Chapter 11, Problem 71A

(a)

Interpretation Introduction

Interpretation:

The electronic configuration which is likely to form negative ion needs to be determined.

Concept Introduction:

An element that by accepting electron will achieve a half-filled or full-filled electronic configuration will likely to form a negative ion.

Answer to Problem 71A

  $(ii){\text{ 1s}}^{2}2s^{2}2p^{6}3s^1$ will form a negative ion.

Explanation of Solution

If the $3s$ orbital accepts one electron to form an anion, then it will achieve full-filled $3s^2$ outer electronic configuration. Hence, it will for an anion.

(b)

Interpretation Introduction

Interpretation:

The electronic configuration which is likely to have low ionization energy needs to be determined.

Concept Introduction:

Removal of an electron that will give the most stable half-filled and full-filled electronic configuration will have the lowest ionization energy.

Answer to Problem 71A

  $(ii){\text{ 1s}}^{2}2s^{2}2p^{6}3s^1$ will have the lowest ionization energy.

Explanation of Solution

When an electron is removed from $3s$ orbital stable noble gas configuration of Neon is attained ${\text{1s}}^{2}2s^{2}2p^6$ .

Now, $(vi)$ also have a similar outer electronic configuration but it is easier to remove an electron from the orbital of a higher quantum number $(n=3)$ rather than orbital of a quantum lower number $(n=2)$ .

(c)

Interpretation Introduction

Interpretation:

The most reactive metal is to be identified.

Concept Introduction:

Metals which easily donate electrons to achieve stable half-filled or full-filled configuration are most reactive. Alkali metals are reactive due to their electronic configuration.

Answer to Problem 71A

  $(ii){\text{ 1s}}^{2}2s^{2}2p^{6}3s^1$ is most reactive of all.

Explanation of Solution

If the $3s$ orbital accepts one electron to form an anion then it will achieve full-filled $3s^2$ outer electronic configuration. When an electron is removed from $3s$ orbital stable noble gas configuration of Neon is achieved ${\text{1s}}^{2}2s^{2}2p^6$ .

Now, $(vi)$ also have similar outer electronic configuration but it is easier to remove or add electron from the orbital of a higher quantum number $(n=3)$ rather than orbital of a quantum lower number $(n=2)$ .

(d)

Interpretation Introduction

Interpretation:

The electronic configuration which is likely to have more unpaired of electrons needs to be determined.

Concept Introduction:

When an electron does not get paired up in orbital but occupies the orbital singly then it is called an unpaired electron.

Answer to Problem 71A

  $(v)1s^{2}2s^{2}2p^3$ have the highest number of unpaired of electrons.

Explanation of Solution

Orbital diagram of $2p$ orbital,

  $2p$

It is obvious from the diagram that there are three unpaired electrons which is highest among the options given.

(e)

Interpretation Introduction

Interpretation:

The metal that will combine with oxygen in $1:1$ ratio needs to be determined.

Concept Introduction:

A metal which can donate two electrons from its valence to complete the octet of oxygen will combine with oxygen in $1:1$ ratio.

Answer to Problem 71A

  $(i)1s^{2}2s^{2}2p^{6}3s^2$ will combine with oxygen in $1:1$ ratio.

Explanation of Solution

Magnesium (Mg) will donate two electrons from its $3s$ orbital to $2p$ of oxygen to form Magnesium oxide (MgO). In this way, both will achieve noble gas neon configuration $1s^{2}2s^{2}2p^6$ .

(f)

Interpretation Introduction

Interpretation:

The formula the compound formed between $(ii)\text{ and (v)}$ needs to be determined.

Concept Introduction:

A stable compound is formed when both the atoms participating in the combination achieves complete octet.

Answer to Problem 71A

The formula of the compound is $N{a}_{3}N$ .

Explanation of Solution

  $(ii)$ have to give total of three electrons from $3s$ to the $2p$ orbital of $\text{(v)}$ so that $\text{(v)}$ can attain noble gas configuration.

So, 3 $Na$ atoms combine with one $N$ to form $N{a}_{3}N$ .

(g)

Interpretation Introduction

Interpretation:

The formula the compound formed between $(i)\text{ and (v)}$ needs to be determined.

Concept Introduction:

A stable compound is formed when both the atoms participating achieves complete octet.

Answer to Problem 71A

The formula of the compound is $M{g}_3{N}_2$ .

Explanation of Solution

Three $Mg$ atoms will donate total of six electrons (each $Mg$ is donating two electrons from its $3s$ orbital) to the two $N$ (each $N$ atom receives three electrons) to form $M{g}_3{N}_2$ .

(h)

Interpretation Introduction

Interpretation:

The order of ionization energy needs to be determined.

Concept Introduction:

Ionization energy is a measure of the difficulty in removing an electron from an atom or ion or the tendency of an atom or ion to lose an electron. It is the energy required. So, this process is endothermic.

Answer to Problem 71A

Order of increase in ionization energy is $(ii)$< $(vi)$< $(i)$< $(iv)$< $(v)$< $(iii)$

Explanation of Solution

  $\begin{array}{l}(i)1s^{2}2s^{2}2p^{6}3s^2\text{ is }Mg\\ (ii)1s^{2}2s^{2}2p^{6}3s^1\text{ is }Na\\ (iii)1s^{2}2s^{2}2p^6\text{ is }Ne\\ (iv)1s^{2}2s^{2}2p^4\text{ is }O\\ (v)1s^{2}2s^{2}2p^3\text{ is }N\\ (vi)1s^{2}2s^{2}2p^1\text{ is }B\end{array}$

• The ionization energy of the elements usually increases throughout the period.
• The ionization energy of the elements usually decreases throughout the group.
• Ionization energies of noble gases are very high because of octet completion. That is why neon has the highest ionization energy of all the elements given.

Considering all facts, the order of increasing ionization energy is $(ii)$< $(vi)$< $(i)$< $(iv)$< $(v)$< $(iii)$