When determining electronic configuration, the answer is made much easier by starting with the next smallest noble gas in brackets. As a result, [Ar] is an appropriate way to incorporate every previous electron before argon.

After argon, vanadium has five other electrons to distribute, and because vanadium is a transitional element, it will fill its 3d subshells before filling the 4p subshells. The 4s subshell is filled first, and the last three electrons are placed into the 3d subshells.

[Ar]3d³4s²

A neutral atom of sodium would contain eleven electrons to balance out the charge of the eleven protons in the nucleus. We are asked, however, for the configuration of a sodium ion. Sodium is an alkali metal, meaning that it will ionize by losing only one electron, gaining a charge of . By losing one electron, sodium drops from having eleven electrons to ten. We will need to select the answer that shows an electron removed form the outermost shell.

Neutral sodium:

Sodium ion:

During alpha decay, an element emits a helium nucleus with 2 neutrons and 2 protons. Thus, the atomic mass of the new element is decreased by four, and the atomic number is decreased by two.

Three subsequent alpha decays result in a new element with an atomic mass of 232 - 3(4) = 220, and a new atomic number of 90 - 3(2) = 84.

Using the periodic table, we find the element with this atomic number is polonium (Po).

In alpha decay, a helium nucleus is emitted, and thus the isotope loses 2 protons and 2 neutrons.

In electron emission, a neutron in the nucleus is converted into a proton and an emitted electron.

In positron emission, a proton in the nucleus is converted into a neutron and an emitted positron.

The given isotope will lose 4 protons and 4 neutrons via alpha emission, gain 1 proton and lose 1 neutron via electron emission, and lose 2 protons and gain 2 neutrons via positron emission. The result is a loss of 5 protons and 8 mass units.

Accounting for the changes in atomic mass and number, we find that the final element is 141-praseodymium.

A Brønsted-Lowry base is any compound that accepts protons in solution. Lewis acids and bases refer to the accepting or donating of an electron pair, respectively.

In terms of the passage, the lone proton can be considered a proton donor and would, therefore, be a Brønsted-Lowry acid. This is not an answer choice.

The third acid-base definition is the Lewis definition, which states that acids are electron acceptors and bases are electron donors. The negative charge on the signifies that it is a Lewis base with available electrons to donate. The proton is accepting these electrons from , and is thus acting as a Lewis acid.

Lewis acids are electron pair acceptors. Molecules and ions that have a full octet cannot act a Lewis acid, therefore  and  are not lewis acids.  is very stable and insoluble and cannot accept an electron pair.  is a well known base and has extremely weak acidity.  is a transition metal ion. Transition metal are known to be Lewis acids because of their positive charge which gives them the ability to accept electron pairs.

One of the easiest ways of determining if a molecule is an electrophile is by the presence of a positive charge. Electrophiles are in need of electrons, therefore they are electron deficient and can be attacked by nucleophiles (compounds that are electron rich). A nucleophile is a compound that provides a pair of electrons to form a new covalent bond. Nucleophiles are electron rich and one of the easiest types of nucleophiles to recognize are ones carrying a negative charge.  is the only option given that contains a negative charge and therefore is not an electrophile.

In the neutralization reaction between NaOH and HCl, NaCl salt is formed. When the solution is evaporated, this salt is left behind.

3.0g of NaCl is equivalent to 0.05mol NaCl. Since the titration is between a strong acid and a strong base, all of the NaOH in the original solution is converted to NaCl in a one-to-one ratio, meaning that mol NaCl = mol NaOH.

We now know that there was 0.05mol NaOH in the 100mL solution, so the concentration must have been .

This question requires use of the simple titration equation M₁V₁ = M₂V₂. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.