Question

Predict the common charge of the silver ion. Explain your reasoning in terms of electron configurations.

Construct a quantum number tree for the principal quantum number n = 4 similar to that depicted for n = 3

Use diagrams similar to Figure 1.12 to determine the number of unpaired electrons in atoms of (a) nitrogen; (b) silicon; (c) iron.

Write the electron confi guration expected for element 113 and the confi gurations for the two cations that it is most likely to form.

Answer 1

1. Predict the common charge of the silver ion. Explain your reasoning in terms of electron configurations.

Silver = Ag

Atomic number = 47

Electronic configuration = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 5s¹ (most stable configuration) or [Kr] 4d¹⁰5s¹

When Silver is ionized the electron is removed from outermost shell (5s¹)

Silver ion = Ag⁺

^2. Quatum number tree for n = 4

n	l	M	Orbital notations	Number of the orbitals in the subshells	Number of the orbitals in the shells
4	0 1 2 3	0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3	4s 4p 4d 4f	1 3 5 7	16

3.Use diagrams similar to Figure 1.12 to determine the number of unpaired electrons in atoms of (a) nitrogen; (b) silicon; (c) iron.

Figure 1.12 not given

(a)    Nitrogen = atomic number 7

Electronic configuration N⁷= 1s2, 2s2, 2p3

Unpaired electron = 7

(b)   Silicon = 14

Electronic configuration Si¹⁴= 1s2, 2s2, 2p6, 3s2, 3p2

Unpaired electron = 2

(c)    Iron = 26

Electronic configuration Fe²⁶= 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6

Unpaired electron = 4

4.Write the electron configuration expected for element 113 and the configurations for the two cations that it is most likely to form.

Element 113 is Nihonium

Electronic configuration of Nh¹¹³ = [Rn] 5f¹⁴, 6d¹⁰, 7s², 7p¹

Cation that will be formed = Nh⁺ and Nh³⁺

Electronic configuration of Nh⁺ = [Rn] 5f¹⁴, 6d¹⁰, 7s²

Electronic configuration of Nh³⁺ = [Rn] 5f¹⁴, 6d¹⁰