Chapter 6, Problem 6.67PAE

(a)

Interpretation Introduction

To determine: of the 5 elements $S,SeandCl$ the one with the largest atomic radius

Explanation of Solution

Atomic radius is the distance between center of nucleus and outermost electron shell.

Atomic radius decreases from left to right in a period and increases from top to bottom in a group of periodic table.

S and Se both belong to some group $16$.

S belongs to third period while Se belongs to fourth period. So, Se is larger than S. $Se>S\cdots \cdots \left(1\right)$

S and Cl both belong to third period. S belongs to group $16$ while Cl belongs to group $17$.

Hence, S has larger size than Cl.

$S>Cl\cdots \cdots \left(2\right)$

On combining equation $\left(1\right)\text{ and }\left(2\right)$ you get

$Se>S>Cl$

Therefore, Se has the largest atomic radius.

(b)

Interpretation Introduction

To determine: the larger radius $Br\text{ or }B{r}^{-}$.

Explanation of Solution

Electronic configuration of Br is $\left[Ar\right]3d^{10}4s^{2}4p^5$

So, electronic configuration of $B{r}^{-}\text{ is }\left[Ar\right]3d^{10}4s^{2}4p^6$ as one electron is added in $4p$ orbital due to presence of more electrons by keeping the same shell, attractive force increases which decreases the atomic radius.

Therefore, the largest radius among $Br\text{ or }B{r}^{-}$ is Br.

(c)

Interpretation Introduction

To determine: Which should have the largest difference between the first and second ionization energy, $\text{Si, Na, P or Mg}$.

Explanation of Solution

Ionization energy $-$ the amount of energy required to remove an electron from normal gaseous state.

Now, electronic configuration of $\text{Si, Na, P or Mg}$ are.

$\begin{array}{l}Si\left(14\right):-1s^{2}2s^{2}2p^{6}3s^{2}3p^2\\ Na\left(11\right):-1s^{2}2s^{2}2p^{6}3s^1\\ P\left(14\right):-1s^{2}2s^{2}2p^{6}3s^{2}3p^3\\ Mg\left(11\right):-1s^{2}2s^{2}2p^{6}3s^2\end{array}$

After first ionization, electronic configuration of ${\text{Si}}^{+}{\text{, Na}}^{+}{\text{, P}}^{+}{\text{ or Mg}}^{+}$ :

$\begin{array}{l}S{i}^{+}\left(13\right):-1s^{2}2s^{2}2p^{6}3s^{2}3p^1\\ N{a}^{+}\left(10\right):-1s^{2}2s^{2}2p^6\\ P^{+}\left(14\right):-1s^{2}2s^{2}2p^{6}3s^{2}3p^2\\ M{g}^{+}\left(11\right):-1s^{2}2s^{2}2p^{6}3s^1\end{array}$

Among all of these electronic configuration, $N{a}^{+}$ has completely filled configuration that is most stable.

Hence, it is so difficult to remove an electron from $N{a}^{+}$. Thus, second ionization energy is so high.

Therefore, the largest difference between the first and second ionization energy among $\text{Si, Na, P or Mg}$ is $Na$.

(d)

Interpretation Introduction

To determine: the element with the largest ionization energy: $N,P\text{ or As}$

Explanation of Solution

Ionization energy decreases from top to bottom in a group.

$N,P\text{ or As}$ All falls in the same group $15$.

So, $\left(\text{Ionization energy of N}\right)>\left(\text{Ionization energy of P}\right)>\left(\text{Ionization energy of As}\right)$.

Therefore, nitrogen has the largest ionization energy.

(e)

Interpretation Introduction

To determine: the following which has the largest radius $O^{2-},N^{3-}{\text{ or F}}^{-}$.

Explanation of Solution

Electronic configurations of $O^{2-}\text{ ion, }{N}^{3-}{\text{ and F}}^{-}$ ions are:

$\begin{array}{l}{O}^{2-}\left(10\right):-1s^{2}2s^{2}2p^6\\ N^{3-}\left(10\right):-1s^{2}2s^{2}2p^6\\ {\text{F}}^{-}\left(10\right):-1s^{2}2s^{2}2p^6\end{array}$

All the three ions have same electronic configuration. Atom with largest negative charge has largest radius.

Therefore, $N^{3-}$ has the largest radius.

Conclusion

Atomic radius increases from top to bottom in a group and right to left in period.

Ionization energy increases from left to right and also it decreases from top to bottom in a group.