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3 years ago

Which of the elements listed below has the smallest atomic radius? Sulfur O Chlorine Oxygen Fluorine

Chemistry

2 answers:

Paha777 [63]3 years ago

6 0

Answer:

Cadmium has larger atomic radius than sulfur.

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.

Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.

Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.

So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.

Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

Maslowich3 years ago

5 0

Answer:

F has the smallest atomic radius.

Explanation:

Atomic radius is one of the perIodic properties defined by the Periodic Table.

Atomich radius is the distance to the nucleus that comprises 95 % of the density electronic load. It can be defined as a function of the internuclear distance. Although this distance varies, depending on whether the atoms are linked by a chemical bond or simply in contact without forming link.

Atomic radius is a periodic property that increases by period (from the top to below) and decreases (from left to right) by group.

In the same group are O and F, so F < O

In the same group, we have S and Cl, so Cl < S

As Cl and F are in the same period, we define the atomic radius Cl > F and

S > Cl (In the same period)

The right way to order the elements are S > Cl > O > F

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A chemist fills a reaction vessel with 7.92 atm nitrogen (N2) gas, 2.02 atm hydrogen (H2) gas, and 2.11 atm ammonia (NH3) gas at

djverab [1.8K]

Answer: The Gibbs free energy of the given reaction is -40 kJ

Explanation:

The given chemical equation follows:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})]$

We are given:

$\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -32.9 kJ/mol = -32900 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$Q_{eq}$ = Ratio of concentration of products and reactants at any time = $\frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}$

$p_{NH_3}=2.11atm$

$p_{N_2}=7.92atm$

$p_{H_2}=2.02atm$

Putting values in above equation, we get:

$\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ$

Hence, the Gibbs free energy of the given reaction is -40 kJ

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Which of the elements listed below has the smallest atomic radius? Sulfur O Chlorine Oxygen Fluorine​

Which of the elements listed below has the smallest atomic radius? Sulfur O Chlorine Oxygen Fluorine