8. how many valence electrons does each of the following atoms have? a. rubidium, z = 37 b. arsenic, z = 33…

8.

Explanation:

Step1: Determine electron - configurations

Use the periodic table and electron - filling rules to find the outermost electrons (valence electrons).

Step2: Identify valence electrons

• a. Rubidium ($Rb$, $Z = 37$) has an electron - configuration of $[Kr]5s^1$. The number of valence electrons is 1.
• b. Arsenic ($As$, $Z = 33$) has an electron - configuration of $[Ar]4s^23d^{10}4p^3$. The number of valence electrons is 5 ($2 + 3$ from the outermost $4s$ and $4p$ sub - shells).
• c. Aluminum ($Al$, $Z = 13$) has an electron - configuration of $[Ne]3s^23p^1$. The number of valence electrons is 3 ($2+1$ from the outermost $3s$ and $3p$ sub - shells).
• d. Nickel ($Ni$, $Z = 28$) has an electron - configuration of $[Ar]4s^23d^8$. The number of valence electrons is 2 (from the outermost $4s$ sub - shell).

Answer:

a. 1 b. 5 c. 3 d. 2

9.

Explanation:

Step1: Write electron - configurations

Use the Aufbau principle to write the electron - configurations of the elements.

Step2: Count 3d electrons

• a. Nickel ($Ni$, $Z = 28$) has an electron - configuration of $[Ar]4s^23d^8$. The number of 3d electrons is 8.
• b. Vanadium ($V$, $Z = 23$) has an electron - configuration of $[Ar]4s^23d^3$. The number of 3d electrons is 3.
• c. Manganese ($Mn$, $Z = 25$) has an electron - configuration of $[Ar]4s^23d^5$. The number of 3d electrons is 5.
• d. Iron ($Fe$, $Z = 26$) has an electron - configuration of $[Ar]4s^23d^6$. The number of 3d electrons is 6.

Answer:

a. 8 b. 3 c. 5 d. 6

10.

Explanation:

Step1: Write electron - configurations

Use the Aufbau principle to write the electron - configurations of the elements.

Step2: Identify the last - filled orbital

• a. Radium ($Ra$, $Z = 88$) has an electron - configuration of $[Rn]7s^2$. The last - filled orbital is the 7s orbital.
• b. Iodine ($I$, $Z = 53$) has an electron - configuration of $[Kr]5s^24d^{10}5p^5$. The last - filled orbital is the 5p orbital.
• c. Gold ($Au$, $Z = 79$) has an electron - configuration of $[Xe]4f^{14}5d^{10}6s^1$. The last - filled orbital is the 6s orbital.
• d. Lead ($Pb$, $Z = 82$) has an electron - configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$. The last - filled orbital is the 6p orbital.

Answer:

a. 7s b. 5p c. 6s d. 6p

11.

Explanation:

Elements in Group VIIA (Group 17) are halogens. They all have 7 valence electrons. They tend to gain one electron to achieve a stable noble - gas electron configuration. They are highly reactive non - metals. They form salts when they react with metals.

Answer:

They all have 7 valence electrons, are highly reactive non - metals, and form salts with metals.

12.

Explanation:

Metals on the periodic table lose electrons most easily. This is because metals have relatively low ionization energies. Their outermost electrons are not tightly held by the nucleus, so it is easier to remove an electron. Non - metals, on the other hand, have a greater tendency to gain electrons to achieve a stable electron configuration.

Answer:

Metals; because they have low ionization energies and their outermost electrons are not tightly held.

13.

Explanation:

Step1: Recall the trend of ionization energy

Ionization energy generally increases across a period and decreases down a group.

Step2: Analyze each set

• a. In the set Cs, K, Li (Group 1 elements), Li has the highest ionization energy because ionization energy decreases down a group.
• b. In the set Ba, Sr, Ca (Group 2 elements), Ca has the highest ionization energy because ionization energy decreases down a group.
• c. In the set I, Br, Cl (Group 17 elements), Cl has the highest ionization energy because ionization energy decreases down a group.
• d. In the set Mg, Si, S (Period 3 elements), S has the highest ionization energy because ionization energy generally increases across a period.

Answer:

a. Li b. Ca c. Cl d. S