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3 years ago

Which of the following has the largest atomic radius: F-, F, CL-, or CL?

2 answers:

6 0

As you go down the table, more electron shells are added to make the radius larger, but as you go to the right, the protons' attraction increases so the radius actually shrinks. Therefore, according to the trend, Cl has a larger radius because it is lower in the table.

Answer: CL has a larger Atomic Radius.

insens350 [35]3 years ago

3 0

Answer: Option (c) is the correct answer.

Explanation:

More is the number of electrons present in an atom more will be its atomic size.

Atomic number of fluorine is 9 which means in neutral state a fluorine atom has only 9 electrons.

But when one electron is gained by a neutral fluorine atom then it changes into $F^{-}$ ion and there will be a total of 10 electrons in it.

On the other hand, atomic number of chlorine is 17 which means in neutral state a chlorine atom has only 7 electrons.

And, when this neutral chlorine atom gains an electron then it will form $Cl^{-}$ ion. Hence, total number of electrons present in a $Cl^{-}$ ion is 18.

Therefore, a $Cl^{-}$ ion has more number of electrons that is, 18 out of the given species.

Thus, we can conclude that atomic radius of $Cl^{-}$ ion is the largest.

You might be interested in

What is the volume of one mole of any gass at STP?

fenix001 [56]

Every mole is 22.4 L at STP

7 0

3 years ago

Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50Ti (5.3%). What is the average atomic

NNADVOKAT [17]

Hey there!:

Isotopes : abundance :

46 Ti 8.0%

47 Ti 7.8 %

48 Ti 73.4 %

49 Ti 5.5 %

50 Ti 5.3 %

Weighted average = ∑ Wa * % / 100

Therefore:

( 46 * 8.0) + (47 * 7.8 ) + (48 * 73.4 ) + ( 49 * 5.5 ) + ( 50*5.3 ) / 100 =

4792.3 / 100

= 47.923 a.m.u

Hope that helps!

7 0

3 years ago

Predict the products of La(s) + O2(aq) ->

mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

Hello there!

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

I hope this helps you to understand better. Enjoy your studies

3 0

2 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T $_2$ ),

P $_1$ V $_1$ / T $_1$ = P $_2$ V $_2$ / T $_2$ ,

T $_2$ = P $_2$ V $_2$ T $_1$ / P $_1$ V $_1$ ,

T $_2$ = 0.73 atm $*$ 1.8 L $*$ 298.15 K / 1 atm $*$ 1.2 L = ( 0.73 $*$ 1.8 $*$ 298.15 / 1 $*$ 1.2 ) K = 326.47425 K,

T $_2$ = 326.47425 K = 53.32425 C

7 0

3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0

3 years ago