Electrovalent compounds do not have

Which of the following compounds has the lowest melting point?

Consider the Born-Haber cycle for the formation of an ionic compound given below and identify the compound $Z$ formed.
$M_{(s)}$ $\xrightarrow{\Delta H_1} M_{(g)}$ $\xrightarrow{\Delta H_2} M^{+}_{(g)}$
$\frac{1}{2} X_{2(g)}$ $\xrightarrow{\Delta H_3} X_{(g)}$ $\xrightarrow{\Delta H_4} X^{-}_{(g)}$
$M^{+}_{(g)} + X^{-}_{(g)} \xrightarrow{\Delta H_5} Z$

The electronic configurations of four elements $L$,$P$,$Q$,and $R$ are given in the brackets:
$L(1s^2, 2s^2 2p^4); Q(1s^2, 2s^2 2p^6, 3s^2 3p^5)$
$P(1s^2, 2s^2 2p^6, 3s^1); R(1s^2, 2s^2 2p^6, 3s^2)$
The formulae of ionic compounds that can be formed between these elements are:

Which of the following halides has the maximum melting point?

Use the following data to calculate $\Delta _{lattice}H^{\theta }$ for $NaBr$. $\Delta _{sub}H^{\theta }$ for sodium metal $= 108.4 \ kJ \ mol^{-1}$,ionization enthalpy of sodium $= 496 \ kJ \ mol^{-1}$,electron gain enthalpy of bromine $= -325 \ kJ \ mol^{-1}$,bond dissociation enthalpy of bromine $= 192 \ kJ \ mol^{-1}$,$\Delta _{f}H^{\theta }$ for $NaBr_{(s)}$ $= -360.1 \ kJ \ mol^{-1}$.