Arrange the following atoms in order of increasing ionization energy: lithium, oxygen, magnesium, strontium, chlorine, and tellurium. Explain your order.

Ionization energy is the energy required to remove the most loosely bound electron from the last shell.

In the periodic table, there are groups and periods.

In a period, the ionization energy increases as we move across a period as the size decreases. The electrons get added to the same shell and the nuclear charge increases.

The valence shell moves nearer to the nucleus.  Thus more energy is required to remove the valence electron.

In a group, the ionization energy decreases from top to bottom as we move down a group.

The size increases as the electrons get added up in the new shell, the electron in the outermost orbital gets far away from the nucleus. And hence, the electron will be easier to remove. Thus, lesser energy is required to remove the valence electron.

The correct order of increasing ionization energy is:

lithium < strontium < magnesium < tellurium < chlorine < oxygen

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lithium < strontium < magnesium < tellurium < chlorine < oxygen  

Explanation:

Ionization energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.

In a period, the ionization energy increases as we move across a period as the size decreases. The electrons get added to the same shell and the nuclear charge increases.The valence shell moves nearer to the nucleus.  Thus more energy is required to remove the valence electron.

In a group , the ionization energy decreases from top to bottom as we move down a group. The size increases as the electrons get added up in the new shell, the electron in the outermost orbital gets far away from the nucleus. And hence, the electron will be easier to remove. Thus, lesser energy is required to remove the valence electron.

The correct order of increasing ionization energy will be:

lithium < strontium < magnesium < tellurium < chlorine < oxygen  

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

possibly, yes.

explanation:

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