Can someone help me with this question

complete the following electronic configuration - Mn and Mn2+

addition of excess sodium hydroxide solution to a solution containing pink hydrated manganese (II) ions results in a sandy or buff coloured precipitate which slowly darkens on exposure to air

exaplin the reactions taking place (4)

i dont have the mark scheme for this particular question for some odd reason

Reply 1

Eau

Mn: [Ar] 3d^5 4s^2

Mn^2+: [Ar] 3d^5

Precipitate forming due to deprotonation of complex ions:

[Mn(H2O)6]^2+(aq) + 2OH-(aq) ---> [Mn(H2O)4(OH)2](s) + 2H2O(l)

"Slowly darkens to air" is the oxidation of Mn2+ to Mn3+.

Reply 2

No Future

Astudentinneed

Mn has 25 e-:
1s2
2s2, 2p6
3s2, 3p6
4s2
3d5

Mn 2+ has 23 e-:
In transition metals, the 4s e- are lost and gained first so the configuration would be:
1s2
2s2, 2p6
3s2, 3p6
4s0
3d5

Reactions

The Mn solution is [Mn(H20)6]2+
If you add OH- you will deprotonate the H20 ligands i.e. remove protons.
So you end up with a precipitate/solid of Mn(OH)2, a buff colour.

Equation: (this is simpler than the one above)
[Mn(H20)6]2+ + 2OH- ---> Mn(OH)2 + 6H20

Explanation:

The H20 ligands are polarised by the positive metal cation so the H is very + and thus is easily removed by the OH-.

Right, if the Mn(OH)2 ppt sits exposed to air for a while, oxygen in the air oxidises the ppt to MnO2 which is a dark brown colour.

Equation:
Mn(OH)2 + [O] ---> MnO2

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Not sure if I made sense, but I hope that helps.

Reply 3

Astudentinneed

i believe so, it's the last part that confuses me

and for the disprotonation can i have

[Mn(h2O)6]2+ + 2OH- -> Mn(H2O)4(OH)2 + 2H2O

Right, if the Mn(OH)2 ppt sits exposed to air for a while, oxygen in the air oxidises the ppt to MnO2 which is a dark brown colour.

Equation:
Mn(OH)2 + [O] ---> MnO2

right so the Mn will go from +2 to +4? why?