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4 years ago

Rank the elements according to highest ionization energy: Be, C, O, Ne, B, Li, F, N

2 answers:

5 0

Given:
Be - Beryllium   -   9,3227
C   - Carbon - 11,2603
O   - Oxygen - 13,6181
Ne - Neon - 21,5645
B   - Boron -   8,298
Li - Lithium -   5,3917
F   - Fluorine - 17,4228
N   - Nitrogen - 14,5341

Arranged from highest ionization energy to lowest ionization energy.

Ne ; F ; N ; O ; C ; Be ; B ; Li

siniylev [52]4 years ago

3 0

The rank of the given elements according to highest ionization energy is as follows:

$\boxed{{\text{Ne}}>{\text{F}}>{\text{N}}>{\text{O}}>{\text{C}}>{\text{Be}}>{\text{B}}>{\text{Li}}}$ .

Further explanation:

Ionization energy:

The amount of energy that is required to remove the most loosely bound valence electrons from the isolated neutral gaseous atom is known as ionization energy. It is represented by IE. Its value is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.

Ionization energy trends in the periodic table:

While moving from left to right in a period, IE increases due to the decrease in the atomic size of the succeeding members. This results in the strong attraction of electrons and hence are difficult to remove.

While moving from top to bottom in a group, IE decreases due to the increase in the atomic size of the succeeding members. This results in the lesser attraction of electrons and hence are easy to remove.

Li, Be, B, C, N, O, F and Ne belong to the same period of the periodic table. Since ionization energy increases while going from left to right in a period, the element present at the leftmost region of the period will have the lowest ionization energy and that present at the rightmost corner will have the highest ionization energy among the rest elements in the same period.

The atomic number of Li is 3 and its electronic configuration is $1{s^2}2{s^1}$ . Its 2s electron can be easily removed, leaving the stable noble gas configuration of helium behind. So its ionization energy is the least among other elements of the same period.

The atomic number of Be is 4 and its electronic configuration is $1{s^2}2{s^2}$ . Its electronic configuration is already fulfilled so it has no tendency to lose any electron so its ionization energy is more than that of Li.

The atomic number of B is 5 and its electronic configuration is $1{s^2}2{s^2}2{p^1}$ . Its 2p electron can be easily removed, leaving completely filled shells behind. So its ionization energy is more than Li but less than that of Be.

The atomic number of C is 6 and its electronic configuration is $1{s^2}2{s^2}2{p^2}$ . It lies to the right of B but to the left of the remaining elements. So its ionization energy is more than B but less than the rest of the elements.

The atomic number of N is 7 and its electronic configuration is $1{s^2}2{s^2}2{p^3}$ . It has a stable half-filled electronic configuration so the electron removal is quite difficult and therefore its ionization energy is higher, even more than that of O.

The atomic number of O is 8 and its electronic configuration is $1{s^2}2{s^2}2{p^4}$ . It lies to the left of F but to the right of N. But N has a stable half-filled configuration. So the ionization energy of O is less than F and N.

The atomic number of F is 9 and its electronic configuration is $1{s^2}2{s^2}2{p^5}$ . It lies to the left of Ne but to the right of the remaining elements. So its ionization energy is less than Ne but more than the rest of the elements.

The atomic number of Ne is 10 and its electronic configuration is $1{s^2}2{s^2}2{p^6}$ . It has a stable fulfilled electronic configuration and therefore it has no tendency to remove an electron. So its ionization energy is the highest among all the elements of the same period.

The rank of the given elements according to highest ionization energy is as follows:

${\text{Ne}}>{\text{F}}>{\text{N}}>{\text{O}}>{\text{C}}>{\text{Be}}>{\text{B}}>{\text{Li}}$

Learn more:

1. Rank the elements according to first ionization energy: brainly.com/question/1550767

2. Write the chemical equation for the first ionization energy of lithium: brainly.com/question/5880605

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: ionization energy, isolated, neutral gaseous atom, valence electrons, Li, Be, B, C, N, O, F, Ne, rank, highest ionization energy.

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Semmy [17]

Answer:

a) the limiting reactant is 02

b) There will remain 0.667 moles of CS2

c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

Explanation:

Step 1: Data given

Number of moles of CS2 = 1.00 mol

Number of moles of O2 = 1.00 mol

Molar mass of O2 = 32 g/mol

Molar mass of CS2 = 76.14 g/mol

Step 2: The balanced equation

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

Step 3: Calculate the limiting reactant

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

O2 is the limiting reactant. It will completely be consumed.(1.00 mol).

CS2 is in excess. There will react 1.00/ 3 = 0.333 moles

There will remain 1.00 - 0.333 = 0.667 moles of CS2

Step 4: Calculate moles of CO2 and SO2

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

For 1.00 mol of O2 we have 1.00 /(3/2) = 0.667 moles of SO2

8 0

3 years ago

If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit

Blizzard [7]

Answer:

For a: The volume of the box is 217.5 mL

For b: The volume of the box is 0.2175 L

Explanation:

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

$V=10.00\times 7.25\times 3.00=217.5cm^3$

For a:

To convert the volume of cuboid into milliliters, we use the conversion factor:

$1mL=1cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL$

Hence, the volume of the box is 217.5 mL

For b:

To convert the volume of cuboid into liters, we use the conversion factor:

$1L=1000cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L$

Hence, the volume of the box is 0.2175 L

3 0

4 years ago

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Morgarella [4.7K]

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Explanation:

There are 2 types of chemical reaction classified on the basis of heat change:

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