Question

periodic table
name: jaydin milladate:
section 1
use the following clues to determine the elements being described and fill in the missing information

1. each atom of this element has 9 protons and 10 neutrons. the atomic number of the element is (1). the mass number is (2). the number of valence electrons is (3). this element can be found in the (4) section of the periodic table. the electron configuration of this atom is (5). this element is classified as a(n) (6). this element is (7), with a symbol of (8).
2. this element is a metalloid of group iv. it has (9) electrons on the third energy level. its name is (10), symbol (11). its atomic number is (12), and its mass number is (13). this element has (14) neutrons.
3. this is a halogen and is found in group (15). it has (16) valence electrons. it has 17 electrons in a neutral atom. the element is (17), symbol (18). it has (19) protons and a mass number of 35. how many neutrons does it have? (20).
4. this element is the largest of period three. it is (21), symbol (22). its atomic number is (23) and it is found in the family (24).
5. each atom of this element has 56 protons and 81 neutrons. the atomic number is (25), and the atomic mass number is (26). the number of valence electrons this atom has is (27) because it is found in group (28). the element is (29), symbol (30).

1.
19
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
28
13.
14.
15.
7
16.
17.
18.
19.
20.
21.
sodium
22.
23.
24.
25.
26.
27.
28.
29.
30.
ba

Explanation:

Problem 1:

Step 1: Determine atomic number

Atomic number = number of protons = 9. So atomic number (Z) = 9.

Step 2: Determine mass number

Mass number (A) = protons + neutrons = 9 + 10 = 19.

Step 3: Determine valence electrons

Element with Z=9 is fluorine (F), group 17, so valence electrons = 7.

Step 4: Period and group

Group 17 (halogens), period 2 (since 2nd energy level is outermost, 2 shells: 2,7).

Step 5: Electron configuration

Fluorine: \(1s^2 2s^2 2p^5\), so electron configuration is \(2,7\).

Step 6: Classification

Non - metal (halogen), symbol F, name fluorine.

Step 1: Identify the metalloid of Group IV

Metalloid in Group IV (14) is germanium (Ge).

Step 2: Determine electrons on third energy level

Germanium has electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^2 3d^{10} 4s^2 4p^2\), electrons on third energy level: \(2 + 6+ 10=18\)? Wait, no, third energy level is \(n = 3\), so \(3s^23p^63d^{10}\), total 18? Wait, no, the question says "8 electrons on the third energy level"? Wait, maybe a mistake, but germanium: atomic number 32, mass number: let's see, germanium has isotopes, but common mass number around 73 - 76. Wait, maybe the problem is about silicon? No, silicon is a metalloid of Group IV, third energy level: \(3s^23p^2\), 4 electrons. Wait, the problem says "8 electrons on the third energy level". Wait, maybe the element is germanium? Wait, no, let's re - evaluate. Group IV (14), metalloid: germanium (Ge). Atomic number 32, mass number: let's assume. Wait, the problem says "8 electrons on the third energy level". Third energy level (\(n = 3\)) can hold up to 18 electrons. If 8 electrons on third level, maybe the element is silicon? No, silicon has 4 on third. Wait, maybe the element is germanium, third level: \(3s^23p^63d^{10}\) (18 electrons), no. Wait, maybe the problem has a typo, but let's proceed. Name: Germanium, symbol: Ge, atomic number: 32, mass number: let's say 73 (common), neutrons: \(73 - 32 = 41\).

Step 1: Identify the halogen

Halogens are in Group 17 (7A). Valence electrons: 7. Atomic number: 17 (chlorine).

Step 2: Mass number and neutrons

Mass number: 35, protons: 17, neutrons: \(35 - 17=18\).

Answer:

1. Atomic number: 9; 2. Mass number: 19; 3. Valence electrons: 7; 4. Period: 2; 5. Electron configuration: 2,7; 6. Classification: non - metal; 7. Name: Fluorine; 8. Symbol: F

Problem 2: