Question

Select the correct order(s).

• A. Bond strength : $N{O}^{-}<NO<N{O}^{+}$
• B. N-O bond angle : $NO{}^{-}<N{O}_2{}^{-}<N{O}_3^{+}$
• C. Thermal stability : $LiF>NaF>KF>RbF>CsF$
• D. 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)

The correct options are
A Bond strength : $N{O}^{-}<NO<N{O}^{+}$
C Thermal stability : $LiF>NaF>KF>RbF>CsF$

$e^{-}$ configuration of $NO$$\to$

${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p^{2}x=2p^{2}y}{\sigma }_{2p_z^2}{\pi }_{2p_x^{\prime }=2p_y^{\prime }}^{\ast }$
Bond order$=\frac{1}{2}$[$e^{-}$is bonding molecular orbital$-e^{-}$is antibonding molecular orbital]

$=\frac{1}{2}[10-5]=2.5$
Electronic configuration of $N{O}^{-}\to$

${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p^{2}x=2p^{2}y}{\sigma }_{2p^2}{\pi }_{2p_x^{\prime }=2p_y^{\prime }}^{\ast }$

Bond order$=\frac{1}{2}.(10-6)=2$

Electronic configuration of $N{O}^{+}\to$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p^{2}x=2p^{2}y}{\sigma }_{2p^{2}z}$

Bond order$=\frac{1}{2}[10-4]=3$

so, inncreasing bond order,

$N{O}^{-}<NO<N{O}^{+}$

So, bonds with higher bond order will have stronger bonds with higher stability.

Thermal stability $LiF>NaF>KF>RbF>CsF$ is due to the high lattice enthalpy. As we move down the group (alkali metal) the lattice enthalpy decrease, thus stability decreases.