Question

The magnetic moment of Cr3+ is similar to
that of
1) Fe2+
2) Fe3+
3) Co3+
4) Co2+​

Answer 1

Answer:

option 4

Explanation:

bcoz number of unpaired electrons is same

n=3 in both Cr+3 and Co+2..

magnetic moment=

$\sqrt{n(n + 2)}$

Answer 2

Answer:

The magnetic moment of Cr³⁺ is similar to that of Co²⁺.

Therefore, the option (4) is correct.

Explanation:

• Large the number of unpaired electrons present in the substance, greater will be paramagnetic character and larger the magnetic moment.

• We can use the formula to find the magnetic moment:

        $\mu=\sqrt{n(n+2)}$

        where n is unpaired electrons.

• The electronic configuration of Cr³⁺: [Ar]3d³4s⁰. The number of unpaired electrons will be three. So, the value of n=3.

        $\mu=\sqrt{3(3+2)} =\sqrt{15}=3.87\;B.M.$

• The electronic configuration of  Fe²⁺: [Ar]3d⁶4s⁰. The number of unpaired electrons will be four. So, the value of n=4.

         $\mu=\sqrt{4(4+2)} =\sqrt{24}=4.89 \;B.M.$

• The electronic configuration of Fe³⁺: [Ar]3d⁵4s⁰. The number of unpaired electrons will be five. So, the value of n=5.
• $\mu=\sqrt{5(5+2)} =\sqrt{35}=5.91\;B.M.$

• The electronic configuration of Co³⁺: [Ar]3d⁶4s⁰. The number of unpaired electrons will be four. So, the value of n=4.
• $\mu=\sqrt{4(4+2)} =\sqrt{24}=4.89\;B.M.$
• The electronic configuration of Co²⁺: [Ar]3d⁷4s⁰. The number of unpaired electrons will be three. So, the value of n=3.
• $\mu=\sqrt{3(3+2)} =\sqrt{15}=3.87\;B.M.$

Therefore, Cr³⁺ and  Co²⁺ has same magnitude of magnetic moment.