Chapter 12, Problem 43CR

Interpretation Introduction

Interpretation:

The smallest ion should be identified.

Concept Introduction:

Energy required to remove an electron from an individual atom in gas phase defined as its ionization energy.

$$

$\text{M} \to {\text{M}}^{+}{}_{\left(\text{g}\right) }{+}_{ }{\text{e}}^{-}$.

Answer to Problem 43CR

a. $\text{S}{\text{e}}^{2-}$ has the smallest ionization energy among b. $\text{B}{\text{r}}^{-}$ c. $\text{S}{\text{r}}^{2+}$ d. $\text{Z}{\text{r}}^{4+}$ and e. $\text{R}{\text{b}}^{+}$.

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^{6}7{\text{s}}^{2}5{\text{f}}^{14}6{\text{d}}^{10}7{\text{p}}^6$

$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^{6}7{\text{s}}^{2}5{\text{f}}^{14}6{\text{d}}^{10}7{\text{p}}^6$

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

$\text{S}\text{e}$, $\text{Z}=34$ →$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^4$

Thus,

$\text{S}{\text{e}}^{2-}$, $\text{Z}=36$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{B}\text{r}$, $\text{Z}=35$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^5$

Thus,

$\text{B}{\text{r}}^{-}$, $\text{Z}=36$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{S}\text{r}$, $\text{Z}=38$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^2$

Thus,

$\text{S}{\text{r}}^{2+}$, $\text{Z}=36$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{Z}\text{r}$, $\text{Z}=40$ →$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{2}5{\text{s}}^{2}4{\text{d}}^2$

Thus,

$\text{Z}{\text{r}}^{4+}$, $\text{Z}=36$ →$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{R}\text{b}$, $\text{Z}=37$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^1$

Thus,

$\text{R}{\text{b}}^{+}$, $\text{Z}=36$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

First ionization energy of neutral atoms vary as follows:

As the energy levels grow up from top to bottom in one group of elements, removal energy of the final electron reduces as attraction of the final electron to the nucleus decreases.

$\text{R}\text{b}$, $\text{S}\text{r}$, $\text{Z}\text{r}$ < $\text{S}\text{e}$, $\text{B}\text{r}$

From left to right of a period ionization energy increases. Thus metals tend to remove electrons easily than nonmetals.

$\text{R}\text{b}<\text{S}\text{r}<\text{Z}\text{r}<\text{S}\text{e}<\text{B}\text{r}$

Ionization energies of different ions vary as follows:

All above five ions have achieved the noble gas configuration of $\text{K}\text{r}$, $\text{Z}=36$. If only considering that the energy needed to remove the most distinct electron should be remain same for all. But different ions have different charges around their ions. Also their atomic sizes varies. Considering all of them, though the electron configuration of above ions are same, the ionization energies modify accordingly.

Positive charge increment of the nucleus tends to pull the electrons towards the nucleus and ions get smaller than the neutral atom and cause more difficult to remove another electron. Opposite happens for anions. More negative charge cause repulsion of most distinct electrons and also the atomic size increases. Because of that it’s more easier to remove another electron from anion than cation whose having same valance electrons.

Ionization energy of anions < Ionization energy of cations

(Have more minus charge) (Have more plus charge)

$\text{S}{\text{e}}^{2-},\text{B}{\text{r}}^{-}<\text{R}{\text{b}}^{+},\text{S}{\text{r}}^{2+},\text{Z}{\text{r}}^{4+}$

$\text{S}{\text{e}}^{2-}$ and $\text{B}{\text{r}}^{-}$ are anions and in the same period. From these two only difference is the ionic size. The most negative ions are larger than all of the others while most positive ions are too small in size. Most distinct electron can be easily removed of the largest anion.

$\text{S}{\text{e}}^{2-}<\text{B}{\text{r}}^{-}$.

Conclusion

Thus, (a) $\text{S}{\text{e}}^{2-}$ has the smallest ionization energy among b. $\text{B}{\text{r}}^{-}$ c. $\text{S}{\text{r}}^{2+}$ d. $\text{Z}{\text{r}}^{4+}$ and e. $\text{R}{\text{b}}^{+}$.