Ionic bond - class-IX

Zinc sulphide is an ionic compound. It is composed of $Zn^{2+} \ and \ S^{2-}$ ions.

Electrovalency is the process of attaining inert gas configuration by transfer of electrons.

$Na^{+}$ is more stable due to completion of octet. $Na$ atom do not have complete octate thus it unstable and try to be stable by losing an electron.

Atomic number of hydrogen $= 1$.

In Ionic bonds, metals lose electron(s) and non-metals gain the electrons. Hence, the transfer of electron(s) takes place from metal to non-metal atom.

Sodium has unstable configuration $1s^2 2s^2p^6 3s^1$ and it achieves stable configuration by loosing one electron as $Na^+ - 1s^2 2s^2p^6$

Aluminium (2,8,3)
oxygen (2,6)
so aluminium oxide is $Al _2O _3$

Only in molten and solution forms electrovalent compounds conduct
electricity because ions flows in molten and solution forms. So, it is not the common characteristic of  electrovalent compound.

An electrovalent bond is formed when a metal atom transfers one or more electrons to a non-metal atom and these compounds are called ionic compounds.

Reaction:

Electrovalency is shown by atoms when it looses or gains electrons in order to attain stability. Electrovalency is seen in ionic bonds. Electrovalency is equal to the number of electrons lost or gained by atom to form ion.

According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by
$F
= \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$ or $F =
\dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

The following factors favour the formation of ionic bond.
The electrostatic attraction between charged ions in the crystal (i.e, lattice energy) should be high.
One of the atoms (metal) should be large in size. Other atom (non metal) should be small in size.
The combining elements should differ by at least 2.0 in electronegativity.
The cation and anion should have inert gas electronic configuration.

The options (B) and (D)  represents true statements.
NaCl is more stable than CsCl because $(r _{Na+} + r _{Cl^-}) < (r _{Cs+} + r _{Cl^-}) $
MgO is more stable than NaCl because the product of $q _1$ and $q _2$ in MgO is nearly four times to that in NaCl.
Note:
According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by $F

= \dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{d^2}$ or $F =

\dfrac {1}{4 \pi \epsilon _0 K} \times \dfrac {q _1q _2}{(r^+  + r^-)^2}$

Electronic configuration of noble gas is $\mathrm{(2,8)}$. It has $\mathrm{8}$ electron in its outermost orbitals.
Element X has $\mathrm{6}$ electrons in its outermost shell. So, it requires $\mathrm{2}$ more electrons to attain the noble gas configuration.
Hence, option $\mathrm{B}$ is the correct answer.

Cation$\rightarrow $positively charged