Question

A. If the following elements were to form an ionic compound, which noble-gas configuration would they most likely attain?

Drag the appropriate elements to their respective bins.

B. An unknown element, X, reacts with lithium to form the compound Li2X. In other compounds this element also can accommodate up to 12 electrons rather than the usual octet. What element could Xbe?
If the following elements were to form an ionic compound, which noble-gas configuration would they most likely attain? Drag the appropriate elements to their respective bins. ► View Available Hint(s) Reset Help Mg SC OM O @ 6A As He Ne Ar

Answer 1

He	Ne	Ar	Kr
Li	Mg , O	Cl,Sc	As , Y

In general, elements ionizes and achieve a E,C of nearest noble gas element.

Li

Atomic number of lithium is 3. Nearest noble gas element is He. Hence, Li losses one electron and achieve E.C of He. E.C of He is 1 s ².

Li ( 1 s ², 2 s ¹ )$\rightarrow$ Li ⁺ ( 1 s ²)

Mg

Atomic number of Mg is 12. Nearest noble gas element is Ne. Hence,Mg losses two electrons and achieve E.C of Ne. E.C of Ne is 1 s ², 2 s ², 2 p ⁶

Mg (1 s ², 2 s ², 2 p ⁶, 3 s ² )$\rightarrow$ Mg ²⁺ (1 s ², 2 s ², 2 p ⁶)

O

Atomic number of oxygen is 8. Nearest noble gas element is Ne. Hence,O gains two electrons and achieve E.C of Ne. E.C of Ne is 1 s ², 2 s ², 2 p ⁶

O (1 s ², 2 s ², 2 p ⁴ ) + 2 e ^- $\rightarrow$ O ^2- (1 s ², 2 s ², 2 p ⁶)

Cl

Atomic number of Cl is 17. Nearest noble gas element is Ar. Hence,Cl gains one electron and achieve E.C of Ar. E.C of Ar is 1 s ², 2 s ², 2 p ⁶ , 3 s ² , 3 p ⁶

Cl ( 1 s ², 2 s ², 2 p ⁶ , 3 s ² , 3 p ⁵) +e ^-   $\rightarrow$ Cl ^- (1 s ², 2 s ², 2 p ⁶)

Sc

Atomic number of Sc is 21. Nearest noble gas element is Ar. Hence,Sc losses three electrons and achieve E.C of Ar. E.C of Ar is 1 s ², 2 s ², 2 p ⁶ , 3 s ² , 3 p ⁶

Sc (1 s ², 2 s ², 2 p ⁶ , 3 s ² , 3 p ⁶, 4 s ² , 3 d ¹)  $\rightarrow$ Sc ³⁺( 1 s ², 2 s ², 2 p ⁶ , 3 s ² , 3 p ⁶ )

As

Atomic number of As is 33. Nearest noble gas element is Kr. It's atomic number is 36. Hence,As gains three electrons and achieve E.C of Kr.

Y

Atomic number of Y is 39. Nearest noble gas element is Kr. It's atomic number is 36. Hence,Y looses three electrons and achieve E.C of Kr.

PART 2

ANSWER : Cl

Element X forms a neutral compound with Li. Means X will be element forming anion i e negatively charged ion.

Hence, Mg and Po ruled out.

O and Cl can form anions.

Now , we given information that X can expand it's octet.

Now, compare O and Cl

O : E.C of O is 1 s ², 2 s ², 2 p ⁴. There are no vacant orbitals in second energy level available for expanding octet. Hence, O can not expand it's octet.

Cl :E.C of Cl is 1 s ², 2 s ², 2 p ⁶ , 3 s 2 , 3 p ⁶, 3 d ⁰

In Cl vacant d orbitals are available for bonding, hence Cl can expand its octet. Hence ANSWER is X = Cl