calculate number of unpaired electrons from magnetic moment for the following: 1) se,v,ti,fe,cr,cu,zn 2) Sc^+3,V^+3,Ti^+2,Cr^+3, Cu^+,Cu^++,Zn^++,Mn^+2,Mn^+7,Fe^+3
Answers
Answer:
Fe – 1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d6
Fe 3+ 1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d5
no. of unpaired electrons in Fe3+ = 5
μs = √n (n+2) BM
= √5 (5+2) = √ 35 = 5.92
Mn – 1s2,2s2,2p6, 3s2, 3p 6, 4s 2, 3d 5
Mn2+ 1s 2,2s 2,2p 6, 3s 2,3p6,4s0,3d5
no. of unpaired electrons in Mn2+ = 5
μs = √n (n+2) BM
= √5 (5+2) = √ 35 = 5.92
Cr – 1s2,2s2,2p6,3s2, 3p 6, 4s 1, 3d 5
Cr 3+ 1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d3
no. of unpaired electrons in Cr3+ = 3
μs = √n (n+2) BM
= √3 (3+2) = √ 15 = 3.87
Zn – 1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d10
Zn 2+ 1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d10
no. of unpaired electrons in Zn 2+ = 0
No magnetic moment
Question 2) Magnetic moment of a compound of vanadium is 1.73 BM. Write the electronic configuration of Vanadium in this compound.
Solution )
μ = 1.73
μs = √n (n+2) BM = 1.73
suppose n = 1, then
μs = √n (n+2) = √1 (1+2)
= √ 3 = 1.73
No. of unpaired electron in Vanadium = 1
23 V – 1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d3
V4+ 1s 2,2s 2,2p 6, 3s 2,3p6,4s0,3d1
V4+ ion has one unpaired electron in ‘d’ sub-shell.
Question 3) Magnetic moment of a compound manganese is 5.92 BM. Find the charge on metal cation in the compound.
Solution )
μ = 5.92
μs = √n (n+2) BM = 5.92
suppose n = 5, then
μs = √n (n+2) = √5 (5+2)
= √ 35 = 5.92
No. of unpaired electron in manganese = 5
25 Mn – 1s 2,2s 2,2p 6,3s 2,3p6,4s2,3d5
Mn 2+ 1s 2,2s 2,2p 6,3s 2,3p6,4s0,3d5
Mn2+ ion has five unpaired electron in ‘d’ sub-shell. So charge on metal cation is +2
Question 4) Calculate the orbital angular momentum for a ‘d’ electron ?
Solution )
orbital angular momentum = √ [l ( l + 1)] h / 2π
For ‘d’ orbital , l = 2
orbital angular momentum = √ [2( 2 + 1)] h / 2π = √ (6) h / 2π Ans.
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