An atom with one valence electron combines with an atom with seven valence electrons. what is likely to happen to the atom with one valence electron?

a.
it will share its valence electron with the other atom.

b.
it will take one electron from the other atom.

c.
it will give up its valence electron to the other atom.

d.
it will take all seven electrons from the other atom.

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.

i think the answer would be reserve.

it would be c. predictability