Question

Q.6 Calculate Spin only magnetic moment of divalent cation of a
transition metal with atomic number 25.​

Answer 1

Answer:

rh^2 (¢=$= 569 nkr) {ask=n=2^e3}

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Answer 2

Answer:

Spin only magnetic moment of divalent cation of a transition metal with atomic number 25 is 5.923M.

Explanation:

Manganese is a transition metal with an atomic number is 25.

The electronic configuration of manganese is $1s^22s^22p^63s^23p^64s^23d^5$.

Electronic configuration of Manganese in the divalent($Mn^{2+}$) state is $1s^22s^22p^63s^23p^63d^5$.

Spin only magnetic moment of an atom is calculated as $\sqrt{n(n+1)}$ in which n is the number of unpaired electrons in d-orbital.

So, the number of unpaired electrons d-orbital is 5.

Spin only magnetic moment of the given transition element is $\sqrt{5(5+1)}$.

Therefore, spin only magnetic moment of the given transition element is 5.923M.