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3 years ago

Which element has a greater electronegativity? fluorine (9) or radium (88)

Chemistry

2 answers:

emmainna [20.7K]3 years ago

7 0

Answer:

Fluorine

General Formulas and Concepts:

Chemistry

Reading a Periodic Table
Periodic Trends
Electronegativity - the tendency for an element to attract an electron to itself
Z-effective and Coulomb's Law, Forces of Attraction

Explanation:

The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.

Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.

However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.

Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.

Lapatulllka [165]3 years ago

6 0

Answer:

Fluorine

Explanation:

Electronegativity increases as you go from left to right across the periodic table and decreases as you go from top to bottom of the periodic table.

Fluorine is in period 3, group 17

Radium is in period 7, group 2

Radium is in period 7 and we know that electronegativity decreases as you move from top to bottom.

Explanation: As you move from top to bottom, you are in higher energy level, which means that your distance from the nucleus is further away.

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Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a

nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

$A->B;\Delta _rG^o=-13 kJ/mol$

$C ->D ;\Delta _rG^o=3.5 kJ/mol$

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

$Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167$

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

$K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249$

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

6 0

3 years ago

Which of the following is a physical property of a substance?? A. Combustion B. Corrosion C. Tendency to tarnish D. Solubility

Fofino [41]

A physical property does not change the substance.

Solubility would be the answer since all of the rest are changing the substance. They all deal with bonds except solubility.

Answer: D. Solubility

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3 years ago

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A geologist conducts an investigation to determine the absolute age of a fossil. She then

ahrayia [7]

Answer:

it improves the accuracy of the results

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3 years ago

Which is a characteristic of mixtures

barxatty [35]

They are three types of mixtures:
-solutions : they are homogeneous mixtures of 2 or more sub. in a single phase.
-suspensions: if the particles in a solvent are so large that they settle out unless constantly tired, the mixture is called a suspension.
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4 years ago

Read 2 more answers

What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3

Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
M acid= 0.129 M
V base= ?
M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

$V base=\frac{0.0137 L*0.129 M}{0.135 M}$

V base=0.0131 L = 13.1 mL

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

4 0

4 years ago