Which of the following sets show isoelectric species?

Which of the following sets represents the collection of isoelectronic species?(a)- $N{{a}^{+}},M{{g}^{2+}}A{{l}^{3+}},C{{l}^{-}}$ (b)- $N{{a}^{+}},C{{a}^{2+}}S{{c}^{3+}},{{F}^{-}}$ (c)- ${{K}^{+}},C{{l}^{-}},C{{a}^{2+}}S{{c}^{3+}}$ (d)- ${{K}^{+}},C{{a}^{2+}},S{{c}^{3+}},{{F}^{-}}$

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Which of the following sets represents the collection of isoelectronic species?(a)- $N{{a}^{+}},M{{g}^{2+}}A{{l}^{3+}},C{{l}^{-}}$ (b)- $N{{a}^{+}},C{{a}^{2+}}S{{c}^{3+}},{{F}^{-}}$ (c)- ${{K}^{+}},C{{l}^{-}},C{{a}^{2+}}S{{c}^{3+}}$ (d)- ${{K}^{+}},C{{a}^{2+}},S{{c}^{3+}},{{F}^{-}}$
The correct option is A K+,Cl−,Ca2+,Sc3+Isoelectronic species are those species which contains same number of electrons. (a) K+=19−1=18 e− Cl−=17+1=18 e− Ca2+=20−2=18 e− Sc3+=21−3=18 e− Thus, all the species are isoelectronic. (b) Ba2+=56−2=54 e− Sr2+=38−2=36 e− K+=19−1=18e− S2−=16+2=18e− Thus, these species are not isoelectronic. (c) N3−=7+3=10e− O2−=8+2=10e− F−=9+1=10 e− S2−=16+2=18 e− Thus, these species are not isoeletronic. (d) Li+=3−1=2 e− Na+=11−1=10 e− Mg2+=12−2=10 e− Ca2+=20−2=18 e− Thus, these species are not isoeletronic.
What are examples of isoelectronic species?
What are isoelectronic sets?
How do you find the isoelectronic set?

Answer

Verified

Hint: Isoelectronic species have the same number of electrons. : In isoelectronic species, as the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases.

Complete answer:
Isoelectronic species or ions are the ions of different elements that have the same number of electrons but differ in the magnitude of the nuclear charge.
Besides ions, a neutral atom may also the same number of electrons hence it is also isoelectronic species.
Example, sulphide ion (${{S}^{2-}}$ ), chloride ion ($C{{l}^{-}}$), argon (Ar), and potassium ion (${{K}^{+}}$ ) are isoelectronic species.
Variation of size among isoelectronic species: As the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radius decreases.
Let us observe all the options one by one:
(a)- $N{{a}^{+}},M{{g}^{2+}}A{{l}^{3+}},C{{l}^{-}}$
Sodium ion has 10 electrons
Magnesium ion has 10 electrons
Aluminum ion has 10 electrons
Chlorine ion has 18 electrons

(b)- $N{{a}^{+}},C{{a}^{2+}}S{{c}^{3+}},{{F}^{-}}$
Sodium ion has 10 electrons
Calcium has 18 electrons
Scandium ion has 18 electrons
Fluorine ion has 10 electrons

(c)- ${{K}^{+}},C{{l}^{-}},C{{a}^{2+}}S{{c}^{3+}}$
Potassium ion has 18 electrons
Chlorine ion has 18 electrons
Calcium ion has 18 electrons
Scandium ion has 18 electrons

(d)- ${{K}^{+}},C{{a}^{2+}},S{{c}^{3+}},{{F}^{-}}$
Potassium ion has 18 electrons
Calcium ion has 18 electrons
Scandium ion has 18 electrons
Fluorine ions have 10 electrons.
Hence, only option (c) has all the ions containing 18 electrons, therefore, ${{K}^{+}},C{{l}^{-}},C{{a}^{2+}}S{{c}^{3+}}$ are isoelectronic species.
So, the correct answer is “Option C”.

