Question

From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: (a)$\mathrm{A}\mathrm{s},\mathrm{H},\mathrm{N},\mathrm{P},\mathrm{S}\mathrm{b}$(b)$\mathrm{C}\mathrm{l},\mathrm{H},\mathrm{P},\mathrm{S},\mathrm{S}\mathrm{i}$(c)$\mathrm{B}\mathrm{r},\mathrm{C}\mathrm{l},\mathrm{G}\mathrm{e},\mathrm{H},\mathrm{S}\mathrm{r}$(d)$\mathrm{C}\mathrm{a},\mathrm{H},\mathrm{K},\mathrm{N},\mathrm{S}\mathrm{i}$(e)$\mathrm{C}\mathrm{l},\mathrm{C}\mathrm{s},\mathrm{G}\mathrm{e},\mathrm{H},\mathrm{S}\mathrm{r}$.

Short answer

The atoms are arranged in their increasing electronegativity order as:

1. $\mathrm{H}<\mathrm{S}\mathrm{b}<\mathrm{A}\mathrm{s}<\mathrm{P}<\mathrm{N}$
2. $\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{P}<\mathrm{S}<\mathrm{C}\mathrm{l}$
3. $\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{B}\mathrm{r}<\mathrm{C}\mathrm{l}$
4. $\mathrm{K}<\mathrm{C}\mathrm{a}<\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{N}$
5. $\mathrm{C}\mathrm{s}<\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{C}\mathrm{l}$

Step-by-step solution

Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

Explanation

As, the addition of a proton has a greater effect than the addition of an electron, the size of the atom decreases as you travel from left to right in the periodic table. Because of the strong force of the nucleus, it will be able to receive electrons more easily as its size decreases.

According to the foregoing argument, electronegativity rises as one moves from left to right in the periodic table.

As you progress from top to bottom in a group, the size of the group grows due to the addition of fresh energy shells. Because the force of attraction of the nucleus is no longer able to attract additional electrons as the size of the nucleus rises, electronegativity will drop as the group size decreases.

Arranging the atom in increasing order of electronegativity

a. The atoms in order of increasing electronegativity is:

$\mathrm{H}<\mathrm{S}\mathrm{b}<\mathrm{A}\mathrm{s}<\mathrm{P}<\mathrm{N}$

Therefore, the order is: $\mathrm{H}<\mathrm{S}\mathrm{b}<\mathrm{A}\mathrm{s}<\mathrm{P}<\mathrm{N}$.

Arranging the atom in increasing order of electronegativity

b. The atoms in order of increasing electronegativity is:

$\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{P}<\mathrm{S}<\mathrm{C}\mathrm{l}$

Therefore, the order is: $\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{P}<\mathrm{S}<\mathrm{C}\mathrm{l}$.

Arranging the atom in increasing order of electronegativity

c. The atoms in order of increasing electronegativity is:

$\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{B}\mathrm{r}<\mathrm{C}\mathrm{l}$

Therefore, the order is: $\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{B}\mathrm{r}<\mathrm{C}\mathrm{l}$.

Arranging the atom in increasing order of electronegativity

d. The atoms in order of increasing electronegativity is:

$\mathrm{K}<\mathrm{C}\mathrm{a}<\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{N}$

Therefore, the order is: $\mathrm{K}<\mathrm{C}\mathrm{a}<\mathrm{H}<\mathrm{S}\mathrm{i}<\mathrm{N}$.

Arranging the atom in increasing order of electronegativity

e. The atoms in order of increasing electronegativity is:

$\mathrm{C}\mathrm{s}<\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{C}\mathrm{l}$

Therefore, the order is: $\mathrm{C}\mathrm{s}<\mathrm{S}\mathrm{r}<\mathrm{H}<\mathrm{G}\mathrm{e}<\mathrm{C}\mathrm{l}$.