Chapter 13, Problem 30E

(a)

Interpretation Introduction

Interpretation:

The electronic configuration for the cations ${\text{Mg}}^{\text{2+}}{\text{, Sn}}^{2+},{\text{ K}}^{+},{\text{ Al}}^{3+},{\text{ Tl}}^{3+},{\text{ As}}^{3+}$ should be written.

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

Answer to Problem 30E

Electronic configuration of ${\text{Mg}}^{\text{2+}}$ = $1s^{2}2s^{2}2p^6$

Electronic configuration of ${\text{Sn}}^{\text{2+}}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}$

Electronic configuration of ${\text{K}}^{+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

Electronic configuration of ${\text{Al}}^{3+}$ = $1s^{2}2s^{2}2p^6$

Electronic configuration of ${\text{Tl}}^{+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}$

Electronic configuration of ${\text{As}}^{3+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$

Explanation of Solution

The number of electrons that the subshells can hold is:

s-block - 2

p-block - 6

d-block - 10

f-block - 14

The increasing order of energy of shells, subshell is:

  

In electronic configuration, the number of electrons present in each sublevel is shown by superscript.

The given cations are: ${\text{Mg}}^{\text{2+}}{\text{, Sn}}^{2+},{\text{ K}}^{+},{\text{ Al}}^{3+},{\text{ Tl}}^{3+},{\text{ As}}^{3+}$

The atomic number of magnesium = 12, after losing two electrons, ${\text{Mg}}^{\text{2+}}$ cation form, thus number of electrons = $12-2=10$

Electronic configuration of ${\text{Mg}}^{\text{2+}}$ = $1s^{2}2s^{2}2p^6$

The atomic number of tin = 50, after losing two electrons, ${\text{Sn}}^{\text{2+}}$ cation form, thus number of electrons = $50-2=48$

Electronic configuration of ${\text{Sn}}^{\text{2+}}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}$

The atomic number of potassium= 19, after losing one electron, ${\text{K}}^{+}$ cation form, thus number of electrons = $19-1=18$

Electronic configuration of ${\text{K}}^{+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

The atomic number of aluminium = 13, after losing three electrons, ${\text{Al}}^{3+}$ cation form, thus number of electrons = $13-3=10$

Electronic configuration of ${\text{Al}}^{3+}$ = $1s^{2}2s^{2}2p^6$

The atomic number of thallium = 81, after losing one electron, ${\text{Tl}}^{+}$ cation form, thus number of electrons = $81-1=80$

Electronic configuration of ${\text{Tl}}^{+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}$

The atomic number of arsenic = 33, after losing three electrons, ${\text{As}}^{3+}$ cation form, thus number of electrons = $33-3=30$

Electronic configuration of ${\text{As}}^{3+}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$

(b)

Interpretation Introduction

Interpretation:

The electronic configuration for anions: ${\text{N}}^{3-},{\text{ O}}^{2-},{\text{ F}}^{-},{\text{ Te}}^{2-}$ should be written.

Concept Introduction:

The distribution of electrons in atom into orbitals is said to be electronic configuration. The electronic configuration for every element present in the periodic table is unique or different. Atomic number is equal to the number of protons, which is further equal to the number of electrons for neutral atom.

Answer to Problem 30E

Electronic configuration of ${\text{N}}^{3-}$ = $1s^{2}2s^{2}2p^6$

Electronic configuration of ${\text{O}}^{2-}$ = $1s^{2}2s^{2}2p^6$

Electronic configuration of ${\text{F}}^{-}$ = $1s^{2}2s^{2}2p^6$

Electronic configuration of ${\text{Te}}^{2-}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$

Explanation of Solution

The number of electrons that the subshells can hold is:

s-block - 2

p-block - 6

d-block - 10

f-block - 14

The increasing order of energy of shells, subshell is:

  

In electronic configuration, the number of electrons present in each sublevel is shown by superscript.

The given anions are: ${\text{N}}^{3-},{\text{ O}}^{2-},{\text{ F}}^{-},{\text{ Te}}^{2-}$

The atomic number of nitrogen= 7, after gaining three electrons, ${\text{N}}^{3-}$ forms. Thus, number of electrons = $7+3=10$

Electronic configuration of ${\text{N}}^{3-}$ = $1s^{2}2s^{2}2p^6$

The atomic number of oxygen = 8, after gaining two electrons, ${\text{O}}^{2-}$ forms. Thus, number of electrons = $8+2=10$

Electronic configuration of ${\text{O}}^{2-}$ = $1s^{2}2s^{2}2p^6$

The atomic number of fluorine = 9, after gaining one electron, ${\text{F}}^{-}$ forms. Thus, number of electrons = $9+1=10$

Electronic configuration of ${\text{F}}^{-}$ = $1s^{2}2s^{2}2p^6$

The atomic number of tellurium = 52, after gaining two electrons, ${\text{Te}}^{2-}$ forms. Thus, number of electrons = $52+2=54$

Electronic configuration of ${\text{Te}}^{2-}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$