Question

which of the following pairs of atoms would you expect to combine chemically to form an ionic compound?
a. li and s \t\td. f and cl
b. o and s \t\te. i and k
c. al and o \t\tf. h and n

how many electrons must an atom of each element lose to attain a noble gas electron configuration?
\ta. ca \t\t\tb. al

Explanation:

Part 1: Ionic Compound Formation

Ionic compounds form between metals (lose electrons) and non - metals (gain electrons).

• Option a: Li (metal, Group 1) and S (non - metal, Group 16). Li loses 1 e⁻, S gains 2 e⁻. Can form ionic compound.
• Option b: O and S are both non - metals (Group 16). Form covalent compounds, not ionic.
• Option c: Al (metal, Group 13) and O (non - metal, Group 16). Al loses 3 e⁻, O gains 2 e⁻. Can form ionic compound.
• Option d: F and Cl are both non - metals (Group 17). Form covalent compounds, not ionic.
• Option e: I (non - metal, Group 17) and K (metal, Group 1). K loses 1 e⁻, I gains 1 e⁻. Can form ionic compound.
• Option f: H and N are non - metals. Form covalent compounds (e.g., NH₃), not ionic.

So the pairs that form ionic compounds are a. Li and S, c. Al and O, e. I and K.

Step 1: Determine Ca's electron configuration

Ca has an atomic number of 20. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$.

Step 2: Noble gas configuration

The nearest noble gas is Ar with configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. To reach this, Ca (in $4s^{2}$) needs to lose 2 electrons.

Step 1: Determine Al's electron configuration

Al has an atomic number of 13. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$.

Step 2: Noble gas configuration

The nearest noble gas is Ne with configuration $1s^{2}2s^{2}2p^{6}$. Al has 3 valence electrons ($3s^{2}3p^{1}$), so it needs to lose 3 electrons to reach Ne's configuration.

Answer:

a. Li and S, c. Al and O, e. I and K

Part 2: Electrons to Lose for Noble Gas Configuration

Sub - question a: Ca