Question

Which among the following is/are correct about chromium ?

• A. Its outermost partially filled orbital electronic configuration is $3d^54s^1$
• B. Total spin of chromium = 3
• C. Cr is paramagnetic and $Cr^{3+}$ is diamagnetic
• D. Magnetic moment of chromium = $\sqrt{48}$ BM

Answer 1

Answer: (A), (B), (D)

(A), (B), (D)

Answer 2

(A) Electronic configuration of Chromium (Cr): Chromium (atomic number 24) is an exception to the Aufbau principle. Its ground state electronic configuration is $[Ar] 3d^5 4s^1$, which provides extra stability due to half-filled d-orbitals. This statement is correct.

(B) Total spin of Chromium: The electronic configuration of Cr is $3d^5 4s^1$. Number of unpaired electrons in 3d = 5 Number of unpaired electrons in 4s = 1 Total number of unpaired electrons (n) = 5 + 1 = 6. Total spin (S) = $n \times (\frac{1}{2})$ = $6 \times (\frac{1}{2}) = 3$. This statement is correct.

(C) Paramagnetism/Diamagnetism of Cr and $Cr^{3+}$:

• Cr: With 6 unpaired electrons ($3d^5 4s^1$), Cr is paramagnetic.
• $Cr^{3+}$: To form $Cr^{3+}$ from Cr ($[Ar] 3d^5 4s^1$), one electron is removed from 4s and two electrons from 3d. So, the configuration becomes $[Ar] 3d^3$. In $3d^3$, there are 3 unpaired electrons. Since it has unpaired electrons, $Cr^{3+}$ is also paramagnetic, not diamagnetic. This statement is incorrect.

(D) Magnetic moment of Chromium: The magnetic moment ($\mu$) is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons. For Cr, n = 6. $\mu = \sqrt{6(6+2)} = \sqrt{6 \times 8} = \sqrt{48}$ BM. This statement is correct.

Core Solution: (A) Cr's configuration is $3d^54s^1$. (B) Cr has 6 unpaired electrons, so total spin = $6 \times \frac{1}{2} = 3$. (C) Cr is paramagnetic. $Cr^{3+}$ is $3d^3$, also paramagnetic (3 unpaired electrons). (D) Magnetic moment of Cr = $\sqrt{6(6+2)} = \sqrt{48}$ BM.