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mr_godi [17]
3 years ago
14

In going across a row of the periodic table, protons and electrons are added and ionization energy generally increases. In going

down a column of the periodic table, protons and electrons are also being added but ionization generally decreases. Explain.
Chemistry
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

Explanation:

Ionization energy:

It is the minimum amount of energy required to remove the electron from isolated gaseous atom to make the ion.

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell.

When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.  Where as,

When we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

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Answer : The correct rate law for the reaction is,

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Expression for rate law for first observation:

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Expression for rate law for second observation:

0.0069=k(0.020)^a(0.010)^b ....(2)

Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

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Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

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\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}

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1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}

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\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}

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