Answer :
Answer:
Option-1 (O²⁻) is the correct answer.
Explanation:
All given anions contains same charge. So, we can ignore the effect of charge on these anions.
As we know all given compounds belongs to same group (Group 6) in periodic table. And from top to bottom along the group the elements are placed as,
Oxygen O
Sulfur S
Selenium Se
Tellurium Te
Hence, moving from top to bottom along the group the number of shells increases. And with increase in number of shell the atomic or ionic radii increases. As Oxygen is present at the top of the group, therefore, it has the smallest radius due to less number of shells.
O²⁻ ion has the smallest radius
Further explanation
In an atom there are levels of energy in the shell and sub shell.
This energy level is expressed in the form of electron configurations.
Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d and f. The maximum number of electrons for each sub shell is
- s: 2 electrons
- p: 6 electrons
- d: 10 electrons and
- f: 14 electrons
Charging electrons in the sub shell uses the following sequence:
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.
Each sub-shell also has an orbital shown in the form of a square in which there are electrons symbolized by half arrows
The atomic radius shows the distance of the atomic nucleus to the electrons in the outer shell
From left to right in the period system the atomic radius gets smaller, while from one group from top to bottom the atomic radius gets longer
The more the number of shells the atom has, the radius of the atom getting longer, but if the number of shells is the same, the larger atomic number has shorter radius because the core charge is greater so that the attraction of the nucleus to the electrons is stronger
In the ions
(1) O²⁻
(2) S²⁻
(3) Se²⁻
(4) Te²⁻
All of these ions are ions from the elements of the VIA group in the periodic system. This group consists of oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po).
The periodic system is arranged in the order of atomic numbers, whereas in this VIA group from top to bottom, the atomic number gets bigger
- Element O
Atomic Number: 8
Electron configuration: 1s² 2s² 2s⁴ (number of shells 2)
- Element S
Atomic Number: 16
Electron configuration: [Ne] 3s² 3p⁴ (number of shells 3)
- Element Se
Atomic Number: 34
Electron configuration: [Ar] 3d¹⁰ 4s²4p⁴ (number of shells 4)
- Element Te
Atomic Number: 52
Electron configuration: [Cr] 4d10 5s² 5p⁴ (number of shells 5)
From the configuration of the electron shows that the lower the shell owned by the elements of the VIA group the greater (from 2 to 5) so that the radius is also getting longer/bigger
Because what is being asked is the radius of the ion, the electron configuration of each element is equally reduced by 2 electrons, but the number of shells remains the same
(1)O²⁻
Electron configuration: 1s² 2s² 2s²
(2) S²⁻
Electron configuration: [Ne] 3s² 3p²
(3) Se²⁻
Electron configuration: [Ar] 3d¹⁰ 4s²4p²
(4) Te²⁻
Electron configuration: [Cr] 4d¹⁰ 5s² 5p²
Because the amount of shell from the Oxygen ion is the smallest, the radius of the O²⁻ is also the smallest
Learn more
electron affinity
brainly.com/question/1440853
Identify the group number in the periodic table
brainly.com/question/2014634
the charge on each ion in the compounds
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Keywords: element, group 6A, the smallest radius
