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4 years ago

How many electrons will a non-metal generally have in its outer shell? 1 - 3 1 - 4 4 - 7 8

Chemistry

2 answers:

Andrews [41]4 years ago

8 0

Answer: The correct answer is 4-7.

Explanation:

A non-metal is defined as the element which gains electron to attain stable electronic configuration and to attain a negative ion which is anion.

$X+ne^-\rightarrow X^{n-}$

Non-metals are the elements which belong to Group 14 to Group 17.

The valence electronic configuration of the elements:

Belonging to Group 14 = $ns^2np^2$

Number of valence electrons = 4

Belonging to Group 16= $ns^2np^3$

Number of valence electrons = 5

Belonging to Group 16 = $ns^2np^4$

Number of valence electrons = 6

Belonging to Group 17 = $ns^2np^5$

Number of valence electrons = 7

Hence, the correct answer is 4-7.

AlladinOne [14]4 years ago

4 0

They generally have 4 - 7 electrons in their outer shell.

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Answer:

Explanation:

C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"?

Based on n+1 rule. Here n=2 (identical beta protons).

2+1=3

So the multiplicity of alpha proton is triplet, .

D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal?

The meaning of the adjective is this: the multiplicity of beta protons is singlet only (no spliting) in absence of alpha proton . But beta protons splits as doublet (n=1) in the presence of alpha proton,

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There are 3 bonds in between alpha and beta protons in a molecule.

F) What is the multiplicity of the Proton-betas' signal?

Following the n+1 rule, here n=1 (1 alpha proton) so 1+1=2. Hence it is a doublet.

7 0

4 years ago

1. The electron arrangement ions X and Y2 are 2, 8, and 2, 8, 8 respectively.

svetoff [14.1K]

The given question is incomplete. the complete question is:

The electron arrangement ions $X^{3+}$ and $Y^{2-}$ are 2, 8, and 2, 8, 8 respectively.

Write the electronic arrangement of the elements X and Y.

Write the formula of the compound that would be formed between X and Y.

Answer: 1. The electronic arrangement for X and Y are 2,8, 3 and 2,8,5 respectively.

The formula of the compound formed is $X_2Y_3$

Explanation:

Electronic configuration is the arangement of electrons in an atom in order of increasing energies.

Cations are formed when electrons are lost by atoms and anions are formed when electrons are gained by atoms.

$X^{3+}:2,8$

$X:2,8,3$

$Y^{3-}:2,8,8$

$Y:2,8,5$

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

Here element X is having an oxidation state of +3 called as $X^{3+}$ cation and $Y^{2-}$ is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $X_2Y_3$

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3 years ago

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3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

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