Write the shorthand electron configurations for the following atoms. Enter full electron configuration in order electrons fill the subshells with number of electrons in the subshells in parentheses. For example, write the electron configuration for Li as [He]2s(1). The answer is case sensitive.

Explanation:

The elements are;

S K, Ti, Sn,

Shorthand electronic configuration basically involves inserting the closest noble gas element before the element in the configuration.

Sulphur (S): This element has 16 atoms. The noble gas before this element is Neon. So the shorthand electronic configuration is written as;

[Ne] 3s² 3p⁴

Potassium (K): This element has 19 electrons, The noble gas before this element is Argon. So the shorthand electronic configuration is written as;

[Ar] 4s¹

Titanium (Ti): This element has 22 electrons. The noble gas before this element is Argon (18) - remaining 4 electrons. The shorthand is given as;

[Ar] 3d² 4s²

Tin (Sn): This element has 50 electrons. The noble gas before this element is Krypton (36) - remaining 14 electrons. The shorthand is given as;

[Kr] 4d¹⁰ 5s² 5p²

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.