Note: The nuclear charge is calculated by the number of electrons + charge on the ion.
For example, In Ar,${{K}^{+}}$ and$C{{a}^{2+}}$ the number of electrons is 18.
The nuclear charge of Ar is 18 because there is no charge on the ion.
The nuclear charge of the ${{K}^{+}}$ ion is 19 because potassium ion has a +1 charge.
The nuclear charge of the $C{{a}^{2+}}$ ion is 20 because calcium ion has a +2 charge.

`K^(+), Ca^(2+), Sc^(3+)` `Na^(+), F^(-), V^(3+)` `K^(+), Cl^(-), Mg^(2+)` `Cr^(3+), Fe^(2+), Co^(3+)`

Answer : A

Solution : `K^(+), Ca^(2+), Sc^(3+)` all contain 18 electrons in rest of species are : `Na^(+)(10), F^(-)(10), V^(3+)(20), Cl^(-)(18), Mg^(2+)(10), Cr^(5+)(21), Fe^(2+)(24) and Co^(3+)(24)`.

`K^(+),Cl^(-),Mg^(2+),Sc^(3+)``K^(+),Ca^(2+),Sc^(3+),Cl^(-)``Na^(+),Ca^(2+),Sc^(3+),F``Na^(+),Mg^(2+),Al^(3+),Cl`

Answer : C

Solution : Isoelectronic species are those which have same electrons. <br> `K^(+)=19-1=18, Ca^(+2)=20-2=18` <br> `Sc^(+3)=121-3=18, Cl^(-)=17+1=18`

`K^(+), Ca^(2+), Sc^(3+), Cl^(-)``Na^(+), Ca^(2+), Sc^(3+), F^(-)``K^(+), Cl^(-), Mg^(2+), Sc^(3+)``Na^(+), Mg^(2+), Al^(3+), Cl^(-)`

Answer : A

Solution : Isoelectronic species are those which have same number of electrons. <br> `K^(+) = 19 -1 = 18 , Ca^(2+) = 20 - 2 = 18` <br> `Sc^(3+) = 21 - 3 = 18 , C^(-) = 17 + 1 = 18`

`K^(+), Ca^(2+), Sc^(3+)``Na^(+), F^(-), V^(3+)``K^(+), Cl^(-), Mg^(2+)`None of these

Answer : A

Solution : `K^(+), Ca^(2+), Sc^(3+)` all contain 18 electrons in rest of species are : `Na^(+) (10), F^(-)(10), V^(3+)(20) Cl^(-)(18), Mg^(2+)(10), Cr^(3+)(21), Fe^(2+) (24) and Co^(3+) (24)`.

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Solution

The correct option is A K+,Cl−,Ca2+,Sc3+Isoelectronic species are those species which contains same number of electrons. (a) K+=19−1=18 e− Cl−=17+1=18 e− Ca2+=20−2=18 e− Sc3+=21−3=18 e− Thus, all the species are isoelectronic. (b) Ba2+=56−2=54 e− Sr2+=38−2=36 e− K+=19−1=18e− S2−=16+2=18e− Thus, these species are not isoelectronic. (c) N3−=7+3=10e− O2−=8+2=10e− F−=9+1=10 e− S2−=16+2=18 e− Thus, these species are not isoeletronic. (d) Li+=3−1=2 e− Na+=11−1=10 e− Mg2+=12−2=10 e− Ca2+=20−2=18 e− Thus, these species are not isoeletronic.

What are examples of isoelectronic species?

Isoelectronic species are elements or ions that have the equal number of electrons. Example: O2−,F−,Mg2+ have 10 electrons.

What are isoelectronic sets?

Isoelectronic refers to two atoms, ions or molecules that have the same electronic structure and the same number of valence electrons. This term means "equal electric" or "equal charge". Isoelectronic chemical species typically display similar chemical properties.

How do you find the isoelectronic set?

Hint: Isoelectronic pairs are those atoms, ions and molecules that have the same number of electrons in them. To find the isoelectronic pairs, we can just add the number of electrons of each atom and also the charge of the ion (if present) to find which of the molecules have the same number of electrons in them.

sets show isoelectric species