄ → KBr + BaSO₄

Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:

The balanced equation is:

BaBr₂ + K₂SO₄ → 2KBr + BaSO₄

From the equation:

1 mole of BaBr₂ produces 2 moles of KBr

∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr

7 0

3 years ago