Which neutral atom of the following elements would have the most unpaired electrons? (A) Titanium (B) Manganese (C) Nickel (D) Zinc

Short Answer

Expert verified
The neutral atom with the most unpaired electrons among Titanium (Ti), Manganese (Mn), Nickel (Ni), and Zinc (Zn) is Manganese (Mn).

Step by step solution

01

Recall the Electron Configurations

Identify the electron configurations of Titanium (Ti), Manganese (Mn), Nickel (Ni), and Zinc (Zn). The configurations are: \nTi: [Ar] 4s2 3d2\nMn: [Ar] 4s2 3d5\nNi: [Ar] 4s2 3d8\nZn: [Ar] 4s2 3d10
02

Applying Hund's Rule

Apply Hund's Rule to the electron configurations. The rule states that, in a subshell, electrons occupy every orbital singly before they occupy orbitals doubly. Hence, unpaired electrons are counted.\nFor Ti: The number of unpaired electrons are 2 from the 3d shell. \nFor Mn: The number of unpaired electrons are 5 from the 3d shell. \nFor Ni: The number of unpaired electrons are 2 from the 3d shell, as 3d8 means 2 electrons are paired and 6 are unpaired. \nFor Zn: The number of unpaired electrons are 0 as all the electrons in the 3d shell are paired.
03

Compare the Number of Unpaired Electrons

From step 2, it's clear that the atom of Manganese (Mn) has the highest number of unpaired electrons, i.e., 5.

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Most popular questions from this chapter

During a chemical reaction, NO(g) gets reduced and no nitrogen- containing compound is oxidized. Which of the following is a possible product of this reaction? (A) \(\mathrm{NO}_{2}(g)\) (B) \(\mathrm{N}_{2}(g)\) (C) \(\mathrm{NO}_{3}^{-}(a q)\) (D) \(\mathrm{NO}_{2}^{-}(a q)\)

A proposed mechanism for a reaction is as follows: $\begin{array}{cc}{\mathrm{NO}_{2}+\mathrm{F}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}+\mathrm{F}} & {\text { Slow step }} \\ {\mathrm{F}+\mathrm{NO}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}} & {\text { Fast step }}\end{array}$ What is the order of the overall reaction? (A) Zero order (B) First order (C) Second order (D) Third order

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Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ If, instead of copper, a nickel bar were to be used, could nickel be plated onto the zinc penny effectively? Why or why not? (A) Yes, nickel’s SRP is greater than that of zinc, which is all that is required for nickel to be reduced at the cathode (B) Yes, nickel is able to take electrons from the \(\mathrm{H}^{+}\) ions in solution, allowing it to be reduced (C) No, nickel's SRP is lower than that of \(\mathrm{H}^{+}\) ions, which means the only product being produced at the cathode would be hydrogen gas (D) No, nickel's SRP is negative, meaning it cannot be reduced in an electrolytic cell

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

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