Question
The calculated spin only magnetic moments of Cr NH 3 6 3 + and CuF 6 3 - in BM , respectively, are (Atomic numbers of Cr and Cu are 24 and 29 , respectively)
JEE Advanced
Chemistry
Coordination Compounds
2021
JEE Advanced 2021 (Paper 1)
JEE Advanced Chemistry Question (2021) — Solution
Options
- A. 3 . 87 and 2 . 84
- B. 4 . 90 and 1 . 73
- C. 3 . 87 and 1 . 73
- D. 4 . 90 and 2 . 84
Answer
A. 3 . 87 and 2 . 84
Step-by-step solution
Spin only magnetic moment can be calculated using the formula μ = n   n + 2 B . M n = number of unpaired electron at central atom. Cr NH 3 6 3 + : Cr 3 + is the central metal ion. Cr 3 + : 3 d 3 configuration It has three unpaired electrons. μ = 3 3 + 2   BM = 3 .87   BM CuF 6 3 − : Cu 3 + is the central metal ion. Cu 3 + : 3 d 8 configuration Fluorine is the weak field ligand. Its electronic configuration is t 2 g 6   e g 2 . Hence, it has two unpaired electrons. μ = 2 2 + 2   BM = 2 .84 BM
Related: Chemistry — Coordination Compounds · All PYQ Banks