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3 years ago

Explain why fluorine has a smaller atomic radius than both oxygen and chlorine

Chemistry

1 answer:

Alex777 [14]3 years ago

5 0

Answer:

1) Has a smaller radius than oxygen because of the increased electromagnetic attraction of the nuclei

2) Has a smaller radius than chlorine because all the electrons of F have lower energy levels and have less repulsion of other electrons and hence are more attracted to the nuclei .

Explanation:

Further the electrons are from the nuclei , the bigger the atomic radius is.

(+) attraction of electrons to the nuclei, (-) repulsion of the electrons away from the nuclei.

1) From O to F:

(+) there is one more proton --> Stronger positive charge of the nuclei means that the electrons are attracted more , then they come closer to it and therefore the radius decreases

(-) There is one more electron --> Every electron is repulsed by others away from the nuclei --> the radius increases. But this effect is not so strong because the new electron is added at the same energy level.

Overall the (+) effect is stronger than the (-) effect --> Radius decreases from O to F

2) From F to Cl

(+) there is one more protons --> Same effect as before

(-) There is one more electron --> Every electron is repulsed by others away from the nuclei. But this time the new electrons have a higher energy level --> Meaning that they are less attracted and hence the radius increases.

And also the other inner layers of electrons (electrons of lower energy levels) repulse this new external layer of electrons more effectively than the case of O --> Strong repulsion effect (called electron shielding effect) --> Radius increases

Overall the (-) effect is stronger than the (+) effect --> Radius increases from F to Cl (or decreases from Cl to F)

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The electron dot diagram for a neutral atom of chlorine (atomic number 17) is shown below.

sergejj [24]

Answer:

A. 35Cl1-

Explanation:

Chlorine needs 1 more electron to have full octet thus will take 1 electron and possess a -1 charge.

7 0

3 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

In ADEF, the measure of F=90°, the measure of ZD=10, and EF = 94 feet. Find the

Flauer [41]

Answer:

answer. the measure. is f=90 the equbaliint is 180

Explanation:

yan ang sagot ko

3 0

3 years ago

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici

Brut [27]

Question:

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

Answer:

The mean activity coefficient for HCl solution is 0.78.

Explanation:

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

$E=E_{0}-0.0514 \mathrm{V} \ln \gamma$