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Anvisha [2.4K]
2 years ago
12

Select all of the correct statements pertaining to the first and second ionization energies of different elements. check all tha

t apply. the first ionization energy of n is smaller than that of p. the second ionization energy of na is larger than that of mg. the first ionization energy of n is larger than that of p. the second ionization energy of na is smaller than that of mg.
Chemistry
2 answers:
sertanlavr [38]2 years ago
8 0
The First Ionization energy of Nitrogen is greater (Not smaller)than that of Phosphorous. This is because going down the group (N and P are in same group) the number of shells increases, the distance of valence electrons from Nucleus increases and hence due to less interaction between nucleus and valence electrons it becomes easy to knock out the electron.

<span>The second ionization energy of Na is larger than that of Mg because after first loss of electron Na has gained Noble Gas Configuration (Stable Configuration) and now requires greater energy to loose both second electron and Noble Gas Configuration. While Mg after second ionization attains Noble Gas Configuration hence it prices less energy.</span>
jekas [21]2 years ago
6 0

The correct statements pertaining to the first and second ionization energies of different elements are as follows:

\boxed{{\text{The first ionization energy of N is larger than that of P}}}

\boxed{{\text{The second ionization energy of Na is larger than that of Mg}}}

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.

Successive ionization energies are evaluated when the electrons are to be removed from successive shells. When the first electron is removed from the isolated neutral gaseous atom, ionization energy is termed as the first ionization energy \left( {{\text{I}}{{\text{E}}_{\text{1}}}} \right). Similarly, when the second electron is removed from the monoatomic cation, ionization energy is called the second ionization energy  \left( {{\text{I}}{{\text{E}}_{\text{2}}}} \right) and so on…

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.

Nitrogen (N) and phosphorus (P) are present in the same group of the periodic table. Phosphorus lies below nitrogen and we know ionization energy decreases from top to bottom in a group. Therefore the first ionization energy of N is larger than that of P.

Sodium and magnesium are present in the same period of the periodic table. The atomic number of Na is 11 and its electronic configuration is  \left[ {{\text{Ne}}} \right]\;3{s^1}. The atomic number of Mg is 12 and its electronic configuration is  \left[ {{\text{Ne}}}\right]\;3{s^1}. Two electrons are to be removed for second ionization energy. Mg will acquire stable noble gas configuration of neon after losing two of its 3s electrons so its second ionization will be less. But Na will have noble gas configuration after removal of one electron and a very large amount of energy is required for removal of its second electron as it has to be removed from stable noble gas configuration of neon. So the second ionization energy of Na is larger than that of Mg.

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, successive ionization energy, IE, IE1, IE2, first ionization energy, second ionization energy, nitrogen, phosphorus, sodium, magnesium.

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MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
2 years ago
2) Of the following, which has the shortest de Broglie wavelength? A) an airplane moving at a velocity of 300 mph B) a helium nu
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Answer:

B) a helium nucleus moving at a velocity of 1000 mph

Explanation:

According to the De Broglie relation

λ= h/mv

h= planks constant

m= mass of the body

v= velocity of the body.

As we can see from De Broglie's relation, the wavelength of matter waves depends on its mass and velocity. Hence, a very small mass moving at a very high velocity will have the greatest De Broglie wavelength.

Of all the options given, helium is the smallest matter. A velocity of 1000mph is quite high hence it will have the greatest De Broglie wavelength.

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Answer:Material

Coefficient

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aluminum

chromium

glass

0.25

0.21

0.12

steel

0.23

titanium

vanadium alloy

0.31

Drag the materials into the correct order from the material that will slide down the ramp the fastest to the material that will slide down the ramp the

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chromium

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2 years ago
What happened to the volume of gas when the syringe was exposed to various temperature conditions? Using the concepts explored i
rewona [7]

Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

\frac {V_1}{T_1}=\frac {V_2}{T_2}

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

<u>Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature. </u>

<u>For example , if the temperature of the gas is reduced to half, the volume also reduced to half. </u>

<u>At -273.15 K, according to Charles's law, it is possible to make the volume of an ideal gas = 0.</u>

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Give a possible explanation for the relative amounts of the isometric methyl nitrobenzoates formed in the nitration reaction. Co
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Answer:

The electrophilic aromatic substitution reaction nitration is used to nitrate methyl benzoate and acetanilide with a nitronium ion. Crystallization was used to purify the product. The melting point was used to determine its purity and the regiochemistry of the products.

Explanation:

Methyl m-Nitrobenzoate is formed in this

reaction rather that ortho/para isomers

because of the ester group of your starting

product of methylbenzoate. The functional

group of ester is a electron withdrawing group

causing nitrobenzene (N02) to become in the

meta position. Thus N02 is a deactivating

group causing itself to be a meta director.

Basically you must look at the substituents

that are attached to your starting benzene ring

in order to figure out whether your reaction

with be ortho/para directors or meta

directors. If the substituents are electron

withdrawing groups then you will be left with

meta as your product but if your substituents

are electron donating groups then your

product will be ortho/para.

4 0
1 year ago
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