Question

1. explain why there is a difference in ionic radius observed when neutral atoms become ions.

use your periodic table to answer questions 4 - 11.

1. which of the following elements would have the smallest atomic radius?

a. fluorine (f) c. bromine (br)
b. iodine (i) d. chlorine (cl)

1. which of the following elements would have the largest atomic radius?

a. copper (cu) c. iron (fe)
b. titanium (ti) d. calcium (ca)

1. which of the following correctly arranges the elements in order of decreasing electronegativity?

a. n, as, sb, p c. te, se, s, o
b. na, k, rb, cs d. al, si, p, s

1. which of the following elements has the highest electronegativity?

a. fluorine (f) c. hydrogen (h)
b. francium (fr) d. oganesson (og)

1. which of the following elements would be considered the most reactive?

a. fluorine (f) c. hydrogen (h)
b. francium (fr) d. oganesson (og)

1. which of the following elements would be less reactive than sulfur (s)?

a. oxygen (o) c. phosphorous (p)
b. chlorine (cl) d. fluorine (f)

1. explain why ionization energy increases to the right within a period.
2. which of the periodic trends is not observed consistently for the noble gases?

a. ionization energy c. atomic radius
b. electronegativity d. reactivity

Explanation:

1. When a neutral atom becomes a cation (loses electrons), the ion - core is pulled in more tightly by the nucleus, reducing the ionic radius. When it becomes an anion (gains electrons), electron - electron repulsions increase, expanding the ionic radius.
2. Atomic radius decreases across a period and increases down a group. Fluorine is highest up and to the right among the options, so it has the smallest atomic radius.
3. Atomic radius increases down a group and from right to left across a period. Calcium is furthest down and to the left among the options, so it has the largest atomic radius.
4. Electronegativity decreases down a group and increases across a period. For option a, N, P, As, Sb is the correct order of decreasing electronegativity (not as given). For option b, Na, K, Rb, Cs (not Ca) would be in decreasing electronegativity order. For option c, O, S, Se, Te is the correct decreasing order (not as given). For option d, S, P, Si, Al is the correct decreasing order of electronegativity.
5. Fluorine is the most electronegative element on the periodic table.
6. Fluorine is highly reactive as a non - metal due to its high electronegativity and desire to gain an electron to achieve a stable octet. Francium is a highly reactive metal as it easily loses its outermost electron. But among non - metals, fluorine is very reactive. Oganesson is a noble gas and very unreactive.
7. Reactivity of non - metals decreases down a group. Phosphorus is less reactive than sulfur as sulfur is more electronegative and closer to achieving a stable octet.
8. Ionization energy increases across a period because the effective nuclear charge increases, making it harder to remove an electron.
9. Noble gases have a full outer shell, so their reactivity is extremely low compared to other elements, and the reactivity trend is not consistent with them.

Answer:

1. Explanation provided above.
2. a. Fluorine (F)
3. d. Calcium (Ca)
4. None of the given options are correct as written. The correct decreasing electronegativity orders are: for non - metals in a group like N, P, As, Sb; for alkali metals Na, K, Rb, Cs; for chalcogens O, S, Se, Te; across a period S, P, Si, Al.
5. a. Fluorine (F)
6. a. Fluorine (F)
7. c. Phosphorous (P)
8. Explanation provided above.
9. d. Reactivity