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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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Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

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Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

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From the stoichiometry of the reaction;

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Answer:

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=(25.0 g)/(17.034 g/mol)

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Now,

Molar mass of O2

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n (O2)=(45.0 g)/(32 g/mol)

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by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

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2 years ago
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Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

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