Question

Write the number of pairs in which size of 1st element or ion is higher as compared to 2nd one. (O, S), (He, Ne), (Kr, Ne), (Na, ${\text{N}}{{\text{a}}^{\text{ + }}}$ ), (Cl, ${\text{C}}{{\text{l}}^{\text{ - }}}$ ), ( ${{\text{I}}^{\text{ - }}}$ , ${\text{C}}{{\text{l}}^{\text{ - }}}$ ), ( ${\text{L}}{{\text{i}}_{{\text{aq}}}}$ , ${\text{Na}}_{{\text{aq}}}^{\text{ + }}$ ), (Li, Na), (Li, ${\text{N}}{{\text{a}}^{\text{ + }}}$ ).

Answer 1

In the above pair of ions or elements are given and we are asked to find out the number of pairs in which the 1st element is greater than 2nd one. We have to see whether the elements in the same row or same column and then only we can decide which one is greater. In the case of ions, we have looked at the nuclear force of attraction.

Complete Step by step solution:
O,S:
Oxygen and sulphur belong to the same group, group 16. But the atomic number of O is 8 and atomic number of S is 16. As the size of elements increases down the group, so OHe, Ne:
Helium and neon belong to the same group, group 18. But the atomic number of He is 2 and the atomic number of Ne is 10. As the size of elements increases down the group, so HeKr, Ne:
Krypton and neon belong to the same group, group 18. But the atomic number of Kr is 36 and the atomic number of Ne is 10. As the size of elements increases down the group, so Kr>Ne.
Na, ${\text{N}}{{\text{a}}^{\text{ + }}}$ :
In ${\text{N}}{{\text{a}}^{\text{ + }}}$ ion, there is loss of 1 electronic, and hence, the nucleus now holds rest electrons more tightly as comparing to Na atoms. So, the size of Na> ${\text{N}}{{\text{a}}^{\text{ + }}}$ .
Cl, ${\text{C}}{{\text{l}}^{\text{ - }}}$ :
In ${\text{C}}{{\text{l}}^{\text{ - }}}$ :ion, there is gain of 1 electronic, and hence, net effect charge between nucleus and surrounding electrons decrease as compared to Cl atoms. So, the size of Cl< ${\text{C}}{{\text{l}}^{\text{ - }}}$ .
${{\text{I}}^{\text{ - }}}$ , ${\text{C}}{{\text{l}}^{\text{ - }}}$ :
Both I and Cl belong to group 17 of the periodic table. The atomic number of Cl is 17 and I is 53. Hence, gain of 1 electron will have negligible effect as compared to addition of new valence shell and therefore, ${{\text{I}}^{\text{ - }}}$ > ${\text{C}}{{\text{l}}^{\text{ - }}}$ .
${\text{L}}{{\text{i}}_{{\text{aq}}}}$ , ${\text{Na}}_{{\text{aq}}}^{\text{ + }}$ :
In aqueous medium, element in the first row shows the diagonal relationship and hence size of Li is equivalent of ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ . So, size of ${\text{L}}{{\text{i}}_{{\text{aq}}}}$ > ${\text{Na}}_{{\text{aq}}}^{\text{ + }}$ .
Li, Na:
Lithium and sodium belong to the same group, group 1. But the atomic number of Li is 3 and the atomic number of Na is 11. As the size of elements increases down the group, so LiLi, ${\text{N}}{{\text{a}}^{\text{ + }}}$ :
Lithium and sodium belong to the same group, group 1. But the atomic number of Li is 3 and the atomic number of Na is 11. As size of elements increase down the group and in ${\text{N}}{{\text{a}}^{\text{ + }}}$ ion, there is loss of 1 electronic, and hence, the nucleus now holds rest electrons more tightly as comparing to Na atoms. This reduction of size due to loss of electrons is less than increase of size due to addition of new valence shell Li< ${\text{N}}{{\text{a}}^{\text{ + }}}$ .
So, the number of pairs in which the size of the 1st element or ion is higher as compared to 2nd one is 4.

Note :
As we move down an element group (column), the size of atoms increases. This is because of gain in additional electron energy shells.
As you move across an element period (row), the overall size of atoms decreases slightly. Even though atoms further to the right have more protons, neutrons, and electrons, the outer electron shell is the same. The increased number of protons exerts a stronger positive charge which results in pulling of the electrons in toward the nucleus.