The ratio of magnetic moments of $F e^{3 +}$ & $C o^{3 +}$ is:

\sqrt{5} : \sqrt{7}

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\sqrt{35} : \sqrt{24}

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7 : 3

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\sqrt{24} : \sqrt{15}

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Solution

The correct option is B $\sqrt{35} : \sqrt{24}$
Magnetic moment = $\sqrt{n (n + 2)}$

$C o^{3 +} \to (A r) 4 s^{0} 3 d^{6}$
No of unpaired electrons, $n = 4$

Magnetic moment $= \sqrt{24}$

$F e^{3 +} \to (A r) 4 s^{0} 3 d^{5}$
No. of unpaired electrons, $n = 5$

Magnetic moment $= \sqrt{35}$

Hence, the ratio is $\sqrt{35} : \sqrt{24}$ .

Therefore, option B is correct.

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Similar questions

Q. The ratio of magnetic moments of

F e (I I I)

and

C o (I I)

is:

Q. Study the pattern and write next step:

24 \times 5 = 120 \times 1

24 \times 15 = 120 \times 3

24 \times 25 = 120 \times 5

24 \times 35 = 120 \times 7

Verify each of the following :

(i) $(\frac{5}{7} \times \frac{12}{13}) \times \frac{7}{18} = \frac{5}{7} \times (\frac{12}{13} \times \frac{7}{18})$
(ii) $\frac{- 13}{24} \times (\frac{- 12}{5} \times \frac{35}{36}) = (\frac{- 13}{24} \times \frac{- 12}{5}) \times \frac{35}{36}$
(iii) $(\frac{- 9}{5} \times \frac{- 10}{3}) \times \frac{21}{- 4} = \frac{- 9}{5} \times (\frac{- 10}{3} \times \frac{21}{- 4})$

Q. Match the metal ions given in Column-I with the spin magnetic moments of the ions given in Column-II and assign the correct code :

\begin{matrix} Column-I & Column-II a . & C o^{3 +} & i . & \sqrt{8} B M b . & C r^{3 +} & i i . & \sqrt{35} B M c . & F e^{3 +} & i i i . & \sqrt{3} B M d . & N i^{2 +} & i v . & \sqrt{24} B M v . & \sqrt{15} B M \end{matrix}

Q. Simplify:

\frac{3^{5} \times 2^{2} \times 7^{- 3}}{7^{- 2} \times 3^{- 3} \times 2^{4}}

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The ratio of magnetic moments of Fe3+ & Co3+ is:

The correct option is B √35:√24Magnetic moment = √n(n+2)Co3+→(Ar)4s03d6No of unpaired electrons, n=4Magnetic moment =√24Fe3+→(Ar)4s03d5No. of unpaired electrons, n=5Magnetic moment =√35Hence, the ratio is √35:√24.Therefore, option B is correct.

The ratio of magnetic moments of $F e^{3 +}$ & $C o^{3 +}$ is: