Question

Number of correct options among the following is/are:
$\left(a\right)I^{-}>B{r}^{-}>C{l}^{-}>F^{-}\text{(Polarizability)}$
$\left(b\right)L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}\text{(Polarization power)}$
$\left(c\right)H_{2}O>H_{2}S>H_{2}Se>H_{2}Te\left(\text{Boiling point}\right)$
$\left(d\right)H_2^{-}>H_2^{+}\text{(order of stability)}$

Answer 1

(a) Polarisability of halide ions increases in the order $F^{-}$< $C{l}^{-}$< $B{r}^{-}$<$I^{-}$. Polarisability is directly proportional to the size of the anion. Larger is the size of the anion, larger is the polarizability and vice versa. Thus, iodide ion has the maximum size, and hence the maximum polarisability.
(b) We know that the size of cations increases from top to bottom in a given group of periodic table. Therefore, the cation size of alkali metals also increases as shown below.
$L{i}^{+}<N{a}^{+}<K^{+}<R{b}^{+}<C{s}^{+}$
Polarizibility of cation increases when size decreases.
Hence, $L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}$.
(c) Boiling point of hydrides of group $16$ elements decreases down the group.
(d) Molecular orbital electronic configuration of $H_2^{+}=\left({\sigma }_{1s}{)}^1\right({\sigma }_{1s}^{\ast }{)}^0$. Molecular orbital electronic configuration of $H_2^{-}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^1$.
Bond order of $H_2^{+}$ is $0.5$ whereas bond order of $H_2^{-}$ is $0.5$. Bond order of both $H_2^{+}$ and $H_2^{-}$ is same but in case of $H_2^{-}$, one electron is present in antibonding molecular orbital. Hence, $H_2^{-}$ is less stable than $H_2^{+}$. Hence (d) is incorrect.