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d1i1m1o1n
2 months ago
7

Which of these elements has the least attraction for electrons ina chemical bond?(1) oxygen (3) nitrogen(2) fluorine (4) chlorin

e
Chemistry
2 answers:
eduard [2.7K]2 months ago
5 0

The element with the lowest attraction for electrons in a chemical bond is \boxed{{\text{(4) chlorine}}}.

Additional Explanation:

Electronegativity:

It refers to an atom's capability to draw shared electrons in a bond towards itself. A more electronegative atom will attract bonding electrons more effectively than its less electronegative counterpart. Consequently, electrons are likely to spend more time around the more electronegative atom, thus acquiring a partial negative charge, while the less electronegative atom gains a partial positive charge.

An element's electronegativity increases as its electronegativity value rises.

Trends of electronegativity in the periodic table:

From left to right, electronegativity increases due to a growing nuclear charge, resulting in a stronger attraction for electrons.

As you move down the groups, electronegativity decreases because the atomic size increases, leading to a weaker attraction between the nucleus and its electrons.

The elements fluorine, oxygen, and nitrogen belong to the same row on the periodic table. As you go to the right, electronegativity increases, hence the sequence is N<O<F. In this grouping, nitrogen possesses the lowest electronegativity.

Comparatively, between chlorine and fluorine, chlorine exhibits a lower electronegativity since electronegativity tapers off as you descend the groups.

Thus, we have two elements, nitrogen and chlorine, with lower electronegativity. Though nitrogen and chlorine have nearly identical electronegativity, because chlorine is larger, its electron is held less tightly by the nucleus. Hence, the element with the least attraction for electrons in a chemical bond is chlorine.

For further information:

1. Which ion was formed by providing the second ionization energy?

2. Determine a chemical equation representing the first ionization energy for lithium:

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Electronegativity

Keywords: Elements with the least attraction for electrons, chemical bond, oxygen, nitrogen, fluorine, chlorine.

Alekssandra [3K]2 months ago
4 0

Electronegativity indicates how strongly an atom attracts electrons in a bond.
You can easily check the values in your reference table!
The values are: oxygen at 3.4, fluorine at 4.0, nitrogen at 3.0, and chlorine at 3.2.
The smallest value corresponds to the least attraction, thus nitrogen (3) is the correct answer.

Hope this clarifies your question!:-)
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lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

500mM = 500mM \times (\frac{1M}{1000M})= 0.5M

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M

10 mL of this solution is further diluted to 250 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M

Convert μM:

2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M

5 mL of this solution is diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M

Convert μM:

0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M

10 mL of this solution is further diluted to 500 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M

10 mL of this solution is further diluted to 1000 mL

M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M

Convert μM:

2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M

Thus, this dilution scheme will yield a 200μM solution.

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