Chapter 6.3, Problem 19PP

(a)

Interpretation Introduction

Interpretation:

The larger ion among the element of period 2, group 1; or the element of period 3, group 18 should be determined.

Concept introduction:

Atomic radii of atoms of elements down the group increase as $Z$ increases because new shells are introduced and therefore the distance between nucleus and valence shell increases. Along a period, size or radii decreases as the electrons are introduced into the same shell and are pulled by higher nuclear charge.

Answer to Problem 19PP

Element of period 2, group 1 that is $\text{Li}$ has higher atomic radius thanthe element of period 3, group 18that is $\text{Ar}$.

Explanation of Solution

As per modern periodic table, the element with period 2, group 1 is $\text{Li}$ and the element of period 3, group 18 is $\text{Ar}$.

$\text{Li}$ has an electronic configuration of $\left[\text{He}\right]2s^1$ with atomic number 3 whereas $\text{Ar}$ has atomic number of 18 with electronic configuration of $\left[\text{Ne}\right]3s^{2}3p^6$.

In a period, as we move to the left, atomic radii decreases as new electrons by unity are added into the same principal energy level and there occurs no electrons between nucleus and valence electron for shielding. So $\text{Ar}$ has lower atomic radii than $\text{Li}$ as valence electrons are pulled by higher nuclear charge than in case of $\text{Li}$.

(b)

Interpretation Introduction

Interpretation:

The larger ion among the element of period 5, group 2; or the element of period 3, group 16 should be determined.

Concept introduction:

Atomic radii of atoms of elements down the group increase as $Z$ increases because new shells are introduced and therefore the distance between nucleus and valence shell increases. Along a period, size or radii decreases as the electrons are introduced into the same shell and are pulled by higher nuclear charge.

Answer to Problem 19PP

Element of period 5, group 2 that is $\text{Sr}$ has higher atomic radius thanthe element of period 3, group 16that is $\text{S}$. This is so because there are 2 new shells introduced in between valence electrons and nucleus in $\text{Sr}$ hence nuclear charge felt by valence electron is much less than in $\text{S}$.

Explanation of Solution

As per modern periodic table, the element with period 5, group 2 is $\text{Sr}$ and the element of period 3, group 16 is $\text{S}$.

$\text{Sr}$ has an electronic configuration of $\left[\text{Kr}\right]2s^2$ with atomic number 38 whereas $\text{S}$ has atomic number of 16 with electronic configuration of $\left[\text{Ne}\right]3s^{2}3p^4$.

In a period, as we move to the left, atomic radii decreases as new electrons by unity are added into the same principal energy level and there occurs no electrons between nucleus and valence electron for shielding, whereas as we go down the group, atomic size increases as nuclear charge on valence electron decreases due to new shell that is introduced between nucleus and valence electrons and hence the intervening electron. So, $\text{Sr}$ have larger atomic radii than $\text{S}$ as in $\text{Sr}$, there are 2 new shells introduced between valence electrons and nucleus, hence nuclear charge felt by valence electron is much less.

(c)

Interpretation Introduction

Interpretation:

The larger ion among the element of period 3, group 14; or the element of period 6, group 15 should be determined.

Concept introduction:

Atomic radii of atoms of elements down the group increase as $Z$ increases because new shells are introduced and therefore the distance between nucleus and valence shell increases. Along a period, size or radii decreases as the electrons are introduced into the same shell and are pulled by higher nuclear charge.

Answer to Problem 19PP

The element of period 3, group 14 that is $\text{Bi}$ have larger atomic radii than the element of period 6, group 15 that is $\text{Si}$ as in $\text{Bi}$ due to higher atomic number, has much higher number of added shells in between valence electrons and nucleus than in $\text{Si}$. So, the nuclear charge on valence shell electrons in $\text{Bi}$ is less in comparison to $\text{Si}$.

Explanation of Solution

As per modern periodic table, the element with period 3, group 14 is $\text{Si}$ and the element of period 6, group 15 is $\text{Bi}$.

$\text{Si}$ has an electronic configuration of $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{2}}$ with atomic number 14 whereas $\text{Bi}$ has atomic number of 83 with electronic configuration of $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{ 5d}}^{\text{10 }}{\text{6s}}^{\text{2}}{\text{ 6p}}^{\text{3}}$.

In a period, as we move to the left, atomic radii decreases as new electrons by unity are added into the same principal energy level and there occurs no electrons between nucleus and valence electron for shielding, whereas as we go down the group, atomic size increases as nuclear charge on valence electron decreases due to new shell that is introduced between nucleus and valence electrons and hence the intervening electron. So $\text{Bi}$ have larger atomic radii than $\text{Si}$ as in $\text{Bi}$ due to higher atomic number has much higher number ofadded shells introduced between valence electrons and nucleus than in $\text{Si}$. So, the nuclear charge on valence shell electrons in $\text{Bi}$ is less in comparison to $\text{Si}$.

(d)

Interpretation Introduction

Interpretation:

The larger ion among the element of period 4, group 18; or the element of period 2, group 16 should be determined.

Concept introduction:

Atomic radii of atoms of elements down the group increase as $Z$ increases because new shells are introduced and therefore the distance between nucleus and valence shell increases. Along a period, size or radii decreases as the electrons are introduced into the same shell and are pulled by higher nuclear charge.

Answer to Problem 19PP

The element of period 4, group 18 that is $\text{Kr}$ have larger atomic radii than the element of period 2, group 16 that is $\text{O}$ as in $\text{Kr}$ due to higher atomic number, it has much higher number of added shells in between valence electrons and nucleus than in $\text{O}$. So, the nuclear charge on valence shell electrons in $\text{Kr}$ is less in comparison to $\text{O}$.

Explanation of Solution

As per modern periodic table, the element with period 4, group 18 is $\text{Kr}$ and the element of period 2, group 16 is $\text{O}$.

$\text{Kr}$ has an electronic configuration of $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$ with atomic number 36 whereas $\text{O}$ has atomic number of 8 with electronic configuration of $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{ 2p}}^4$.

In a period, as we move to the left, atomic radii decreases as new electrons by unity are added into the same principal energy level and there occurs no electrons between nucleus and valence electron for shielding, whereas as we go down the group, atomic size increases as nuclear charge on valence electron decreases due to new shell that is introduced between nucleus and valence electrons and hence the intervening electron. So, $\text{Kr}$ have larger atomic radii than $\text{O}$ as in $\text{Kr}$ due to higher atomic number, it has much higher number of added shells introduced between valence electrons and nucleus than in $\text{O}$. So, the nuclear charge on valence shell electrons in $\text{Kr}$ is less in comparison to $\text{O}$.