Question

The spin-multiplicity of Fe3+ (Ec=[Ar]3d5) in its ground state
A)6
B)2
C)3
D)4

Answer 1

Answer:

option c(3)is your answer.....

Explanation:

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Answer 2

The correct answer is 5.

Given: Ion = $Fe^{+3}$.

To Find: Spin multiplicity of $Fe^{+3}$ in ground state.

Solution:

Spin multiplicity = 2s + 1 (s = n*$\frac{1}{2}$ and n is number of unpaired electrons.).

Electronic configuration of $Fe^{+3} = [Ar]3d^{5}$.

We have to calculate spin multiplicity in ground state. So in ground state 3 electrons will be more than in $Fe^{+3}$. SO add 3 electrons in electronic configuration of $Fe^{+3}$.

Fe = $[Ar]3d^{6}4s^{2}$.

Now as we can see that both the electrons in s are paired as s contains only 1 shell.

In d shell, 5 shells are there. Firstly fill 1 electron in each shell.

Now we are left with one more electron. So pair it with any one electron.

So, according to the electron filling. We have 4 unpaired electrons.

n = 4

s = $n*\frac{1}{2} =4*\frac{1}{2}=2$

Spin multiplicity = 2s + 1  = 2(2) + 1 = 5

Hence, the spin multiplicity of $Fe^{+3}$ in ground state is 5.

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