Chapter 11, Problem 76AE

(a)

Interpretation Introduction

Interpretation:

The rank has to be given to the given compounds, $NaO,CO,NO$ in the order of increasing polarity.

Concept Introduction:

An electronegativity of an atom gives information about the strength that the atom has for attracting a pair of shared electrons.  The greater the attraction for a pair of shared electrons when the electronegativity value is high.  Consequently, when two different electronegative atoms share a pair of electrons, one atom becomes partially positive and the other atom becomes partially negative due to the attractive forces for the pair of electrons.

The larger is the electronegativity difference between the atoms, the more polar will be the bond between the atoms.

Explanation of Solution

Given molecules,

        $NaO,CO,NO$

The electronegativity value of $Na$ is $0.9$

The electronegativity value of $C$ is $2.5$

The electronegativity value of $N$ is $3.0$

The electronegativity value of $O$ is $3.5$

The difference in electronegativity between $Na$ and $O$ in $NaO$ is $2.6$

The difference in electronegativity between $C$ and $O$ in $CO$ is $1.0$.

The difference in electronegativity between $N$ and $O$ in $NO$ is $0.5$

Therefore the more polar compound will be $NaO$, which has highest electronegativity difference. The less polar compound will be $NO$, which has lowest electronegativity difference.

Hence the compounds can be ranked in the increasing order of polarity is $NO<CO<NaO$.

(b)

Interpretation Introduction

Interpretation:

The rank has to be given to the given compounds, $CO,SiO,GeO$ in the order of increasing polarity.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given molecules,

        $CO,SiO,GeO$

The electronegativity value of $C$ is $2.5$

The electronegativity value of $Si$ is $1.8$

The electronegativity value of $Ge$ is $1.8$

The electronegativity value of $O$ is $3.5$

The difference in electronegativity between $C$ and $O$ in $CO$ is $1.0$.

The difference in electronegativity between $Si$ and $O$ in $SiO$ is $1.7$

The difference in electronegativity between $Ge$ and $O$ in $GeO$ is $1.7$

Therefore the more polar compounds will be $SiO$ and $GeO$, which has highest electronegativity difference. Among $SiO$ and $GeO$, germanium is larger in size and hence greater will be the bond length between germanium and oxygen. So more polar will be $GeO$ than $SiO$.

The less polar compound will be $CO$, which has lowest electronegativity difference.

Hence the compounds can be ranked in the increasing order of polarity is $CO<SiO<GeO$.

(c)

Interpretation Introduction

Interpretation:

The rank has to be given to the given compounds, $BO,BS,BSe$ in the order of increasing polarity.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given molecules,

        $BO,BS,BSe$

The electronegativity value of $B$ is $2.0$

The electronegativity value of $Se$ is $2.4$

The electronegativity value of $S$ is $2.5$

The electronegativity value of $O$ is $3.5$

The difference in electronegativity between $B$ and $O$ in $BO$ is $1.5$

The difference in electronegativity between $B$ and $S$ in $BS$ is $0.5$.

The difference in electronegativity between $Se$ and $B$ in $BSe$ is $0.4$

Therefore the more polar compound will be $BO$, which has highest electronegativity difference as compared to other molecules. The less polar compound will be $BSe$, which has lowest electronegativity difference.

Hence the compounds can be ranked in the increasing order of polarity is $BSe<BS<BO$.