Spin only magnetic moment of the compound Hg - Co SCN 4- isA- -8B- -3- -24D- -15

The correct option is D √(15) Hg[Co(SCN)_4], Co(Z = 27) ⇒ 3d^7 4s^2, Co^2+ = 3d^7 Unpaired electrons, n = 3 Magnetic moment =√(n(n+2)) = √(3(3+2)) = √(15) BM ...

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Standard XII

Chemistry

Magnetic Moment Examples

Spin only mag...

Question

Spin only magnetic moment of the compound $H g [C o (S C N)_{4}]$ is

$\sqrt{3}$

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$\sqrt{15}$

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$\sqrt{24}$

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$\sqrt{8}$

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Solution

The correct option is B
$\sqrt{15}$

$H g [C o (S C N)_{4}]$ , Co(Z = 27) $\Rightarrow 3 d^{7} 4 s^{2}, C o^{2 +} = 3 d^{7}$

Unpaired electrons, n = 3

Magnetic moment $= \sqrt{n (n + 2)} = \sqrt{3 (3 + 2)} = \sqrt{15}$ BM

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Spin only magnetic moment of the compound Hg[Co(SCN)4] is

The correct option is B √15 Hg[Co(SCN)4], Co(Z = 27) ⇒3d7 4s2, Co2+=3d7 Unpaired electrons, n = 3 Magnetic moment =√n(n+2)=√3(3+2)=√15 BM

Spin only magnetic moment of the compound $H g [C o (S C N)_{4}]$ is

The correct option is B
$\sqrt{15}$

$H g [C o (S C N)_{4}]$ , Co(Z = 27) $\Rightarrow 3 d^{7} 4 s^{2}, C o^{2 +} = 3 d^{7}$

Unpaired electrons, n = 3

Magnetic moment $= \sqrt{n (n + 2)} = \sqrt{3 (3 + 2)} = \sqrt{15}$ BM