NEETChemistryd and f Block Elements
Match List I with List II: List I (Ion) List II (Spin only magnetic moment in BM) A. Ti ²⁺ I. 5.92 B. Cr ³⁺ II. 1.73 C. Mn ²⁺ III. 3.87 D. Cu ²⁺ IV. 2.84 Choose the correct answer from the options given below:
Options
- AA-IV, B-III, C-I, D-II
- BA-III, B-IV, C-I, D-II
- CA-IV, B-I, C-III, D-II
- DA-II, B-IV, C-I, D-III
Correct answer
A. A-IV, B-III, C-I, D-II
Step-by-step solution
The spin-only magnetic moment is calculated using = n(n+2) BM , where n is the number of unpaired electrons. A. Ti ²⁺ ( Z=22 ): Electronic configuration is [ Ar ] 3d^2 . It has n=2 unpaired electrons. = 2(2+2) = 8 2.84 BM . (A matches IV) B. Cr ³⁺ ( Z=24 ): Electronic configuration is [ Ar ] 3d^3 . It has n=3 unpaired electrons. = 3(3+2) = 15 3.87 BM . (B matches III) C. Mn ²⁺ ( Z=25 ): Electronic configuration is [ Ar ] 3d^5 . It has n=5 unpaired electrons. = 5(5+2) = 35 5.92 BM . (C matches I) D. Cu ²⁺ ( Z=29 ):