Question

question 5. how many electrons must an atom of each element lose to attain a noble - gas electron configuration?
a. potassium (k).

b. aluminum (al).

c. strontium (sr).

d. cesium (cs).

e. barium (ba).

Explanation:

Step1: Analyze Potassium (K)

Potassium is in Group 1 of the periodic table. Atoms in Group 1 have 1 valence electron. To attain a noble - gas configuration (which has a full outer shell), they lose their 1 valence electron. So potassium (K) loses 1 electron.

Step2: Analyze Aluminum (Al)

Aluminum is in Group 13 of the periodic table. The number of valence electrons for Group 13 elements is 3 (since the group number for main - group elements (excluding transition metals) can be used to determine valence electrons: for Groups 1 - 2 and 13 - 18, the number of valence electrons is equal to the group number for Groups 1 - 2 and group number - 10 for Groups 13 - 18). So aluminum has 3 valence electrons. To attain a noble - gas configuration, it loses 3 electrons.

Step3: Analyze Strontium (Sr)

Strontium is in Group 2 of the periodic table. Atoms in Group 2 have 2 valence electrons. To attain a noble - gas configuration, they lose their 2 valence electrons. So strontium (Sr) loses 2 electrons.

Step4: Analyze Cesium (Cs)

Cesium is in Group 1 of the periodic table. Similar to potassium, it has 1 valence electron. To attain a noble - gas configuration, it loses 1 electron.

Step5: Analyze Barium (Ba)

Barium is in Group 2 of the periodic table. Similar to strontium, it has 2 valence electrons. To attain a noble - gas configuration, it loses 2 electrons.

Answer:

a. Potassium (K): 1
b. Aluminum (Al): 3
c. Strontium (Sr): 2
d. Cesium (Cs): 1
e. Barium (Ba): 2