Question

Write the number of pairs in which size of $I^{st}$ element or ion is higher as compared to $I{I}^{nd}$ out of following nine pairs.
$(O,S),(He,Ne),(Kr,Ne),(Na,{\left(Na\right)}^{+}),(Cl,{\left(Cl\right)}^{-}),(I^{-},{\left(Cl\right)}^{-}),{\left(Li\right)}^{+}\left(aq\right),{\left(Na\right)}^{+}\left(aq\right),(Li,Na),\left(Li,{\left(Na\right)}^{+}\right)$ .

Answer 1

Answer: 4

The atomic size decreases in period and due to the fact that the number of shells remains the same in a period but nuclear charge increases gradually as the atomic number increases. This increases the force of attraction towards the nucleus which brings a decrease in size.
The atomic size increases down the group because electrons are added to the next energy shell.
The size of cation is always less than the parent atom because cation has more effective nuclear charge as compared to the parent atom.
The size of the hydrated ion depends on the ionic size. Smaller the ion more will be the binding with a water molecule, greater will be the hydrated radii.
So correct order is- $S>O,Ne>He,Kr>Ne,Na>N{a}^{+},C{l}^{-}>Cl,I^{-}>C{l}^{-},Na>Li,N{a}^{+}>Li$ and ${\left(Li\right)}^{+}\left(aq\right)>{\left(Na\right)}^{+}\left(aq\right)$ .
(hydrated radius).