Question

1. write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic.

a. v5+
b. cr3+
c. ni2+
d. fe3+

1. write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic.

Explanation:

To solve this, we analyze each ion by determining its electron configuration, then constructing the orbital diagram and checking for unpaired electrons (paramagnetic has unpaired, diamagnetic does not).

Part 109 (a. \( \boldsymbol{V^{5+}} \))

Step 1: Electron Configuration of V and \( V^{5+} \)

Vanadium (V) has atomic number 23, so electron configuration: \( [Ar] 3d^3 4s^2 \).
\( V^{5+} \) loses 5 electrons (3 from \( 3d \), 2 from \( 4s \)): \( [Ar] \) (or \( 1s^2 2s^2 2p^6 3s^2 3p^6 \)).

Step 2: Orbital Diagram

All orbitals (1s, 2s, 2p, 3s, 3p) are fully paired (no unpaired electrons).

Step 3: Magnetic Behavior

No unpaired electrons → diamagnetic.

Part 109 (b. \( \boldsymbol{Cr^{3+}} \))

Step 1: Electron Configuration of Cr and \( Cr^{3+} \)

Chromium (Cr) has atomic number 24, electron configuration: \( [Ar] 3d^5 4s^1 \) (exception: half-filled \( 3d \) is stable).
\( Cr^{3+} \) loses 3 electrons (1 from \( 4s \), 2 from \( 3d \)): \( [Ar] 3d^3 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow \) \( \uparrow \) \( \uparrow \) (three unpaired electrons).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 109 (c. \( \boldsymbol{Ni^{2+}} \))

Step 1: Electron Configuration of Ni and \( Ni^{2+} \)

Nickel (Ni) has atomic number 28, electron configuration: \( [Ar] 3d^8 4s^2 \).
\( Ni^{2+} \) loses 2 electrons (from \( 4s \)): \( [Ar] 3d^8 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow\downarrow \) \( \uparrow\downarrow \) \( \uparrow\downarrow \) \( \uparrow \) \( \uparrow \) (two unpaired electrons).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 109 (d. \( \boldsymbol{Fe^{3+}} \))

Step 1: Electron Configuration of Fe and \( Fe^{3+} \)

Iron (Fe) has atomic number 26, electron configuration: \( [Ar] 3d^6 4s^2 \).
\( Fe^{3+} \) loses 3 electrons (2 from \( 4s \), 1 from \( 3d \)): \( [Ar] 3d^5 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow \) \( \uparrow \) \( \uparrow \) \( \uparrow \) \( \uparrow \) (five unpaired electrons, half-filled \( 3d \)).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 110 (General Approach)

For each ion in 110, follow the same steps:

1. Determine the atomic number and electron configuration of the neutral atom.
2. Subtract electrons to form the ion (lose from \( 4s \) first, then \( 3d \) for transition metals).
3. Draw the orbital diagram (fill orbitals with electrons, following Hund’s rule and Pauli exclusion).
4. Check for unpaired electrons:

• Diamagnetic: No unpaired electrons.
• Paramagnetic: At least one unpaired electron.

Final Answers (109)

a. \( V^{5+} \): Diamagnetic
b. \( Cr^{3+} \): Paramagnetic
c. \( Ni^{2+} \): Paramagnetic
d. \( Fe^{3+} \): Paramagnetic

(For 110, apply the same method to each ion provided in the problem.)

Answer:

To solve this, we analyze each ion by determining its electron configuration, then constructing the orbital diagram and checking for unpaired electrons (paramagnetic has unpaired, diamagnetic does not).

Part 109 (a. \( \boldsymbol{V^{5+}} \))

Step 1: Electron Configuration of V and \( V^{5+} \)

Vanadium (V) has atomic number 23, so electron configuration: \( [Ar] 3d^3 4s^2 \).
\( V^{5+} \) loses 5 electrons (3 from \( 3d \), 2 from \( 4s \)): \( [Ar] \) (or \( 1s^2 2s^2 2p^6 3s^2 3p^6 \)).

Step 2: Orbital Diagram

All orbitals (1s, 2s, 2p, 3s, 3p) are fully paired (no unpaired electrons).

Step 3: Magnetic Behavior

No unpaired electrons → diamagnetic.

Part 109 (b. \( \boldsymbol{Cr^{3+}} \))

Step 1: Electron Configuration of Cr and \( Cr^{3+} \)

Chromium (Cr) has atomic number 24, electron configuration: \( [Ar] 3d^5 4s^1 \) (exception: half-filled \( 3d \) is stable).
\( Cr^{3+} \) loses 3 electrons (1 from \( 4s \), 2 from \( 3d \)): \( [Ar] 3d^3 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow \) \( \uparrow \) \( \uparrow \) (three unpaired electrons).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 109 (c. \( \boldsymbol{Ni^{2+}} \))

Step 1: Electron Configuration of Ni and \( Ni^{2+} \)

Nickel (Ni) has atomic number 28, electron configuration: \( [Ar] 3d^8 4s^2 \).
\( Ni^{2+} \) loses 2 electrons (from \( 4s \)): \( [Ar] 3d^8 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow\downarrow \) \( \uparrow\downarrow \) \( \uparrow\downarrow \) \( \uparrow \) \( \uparrow \) (two unpaired electrons).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 109 (d. \( \boldsymbol{Fe^{3+}} \))

Step 1: Electron Configuration of Fe and \( Fe^{3+} \)

Iron (Fe) has atomic number 26, electron configuration: \( [Ar] 3d^6 4s^2 \).
\( Fe^{3+} \) loses 3 electrons (2 from \( 4s \), 1 from \( 3d \)): \( [Ar] 3d^5 \).

Step 2: Orbital Diagram

\( 3d \) orbitals: \( \uparrow \) \( \uparrow \) \( \uparrow \) \( \uparrow \) \( \uparrow \) (five unpaired electrons, half-filled \( 3d \)).

Step 3: Magnetic Behavior

Unpaired electrons → paramagnetic.

Part 110 (General Approach)

For each ion in 110, follow the same steps:

1. Determine the atomic number and electron configuration of the neutral atom.
2. Subtract electrons to form the ion (lose from \( 4s \) first, then \( 3d \) for transition metals).
3. Draw the orbital diagram (fill orbitals with electrons, following Hund’s rule and Pauli exclusion).
4. Check for unpaired electrons:

• Diamagnetic: No unpaired electrons.
• Paramagnetic: At least one unpaired electron.

Final Answers (109)

a. \( V^{5+} \): Diamagnetic
b. \( Cr^{3+} \): Paramagnetic
c. \( Ni^{2+} \): Paramagnetic
d. \( Fe^{3+} \): Paramagnetic

(For 110, apply the same method to each ion provided in the problem.)