The ratio of magnetic moments of Fe(III) and Co(III) 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 $\sqrt{n (n + 2)}$

Electronic configuration of $C o^{3 +} i s (A r) 3 d^{6}$ is $\sqrt{24}$

Electronic configuration of $F e^{3 +} i s (A r) 3 d^{5}$ is $\sqrt{35}$

Option (B) is right.

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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

Q. The ratio of magnetic moments of

F e (I I I)

and

C o (I I)

is:

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. 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 Fe(III) and Co(III) is :

The correct option is B √35:√24Magnetic moment √n(n+2)Electronic configuration of Co3+is(Ar)3d6 is √24Electronic configuration of Fe3+is(Ar)3d5 is √35Option (B) is right.

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

Electronic configuration of $C o^{3 +} i s (A r) 3 d^{6}$ is $\sqrt{24}$

Electronic configuration of $F e^{3 +} i s (A r) 3 d^{5}$ is $\sqrt{35}$

Option (B) is right.