Question

Set of isoelectronic ions among the following is:

A ${\text{N}}{{\text{a}}^ + },{\text{C}}{{\text{l}}^ - }\,,\,{{\text{O}}^ - }$
B ${{\text{K}}^ + },{\text{C}}{{\text{a}}^{2 + }}\,,\,{{\text{F}}^ - }$
C ${\text{C}}{{\text{l}}^ - }\,,\,{{\text{K}}^ + },{{\text{S}}^{2 - }}$
D ${{\text{H}}^ + }\,,\,{\text{B}}{{\text{e}}^{2 + }},{\text{N}}{{\text{a}}^ + }$

Answer 1

Hint:To determine the answer we should know about isoelectronic species and atomic number of each given element. The species having the same number of electrons are known as isoelectronic species. The total count of the electrons of a species is the sum of atomic number and negative charge and subtraction of positive charge.

Complete step-by-step solution:We will determine the total electrons in each of the given species as follows:

For a neutral atom, the atomic number gives the total electron count.

For a positively charged atom, we will subtract the positive charge from the atomic number to get the total electron count.

For a negative charged atom, we will add the negative charge to the atomic number to get the total electron count.

SET-A: ${\text{N}}{{\text{a}}^ + },{\text{C}}{{\text{l}}^ - }\,,\,{{\text{O}}^ - }$

The atomic number of sodium is$11$, chlorine is $17$ and oxygen is $8$.

Total electrons in ${\text{N}}{{\text{a}}^{\text{ + }}}$,
${\text{N}}{{\text{a}}^{\text{ + }}}\, = \,11 - 1$
${\text{N}}{{\text{a}}^{\text{ + }}}\, = \,10$

Total electrons in${\text{C}}{{\text{l}}^ - }$,
${\text{C}}{{\text{l}}^ - }\, = \,17 + 1$
${\text{C}}{{\text{l}}^ - } = 18$

Total electrons in $\,{{\text{O}}^ - }$,
$\,{{\text{O}}^ - } = \,8 + 1$
$\,{{\text{O}}^ - }\, = \,9$

All the ions of SET-A have different electron count. So, the SET-A does not represent the isoelectronic ions.

SET-B

${{\text{K}}^ + },{\text{C}}{{\text{a}}^{2 + }}\,,\,{{\text{F}}^ - }$

The atomic number of potassium is$19$, calcium is $20$ and fluorine is $9$.

Total electrons in ${{\text{K}}^ + }$,
${{\text{K}}^ + }\, = \,19 - 1$
${{\text{K}}^ + }\, = \,18$

Total electrons in ${\text{C}}{{\text{a}}^{2 + }}$,
${\text{C}}{{\text{a}}^{2 + }} = \,20 - 2$
${\text{C}}{{\text{a}}^{2 + }}\, = \,18$

Total electrons in ${{\text{F}}^ - }$,
${{\text{F}}^ - } = \,9 + 1$
${{\text{F}}^ - }\, = \,10$

All the ions of SET-B have a different electron count. So, the SET-B does not represents the isoelectronic ions.

SET-C

${\text{C}}{{\text{l}}^ - }\,,\,{{\text{K}}^ + },{{\text{S}}^{2 - }}$

The atomic number of chlorine is$17$, potassium is $19$ and sulphur is $16$.

Total electrons in \[{\text{C}}{{\text{l}}^ - }\],
${\text{C}}{{\text{l}}^ - }\, = \,17 + 1$
${\text{C}}{{\text{l}}^ - }\, = \,18$

Total electrons in ${{\text{K}}^ + }$,
${{\text{K}}^ + }\, = \,19 - 1$
${{\text{K}}^ + }\, = \,18$

Total electrons in${{\text{S}}^{2 - }}$,
${{\text{S}}^{2 - }} = \,16 + 2$
${{\text{S}}^{2 - }}\, = \,18$

All the ions of SET-C have the same electron count. So, the SET-C represents the isoelectronic ions.

SET-D

${{\text{H}}^ + }\,,\,{\text{B}}{{\text{e}}^{2 + }},{\text{N}}{{\text{a}}^ + }$

The atomic number of hydrogen is$1$, beryllium is $4$ and sodium is $11$.

Total electrons in ${{\text{H}}^ + }$,
${{\text{H}}^ + }\, = 1 - 1$
${{\text{H}}^ + }\, = \,0$

Total electrons in ${\text{B}}{{\text{e}}^{2 + }}$,
${\text{B}}{{\text{e}}^{2 + }}\, = \,4 - 2$
${\text{B}}{{\text{e}}^{2 + }} = \,2$

Total electrons in ${\text{N}}{{\text{a}}^{\text{ + }}}$,
${\text{N}}{{\text{a}}^{\text{ + }}}\, = \,11 - 1$
${\text{N}}{{\text{a}}^{\text{ + }}}\, = \,10$

All the ions of SET-D have a different electron count. So, the SET-D does not represent the isoelectronic ions.

So, the isoelectronic ions among the following is ${\text{C}}{{\text{l}}^ - }\,,\,{{\text{K}}^ + },{{\text{S}}^{2 - }}$.

Therefore, option (C), ${\text{C}}{{\text{l}}^ - }\,,\,{{\text{K}}^ + },{{\text{S}}^{2 - }}$is correct.

Note: Isoelectronic species have different protons (atomic number) and mass number only the electrons number are the same. Positive change means the electron is donated by the atom to form cation. Negative change means the electron is accepted by the atom to form an anion. The subscript represents the number of atoms, so the atomic number should be added as many times as denoted in the subscript. Superscript represents the charge that will be added in case of negative charge and subtracted in case of a positive charge.