Question

calculat in BM manetic moment of v4+​

Answer 1

Vanadium has atomic number 23

it's electronic valence configuration is 4s²3d3 if it loses 4 electrons to become V⁴ then it will have 1 electron in its valence 4s1 3d^0 so it has one electron unpaired in its valence shell. Magnetic Moment=

$\sqrt{n(n + 2)}$

so it's √(1)(1+2)=√3=1.732 BM

Answer 2

The spin only magnetic moment is 1.73 BM

Explanation:

The electronic configuration of $V$ is, $1s^22s^22p^63s^23p^64s^23d^3$

The electronic configuration of $V^{4+}$ is, $1s^22s^22p^63s^23p^63d^1$

The number of unpaired electrons in $V^{4+}$ is 1

Formula used for spin only magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the magnetic moment.

$\mu=\sqrt{1(1+2)}=1.73BM$

Therefore, the spin only magnetic moment is 1.73 BM

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