![]() So, if EN values are not available to you, use the aforementioned trends, but keep an eye out for transition metals, since a lot of them do not follow these trends exactly. In the third period the 3 s subshell is filling for Na and Mg, and therefore Al, Si, P, S, Cl, and Ar. Option e) ends on the right element, fluorine ( 4.0), but starts off in order of decreasing EN values - lithium ( 0.98), sodium ( 0.93), and potassium ( 0.82). Across the second period Li and Be have distinguishing electrons in the 2 s subshell, and electrons are being added to the 2 p subshell in the atoms from B to Ne. Option d) is correct, since it progresses through tin ( 1.96), arsenic ( 2.18), phosphorus ( 2.19), and sulfur ( 2.58). Likewise, option c) starts off well, with zirconium ( 1.33), vanadium ( 1.63), and niobium ( 1.6), but ends on tantalum ( 1.5), which has a lower EN value than niobium. Option b) starts off well, with rubidium (EN = 0.82), calcium ( 1.00) and scandium ( 1.36), but then ends with cesium ( 0.79), which is the least electronegative of the group. It has 3 each of protons, neutrons and electrons, and represents that element Lithium (Li). Option a) is eliminated from the start, since fluorine is actually the most electronegative element in the periodic table. This problem makes use of the periodic trends in electronegativity, which can be described like thisĪs you move up a group, electronegativity increases likewise, electronegativity increases when moving from left to right across a period.
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