As per the question the first atom has 1 valence electron.
The second atom has 7 valence electrons.
The two atoms come closer and combined to form a substance.
Atoms having one valence electrons will be considered as electropositive and the atoms having seven valence electrons are considered as electronegative .
As per the octet role,every atom tends to possess eight electrons in their valence shells to satisfy the inert gas configuration to be stable.
Here the atoms having one valence electrons will lose one atom and atoms having seven electrons will accept this electron .In this way both the atoms will satisfy inert gas configuration. The atom which will lose electrons tends to form cation and the atom which will accept electrons tends to form anion.So there is the chance of formation of ionic bond.
The electronegative atom can not emit seven electrons as it will require more ionization energy.The atom having one atom in its valence shell can not accept one electron from the other atom as by doing so they will lose their stability.
There will be no sharing of electrons here.It is because the atoms having one atom in their valence shell has low ionization energy and atoms having seven electrons in their valence shells have high electron affinity.Here ionic bond is favorable not covalent bond.
Hence the option C is right.
