Choose the atom in each set that has the higher first ionization energy. a. alors b. as or sb c. n or si

The atom with the highest first ionization energy in each pair is:

a) S

b) As

c) N

Explanation:

Ionization energy is the amount of energy required to remove the last electron from a neutral atom in the gas state. First ionization energy (1 I.E.) refers to the first electron being removed, while second ionization energy (2 I. E.) refers to the energy required to remove a second electron.

A(g) + 1 I.E. ⇒ A⁺(g) + 1 e⁻

A⁺(g) + 2 I.E. ⇒ A²⁺(g) + 1 e⁻

Metals lose their electrons pretty easily so the amount of energy required for this process is low (low ionization energies). On the other hand, non-metals tend to gain electrons rather than lose them so it is necessary to provide them with much energy to take an electron (high ionization energies).

According to this, ionization energy increases in the periodic table from the bottom to the top and from the left to the right. This is what we need to know to solve the problem.

In a) S is to the right of Al, so it has a greater ionization energy.

I.E. (Al): 577.6 kJ/mol < I.E (S): 999.6 kJ/mol.

In b) As is on top of Sb, so it has a greater ionization energy.

I.E. (Sb): 834 kJ/mol < I.E. (As): 947.0 kJ/mol.

In c) N is to the right and above Si, so it has a greater ionization energy.

I.E. (Si): 786.5 kJ/mol < I.E. (N): 1,402.3 kJ/mol.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole