Question

Config of mn+6 in mno4 2-

Answer 1

[Ar] 3d^1

Answer 2

To determine the electron configuration of Mn in $\text{MnO}_4^{2-}$, we follow these steps:

1. Determine the oxidation state of Manganese (Mn) in $\text{MnO}_4^{2-}$:

Let the oxidation state of Mn be $x$. The oxidation state of oxygen (O) in most compounds is $-2$. The overall charge of the ion is $-2$. So, we set up the equation: $x + 4(-2) = -2$ $x - 8 = -2$ $x = -2 + 8$ $x = +6$ Therefore, manganese is in the $+6$ oxidation state, meaning it is $\text{Mn}^{6+}$.

2. Write the electron configuration of a neutral Manganese (Mn) atom:

Manganese (Mn) has an atomic number of 25. Its ground state electron configuration is: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$ Using the noble gas notation (Argon, Ar, has 18 electrons: $1s^2 2s^2 2p^6 3s^2 3p^6$): $[\text{Ar}] 3d^5 4s^2$

3. Determine the electron configuration of the $\text{Mn}^{6+}$ ion:

To form a cation, electrons are removed from the neutral atom. For transition metals, electrons are removed from the orbital with the highest principal quantum number (n) first, even if it was filled last. In this case, electrons are removed from the $4s$ orbital before the $3d$ orbital. We need to remove a total of 6 electrons from the neutral Mn atom.

• First, remove the 2 electrons from the $4s$ orbital: $\text{Mn}^{2+}$: $[\text{Ar}] 3d^5$
• We still need to remove $6 - 2 = 4$ more electrons. These will be removed from the $3d$ orbital.

Original $3d$ electrons = 5. After removing 4 electrons from $3d$: $5 - 4 = 1$. Therefore, the electron configuration of $\text{Mn}^{6+}$ is: $[\text{Ar}] 3d^1$

The electron configuration of Mn in $\text{MnO}_4^{2-}$ (which is $\text{Mn}^{6+}$) is $[\text{Ar}] 3d^1$.