JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The magnetic moment of a transition metal compound has been calculated to be 3.87 BM. The metal ion is
Given: $$\mu = 3.87$$ BM. Using $$\mu = \sqrt{n(n+2)}$$ BM:
For $$n = 3$$: $$\mu = \sqrt{3 \times 5} = \sqrt{15} = 3.87$$ BM ✓
So the ion has 3 unpaired electrons.
$$V^{2+}$$: [Ar] 3d³ — has 3 unpaired electrons ✓
$$Cr^{2+}$$: [Ar] 3d⁴ — has 4 unpaired (high spin). $$Mn^{2+}$$: 3d⁵ — 5 unpaired. $$Ti^{2+}$$: 3d² — 2 unpaired.
The metal ion is $$V^{2+}$$.
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.