The magnetic moment of a transition metal ion is found to be 3.87 Bohr Magneton (BM). The number of β d and f Block Elements Chemistry Question
Question
The magnetic moment of a transition metal ion is found to be 3.87 Bohr Magneton (BM). The number of unpaired electrons present in it is :
Answer: 3. $\MU_{EFF}$ = 3.87 B.M. = $\SQRT{N(N+2)}$. HENCE VALUE OF $N$ I.E., NO. OF UNPAIRED ELECTRONS = 3
π‘ Solution & Explanation
Step 1: When silver halides are fused with sodium carbonate, they undergo a displacement reaction that initially produces silver carbonate. Step 2: At the high temperatures of fusion, silver carbonate is highly unstable and decomposes to silver oxide, which further decomposes to metallic silver (Ag) and oxygen. Step 3: The overall reaction is: 4AgCl + 2Na2CO3 ---> 4NaCl + 2CO2 + O2 + 4Ag. Hence, metallic silver (Ag) is formed, matching option (c).
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