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Question

Chemistry Question on d -and f -Block Elements

Magnetic moment of Cr2+Cr^{2+} is nearest to

A

Fe2+Fe^{2+}

B

Mn2+Mn^{2+}

C

Co2+Co^{2+}

D

Ni2+Ni^{2+}

Answer

Fe2+Fe^{2+}

Explanation

Solution

Cr = 3d, No. of unpaired electrons (n) = 4 Magnetic moment =n(n+2)=\sqrt{n\left(n+2\right)} BM =4(4+2)=24=4.89\quad\quad\quad\quad=\sqrt{4\left(4+2\right)}=\sqrt{24}=4.89 BM Fe = 3d, No. of unpaired electrons (n) = 4 Magnetic moment =4(4+2)=\sqrt{4\left(4+2\right)} BM =24=4.89\quad\quad\quad\quad\quad\quad\quad=\sqrt{24}=4.89 BM Mn = 3d, No. of unpaired electrons (n) = 5 Magnetic moment =5(5+2)=\sqrt{5\left(5+2\right)} BM =35=5.91\quad\quad\quad\quad\quad\quad\quad=\sqrt{35}=5.91 BM Co = 3d, No. of unpaired electrons (n) = 3 Magnetic moment =3(3+2)=\sqrt{3\left(3+2\right)} BM =15=3.87\quad\quad\quad\quad\quad\quad\quad=\sqrt{15}=3.87 BM Ni = 3d, No. of unpaired electrons (n) = 2 Magnetic moment =2(2+2)=\sqrt{2\left(2+2\right)} BM =8=2.82\quad\quad\quad\quad\quad\quad\quad=\sqrt{8}=2.82 BM