I need help understanding these questions (g) A different element, L, is located in the same row of the periodic table as element E, but L has roughly half the atomic mass of E. Is the atomic radius of L lessthan, equal to, or greater than the atomic radius of E? Justify your answer using principles of atomic structure.ВIx2 X2Ω0 / 10000 Word Limit(h) The ionic radii of elements E and a different metallic element, M, are shown in the following table. Both elements form oxides, E20 and MO. If lattice energy isdefined as the energy required to separate an ionic solid into individual separate gaseous ions, would the lattice energy of MO be less than, equal to, or greater thanthe lattice energy of the oxide E2O? Justify your answer in terms of Coulomb's law.lonic radiusElement(pm)E170M110B I U x? X2Ω!!

(g) The elements L and E lies on the same period.L is said to have roughly half the atomic mass of E. We know that as we move towards right in a period, the atomic mass of the elements increases, this suggests that L comes first as we move along the period containing L and E. On moving across a period, the atomic radius decreases, that's because more and more electrons added to the same principal shell, which results in an increased nuclear charge. This leads to the electrons in the atom being strongly pulled to the nucleus. Thus, the atomic radius of L is greater than the atomic radius of E.(h) As per the formula of the oxide, E2O, the valency of E should be 1. That's because the valency of 'O' is 2, and in writing a chemical formula, the valencies are cross-multiplied. As per the formula of the oxide, MO, the valency of M should be the same as that of O. Thus, valency of M is 2. As per the Coulomb's law, the electrostatic forces between two ions is given as: F=kq1q2r2where,k=Coulomb's constantq1 and q2 are the magnitude of charges on the ionsr is the distance between ionsFor E2O, using Coulomb's law, Charge on E=+1Charge on O=-2FE2O=k1×2(170+radius of O)2FE2O=2 k(170+radius of O)2 (1) For MO, using Coulomb's law, Charge on E=+2Charge on O=-2FMO=k2×2(110+radius of O)2FMO=4 k(110+radius of O)2 (2)Comparing (1) and (2), FMO > FE2O Thus, the electrostatic forces in MO are greater than that in E2O, thus it is comparatively difficult to separate MO into their ions that E2O. Thus, lattice energy of MO is greater than the lattice energy of E2O.