All categories

16 days ago

Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?

Chemistry

2 answers:

lions [2.9K]16 days ago

8 0

does not conform to the octet rule.

Further Details:

The octet rule: establishes that an atom must attain an outer shell of eight electrons for stability (octet signifies a group of eight). This principle was elucidated by Kossel and Lewis.

Atoms of identical or differing elements possessing an incomplete electronic configuration, with fewer than 8 electrons, exhibit instability and thus bond to form stable molecules with complete octets. They will either share or gain/loss electrons to bond effectively.

In ${\mathbf{ClO}}$ , Chlorine (Cl) has 7 electrons in its outer shell while oxygen has 6. Consequently, they share 2 electrons each to achieve stability with their octets. This process leaves the chlorine atom with ten electrons in its outer shell, thus violating the octet rule.

In ${\mathbf{Cl}}{{\mathbf{O}}^ - }$ , Chlorine (Cl) holds 7 electrons in its outer shell, while the oxygen ion carries a negative charge, indicating it too has 7 outer shell electrons. They then share one electron each to fill their octets and achieve stability. All atoms maintain eight electrons after that interaction thus adhering to the octet rule.

In ${\mathbf{ClO}}_2^ -$ , Chlorine (Cl) with its 7 outer shell electrons has two oxygens attached, one negatively charged and one neutral. Chlorine combines its electrons: two with neutral oxygen (forming a double bond) and one with the negatively charged oxygen (creating a single bond). Following this, the Chlorine holds ten electrons in its outer shell and fails to observe the octet rule.

In ${\mathbf{ClO}}_3^ -$ , Chlorine (Cl), possessing seven outer shell electrons and connected to four oxygens, also ends up with several additional electrons from those bonds, thus not following the octet rule due to having fourteen electrons.

Discover more: ${\mathbf{ClO}}_4^ -$ $\left( {{{\text{O}}^ - }} \right)$

The octet rule clarifies that:

2. The number of covalent bonds formed by nitrogen:

Response summary:

Grade: Secondary School Subject: Chemistry

Chapter:

Chemical Bonding

Keywords: octet rule, number of electrons, chlorine, oxygen, bonds

alisha [2.9K]16 days ago

3 0

The proper response includes ClO, ClO3-, ClO- and ClO4-

Kossel and Lewis proposed an important chemical bonding theory in 1916 referred to as the electronic theory of chemical bonding. By this concept, atoms bond through the transfer of valence electrons (either gaining or losing them) or by sharing valence electrons to attain a full shell of eight (the octet rule).

In ClO2-, oxygen attains 8 electrons in its valence shell, sharing one electron with chlorine to finish Cl’s octet. In the other four compounds, we can see they have either more or less than 8 electrons in the outer shell of oxygen, indicating that ClO, ClO3-, ClO- and ClO4- are not compliant with the octet rule.

You might be interested in

The electron configuration of an element is 1s22s22p4. Describe what most likely happens when two atoms of this element move tow

castortr0y [3046]

Answer:

Cuando dos átomos se acercan entre sí, se genera un compuesto al compartir pares de electrones que cada uno de los átomos aporta, permitiéndoles alcanzar los 8 electrones de valencia (octeto) en su capa externa.

Explanation:

La configuración electrónica del elemento puede escribirse de la siguiente manera;

1s²2s²2p⁴

La configuración electrónica dada es equivalente a la del oxígeno, por lo tanto, tenemos;

El número de electrones en la capa de valencia = 2 + 4 = 6 electrones

Por consiguiente, cada átomo necesita 2 electrones para completar sus 8 electrones (octeto) en la capa externa.

Al acercarse los dos átomos, reaccionan y se combinan para formar un compuesto al compartir 4 electrones, 2 de cada átomo, de modo que cada átomo obtenga 2 electrones adicionales en su órbita externa en el nuevo compuesto y así se logre la configuración estable de octeto para cada uno de los átomos en el compuesto recién formado.

4 0

2 months ago

Read 2 more answers

From this list, choose all of the ionic compounds. Check all that apply. View Available Hint(s) Check all that apply. POCl3 KOCH

lions [2927]

Answer: $KOCH_2CH_3$ , $CH_3CH_2COONa$ , NaOH are all ionic compounds.

Explanation:

Ionic compounds form when one atom transfers its valence electrons to another atom, resulting in partial opposite charges on the participating atoms, which creates a strong attractive force between them.

An ionic bond is typically established between a metal and a non-metal.

For instance, $KOCH_2CH_3$ , $CH_3CH_2COONa$ , NaOH are all ionic compounds.

Conversely, a covalent compound arises through the sharing of electrons among the atoms involved, where generally a non-metal bonds with the same or another non-metal.

For example, $POCl_{3}$ , $SOCl_{2}$ etc are all covalent compounds.

Therefore, we can affirm that $KOCH_2CH_3$ , [tex]CH_3CH_2COONa, NaOH are indeed ionic compounds.

4 0

1 month ago

A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained

VMariaS [2998]

Answer: Yes, there is sufficient sodium carbonate available.

Explanation:

In this scenario, according to the specified reaction:

Using stoichiometry, one can figure out the grams of sodium carbonate required to neutralize 1,665 g of sulfuric acid as outlined below:

$H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)$

Hence, the amount on hand is 2.0 kg, which leaves 0.2 kg as surplus, therefore:

A. Yes, there is sufficient sodium carbonate available.

$1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3$ Best regards.

5 0

24 days ago

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

KiRa [2933]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × $\frac{1mol}{159,609g}$ = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × $\frac{1mol}{18,02g}$ = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0

1 month ago

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

lions [2927]

In a 100 g sample of the compound, there are 63.57 g of carbon, 6 g of hydrogen, 9.267 g of nitrogen, and 21.17 g of oxygen. First, convert these masses into moles (n) using the formula n = m/M, where M is the molar mass from the periodic table. For carbon: 63.57 g C -> 63.57 g C / 12.01 g/mol = 5.29 moles C. For hydrogen: 6 g H -> 6 g H / 1.008 g/mol = 5.95 moles H. For nitrogen: 9.267 g N -> 9.267 g N / 14.01 g/mol = 0.6615 moles N. For oxygen: 21.17 g O -> 21.17 g O / 16.00 g/mol = 1.32 moles O. Thus, the mole ratio looks like this: C 5.29 H 5.95 N 0.6615 O 1.32. Now, divide each value by the smallest number (1.32): C 4 H 4.5 N 0.5 O 1. To eliminate fractions, multiply all values by 2, yielding C8H9N1O2. Now, all numbers are integers! Hence, the empirical formula is C8H9NO2. Although the empirical formula isn't always the same as the molecular formula, in this instance, it corresponds to acetaminophen.

5 0

17 days ago

Read 2 more answers