Ionic bond is formed due to the transfer of electrons from one atom to another so that all atoms involved in the bond would become stable (with 8 electrons in the outermost level)
Now, for bromine, it has 35 electrons. This means that bromine has 7 valence electrons in the outermost level. Therefore, bromine needs to gain one electron in order to become stable.
Bromine can react with elements from:
group 1: each element in group 1 needs to lose one electron to become stable. Therefore, one bromine atom can form an ionic bond when combined with an atom of an element from group 1 (element in group 1 loses its electron for bromine atom).
group 2: each element in group 2 needs to lose two electrons to become stable. Therefore, two bromine atoms can form ionic bonds when combined with an atom of an element from group 2 (element in group 2 loses two electrons, one for each bromine atom).
group 3: each element in group 3 needs to lose three electrons to become stable. Therefore, three bromine atoms can form ionic bonds when combined with an atom of an element from group 1 (element in group 3 loses three electrons, one for each bromine atom).
Since no choices are given , I cannot tell the exact choice. But the correct one would be the element from either group 1 , 2 or 3.
According to the balanced equation:
C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4 H2O (g)
aA + bB ↔ cC + dD
according to K formula when:
K = concentration of the products / concentration of the reactants
K = [C]^c[D]^d / [A]^a[B]^b
when [A],[B],[C]and[D] is the concentrations
and a,b,c and d is the no of moles
∴ K = [CO2]^3[H2O]^4 / [C3H8] [ O2]^5
Answer:
C.
Explanation:
I believe that it should be A and B.
