Question

Write the symbol of each element described below. (a) largest atomic radius in Group 1 (b) smallest atomic radius in period 3 (c) largest first ionization energy in Group 2 (d) most electronegative in Group 16 (e) element(s) in period 2 with no unpaired p electron (f) abbreviated electron configuration is \([\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{3}\) (g) \(\mathrm{A}+2\) ion with abbreviated electron configuration \([\mathrm{Ar}] 3 \mathrm{~d}^{5}\) (h) A transition metal in period 4 forming a +2 ion with no unpaired electrons

Short answer

Question: Based on the trends in the periodic table, identify the following elements: a) The element with the largest atomic radius in Group 1 b) The element with the smallest atomic radius in period 3 c) The element in Group 2 with the largest first ionization energy d) The most electronegative element in Group 16 e) The element(s) in period 2 with no unpaired p electron f) The element with the abbreviated electron configuration \([\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{3}\) g) The \(\mathrm{A}+2\) ion with an abbreviated electron configuration \([\mathrm{Ar}] 3 \mathrm{~d}^{5}\) h) A transition metal in period 4 forming a +2 ion with no unpaired electrons Answer: a) Francium (Fr) b) Argon (Ar) c) Beryllium (Be) d) Oxygen (O) e) Neon (Ne) f) Vanadium (V) g) Manganese (Mn) h) Zinc (Zn)

Step-by-step solution

a) Largest atomic radius in Group 1

In group 1, as we go down the group, the atomic radius increases due to an increase in the number of electron shells. The largest atomic radius is found at the bottom of the group, which is francium (Fr).

b) Smallest atomic radius in period 3

In period 3, as we go from left to right, the atomic radius decreases due to an increase in the nuclear charge. Therefore, the smallest atomic radius is found at the far right of the period, which is argon (Ar).

c) Largest first ionization energy in Group 2

In group 2, the first ionization energy increases as we go up the group because it is easier to remove an electron from elements with larger atomic radii. Therefore, the largest first ionization energy is found at the top of the group, which is beryllium (Be).

d) Most electronegative element in Group 16

Electronegativity increases as we go up a group and to the right across a period. For group 16, the most electronegative element will be the one at the top of the group, which is oxygen (O).

e) Element(s) in period 2 with no unpaired p electron

In period 2, p orbitals appear in the elements starting from group 13, which is boron (B). However, to have no unpaired p electron, the p subshell must be completely filled. The filled p subshell can be found in group 18, which is neon (Ne).

f) Abbreviated electron configuration \([\mathrm{Ar}] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{3}\)

The given electron configuration represents an element with 2 electrons in its 4s orbital and 3 electrons in its 3d orbital, after the 18 electrons of argon. Counting the total number of electrons, we have 18+2+3 = 23 electrons, corresponding to the element vanadium (V).

g) \(\mathrm{A}+2\) ion with abbreviated electron configuration \([\mathrm{Ar}] 3 \mathrm{~d}^{5}\)

The given electron configuration represents an ion, after gaining 2 electrons by losing 2 from its 4s orbital, leaving 5 electrons in its 3d orbital. Since the configuration has a total of 18+5=23 electrons, the neutral element should have 23+2=25 electrons, corresponding to manganese (Mn).

h) A transition metal in period 4 forming a +2 ion with no unpaired electrons

In period 4, the transition metals are found in the 3d series. To have a +2 ion with no unpaired electrons, the element must have a completely filled d subshell after losing the two 4s electrons. This is the case with zinc (Zn), which has the electron configuration \([\mathrm{Ar}] 3\mathrm{~d}^{10} 4\mathrm{~s}^{2}\). After losing two 4s electrons, Zn still has a completely filled d subshell.