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3 years ago

Which ion has the smallest radius? (1) O2- (2) S2- (3) Se2- (4) Te2-

Chemistry

2 answers:

podryga [215]3 years ago

5 0

Answer:

Option-1 (O²⁻) is the correct answer.

Explanation:

All given anions contains same charge. So, we can ignore the effect of charge on these anions.

As we know all given compounds belongs to same group (Group 6) in periodic table. And from top to bottom along the group the elements are placed as,

Oxygen O

Sulfur S

Selenium Se

Tellurium Te

Hence, moving from top to bottom along the group the number of shells increases. And with increase in number of shell the atomic or ionic radii increases. As Oxygen is present at the top of the group, therefore, it has the smallest radius due to less number of shells.

mel-nik [20]3 years ago

3 0

O²⁻ ion has the smallest radius

<h3>Further explanation </h3>

In an atom there are levels of energy in the shell and sub shell.

This energy level is expressed in the form of electron configurations.

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d and f. The maximum number of electrons for each sub shell is

s: 2 electrons
p: 6 electrons
d: 10 electrons and
f: 14 electrons

Charging electrons in the sub shell uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Each sub-shell also has an orbital shown in the form of a square in which there are electrons symbolized by half arrows

The atomic radius shows the distance of the atomic nucleus to the electrons in the outer shell

From left to right in the period system the atomic radius gets smaller, while from one group from top to bottom the atomic radius gets longer

The more the number of shells the atom has, the radius of the atom getting longer, but if the number of shells is the same, the larger atomic number has shorter radius because the core charge is greater so that the attraction of the nucleus to the electrons is stronger

In the ions

(1) O²⁻

(2) S²⁻

(3) Se²⁻

(4) Te²⁻

All of these ions are ions from the elements of the VIA group in the periodic system. This group consists of oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po).

The periodic system is arranged in the order of atomic numbers, whereas in this VIA group from top to bottom, the atomic number gets bigger

Element O

Atomic Number: 8

Electron configuration: 1s² 2s² 2s⁴ (number of shells 2)

Element S

Atomic Number: 16

Electron configuration: [Ne] 3s² 3p⁴ (number of shells 3)

Element Se

Atomic Number: 34

Electron configuration: [Ar] 3d¹⁰ 4s²4p⁴ (number of shells 4)

Element Te

Atomic Number: 52

Electron configuration: [Cr] 4d10 5s² 5p⁴ (number of shells 5)

From the configuration of the electron shows that the lower the shell owned by the elements of the VIA group the greater (from 2 to 5) so that the radius is also getting longer/bigger

Because what is being asked is the radius of the ion, the electron configuration of each element is equally reduced by 2 electrons, but the number of shells remains the same

(1)O²⁻

Electron configuration: 1s² 2s² 2s²

(2) S²⁻

Electron configuration: [Ne] 3s² 3p²

(3) Se²⁻

Electron configuration: [Ar] 3d¹⁰ 4s²4p²

(4) Te²⁻

Electron configuration: [Cr] 4d¹⁰ 5s² 5p²

Because the amount of shell from the Oxygen ion is the smallest, the radius of the O²⁻ is also the smallest

<h3>Learn more </h3>

electron affinity

brainly.com/question/1440853

Identify the group number in the periodic table

brainly.com/question/2014634

the charge on each ion in the compounds

brainly.com/question/5880856

Keywords: element, group 6A, the smallest radius

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A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite

zheka24 [161]

Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

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3 years ago

Which statement is a scientific conclusion?

dolphi86 [110]

Answer:

Explanation:

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3 years ago

How many moles are in 8.63 x 103 atoms of Li?

Ira Lisetskai [31]

<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})$
Multiply/Divide: $\displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

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3 years ago

Mr. Chem S. Tree added water to 250. ML of a 2.50 M NaOH solution, until the final volume was 500. ML. What is the new molarity

tresset_1 [31]

Answer:

molarity of diluted solution = 1.25 M

Explanation:

Using,

C1V1 (Stock solution) = C2V2 (dilute solution)

given that

C1 = 2.50M

V1 = 250ML

C2 = ?

V2 = 500ML

2.50 M x 250 mL = C2 x 500 mL

C2 = (2.50 M x 250 mL) / 500 mL

C2 = 1.25 M

Hence, molarity of diluted solution = 1.25 M

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Shkiper50 [21]

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