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QJEEM23Ch1de4: JEE Main 2023 Chemistry - Class 11 JEEM23Ch1de4 Solution: The magnetic moment of a transition metal compound has been…

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is

Multiple Choice Options

  • A Ti 2+^{2+}
  • B V 2+^{2+}
  • C Cr 2+^{2+}
  • D Mn 2+^{2+}

Step-by-Step Verified Solution

amp;Cr+2:[Ar],3 d4,4 s0n=4,μ=4(4+2)=24=4.89BMamp;Mn+2:[Ar],3 d5,4 s0n=5,μ=5(5+2)=35=5.91BMamp;V+2:[Ar],3 d3,4 s0n=3,μ=3(3+2)=15=3.87BMamp;Ti+2:[Ar],3 d2,4 s0n=2,μ=2(2+2)=8=2.82BM\begin{aligned}& \mathrm{Cr}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^4, 4 \mathrm{~s}^0 \mathrm{n}=4, \mu=\sqrt{4(4+2)}=\sqrt{24} =4.89 \mathrm{BM} \\\\& \mathrm{Mn}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^5, 4 \mathrm{~s}^0 \mathrm{n}=5, \mu=\sqrt{5(5+2)}=\sqrt{35} =5.91 \mathrm{BM} \\\\& \mathrm{V}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^3, 4 \mathrm{~s}^0 \mathrm{n}=3, \mu=\sqrt{3(3+2)}=\sqrt{15} =3.87 \mathrm{BM} \\\\& \mathrm{Ti}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^2, 4 \mathrm{~s}^0 \mathrm{n}=2, \mu=\sqrt{2(2+2)}=\sqrt{8} =2.82 \mathrm{BM}\end{aligned}

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