Question

Select the correct order:

• A. Bond strength : $N{O}^{-}<NO<N{O}^{+}$
• B. $N-O$ bond angle : $N{O}_2^{+}<N{O}_2^{-}<N{O}_3^{-}$
• C. Thermal stability : $LiF>NaF>KF>RbF>CsF$
• D. Hydrated size : $B{e}^{2+}\left(aq\right)<M{g}^{2+}\left(aq\right)<C{a}^{2+}\left(aq\right)<S{r}^{2+}\left(aq\right)<B{a}^{2+}\left(aq\right)$

Answer 1

Answer: (A), (C)

Thermal stability : $LiF>NaF>KF>RbF>CsF$

To start with, let us take a look at the MO diagram for nitric oxide and extrapolate:

In the $N{O}^{+}$ cation, the last, highest-energy electron won't be
found in the anti-bonding MO. The presence of or addition of
electrons to the anti-bonding orbital(s) weakens the bond strength.
In other words, bond order decreases. Hence, in terms of bond strength,
we have:
$N{O}^{-}<NO<N{O}^{+}$

We know that $N{O}_2$ is a paramagnetic substance with an unpaired electron (radical).
$N{O}_2^{+}$ loses that lone electron of nitrogen dioxide and is thus linear. It is isostructural
with $C{O}_2$. $N{O}_2^{-}$ has a lone pair (no longer paramagnetic) and thus will have a
smaller O-N-O angle than $N{O}_2$. Nitrate anion is trigonal planar with the negative charge
equally shared by all three oxygen atoms.

The fluoride anion is a hard anion and thus forms more stable compounds with cations
that possess a good deal of polarizing power. Hence, thermal stability decreases as:
$LiF>NaF>KF>RbF>CsF$

Of the alkaline earth dipositive cations, $B{e}^{2+}$ has the highest charge density. Hence,
its hydration sphere is the largest.