Spin magnetic moment of X k - Z -26 is -24 BM-Hence- the number of unpaired electrons and value of k -A- 0-2B- 4-2C- 2-4D- 3-1

The correct option is B 4,2 Spin magnetic moment =√(n(n+2)) √(n(n+2))=√(24) (given) ⇒ n=4 No. of unpaired electrons = 4 If it is X^2+, configuration will be, 1 ...

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Standard XII

Chemistry

Magnetic Properties

Spin magnetic...

Question

Spin magnetic moment of $X^{k +} (Z = 26)$ is $\sqrt{24}$ BM.Hence, the number of unpaired electrons and value of k:

4,2

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Solution

The correct option is A
4,2

Spin magnetic moment $= \sqrt{n (n + 2)}$

$\sqrt{n (n + 2)} = \sqrt{24}$ (given)

$\Rightarrow n = 4$

No. of unpaired electrons = 4

If it is $X^{2 +}$ , configuration will be,

$1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{6}$

No. of unpaired electrons = 4

Hence value of k is 2 in $X^{k +} .$

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Spin magnetic moment of Xk+(Z=26) is √24BM.Hence, the number of unpaired electrons and value of k:

The correct option is A 4,2 Spin magnetic moment =√n(n+2) √n(n+2)=√24 (given) ⇒n=4 No. of unpaired electrons = 4 If it is X2+, configuration will be, 1s2,2s2,2p6,3s2,3p6,3d6 No. of unpaired electrons = 4 Hence value of k is 2 in Xk+.

Spin magnetic moment of $X^{k +} (Z = 26)$ is $\sqrt{24}$ BM.Hence, the number of unpaired electrons and value of k: