Question

Choose the correct matching of transition metal ion and magnetic moment from the codes given below (At. No: Ti = 22, V = 23, Fe = 26) Transition element
• [(a)] Titanium (III) (1) 4.9
• [(b)] Vanadium (II) (2) 1.73
• [(c)] Iron (II) (3) 3.87

• A. a-(2), b-(3), c-(1)
• B. a-(2), b-(1), c-(3)
• C. a-(1), b-(2), c-(3)
• D. a-(1), b-(3), c-(2)
• E. a-(3), b-(2), c-(1)

Hint:

Remember: $n = 1,2,3,4,5$ → $\mu = 1.73, 2.83, 3.87, 4.9, 5.92$

Answer 1

The Correct Option is A

Concept: Magnetic moment formula
\[ \mu = \sqrt{n(n+2)} \text{ BM} \] where $n$ = number of unpaired electrons ---

Step 1: Titanium (III)
\[ \text{Ti} = 3d^2 4s^2 \Rightarrow \text{Ti}^{3+} = 3d^1 \] \[ n = 1 \] \[ \mu = \sqrt{1(3)} = \sqrt{3} = 1.73 \] \[ \Rightarrow (2) \] ---

Step 2: Vanadium (II)
\[ \text{V} = 3d^3 4s^2 \Rightarrow \text{V}^{2+} = 3d^3 \] \[ n = 3 \] \[ \mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \] \[ \Rightarrow (3) \] ---

Step 3: Iron (II)
\[ \text{Fe} = 3d^6 4s^2 \Rightarrow \text{Fe}^{2+} = 3d^6 \] \[ n = 4 \] \[ \mu = \sqrt{4(6)} = \sqrt{24} \approx 4.9 \] \[ \Rightarrow (1) \] --- Final Matching:
\[ a \rightarrow (2), \quad b \rightarrow (3), \quad c \rightarrow (1) \] \[ \boxed{\text{Option (A)}} \]