We know that the hyberdization of valence shell of Mn is 3d5 4s2, 4s2 is full and 3d5 is half and it must be very stable but its not and have 7 oxidation state(( Mn+1,Mn+2,. Mn+7)),WHY?
Thanks for helping

First, it's "hybridization" of valence shells.

I'm having trouble understanding your sentence structure, but I'll try to guess at what you're asking.

Manganese can exist in all oxidation states from +1 to +7. KMnO4 is the highest oxidation state of Mn. It is a +7 oxidation state. To form the +7 oxidation state, you need to lose the 3d5 electrons and 4s2 electrons. It is stable (it's a strong oxidizer, but it's stable at ordinary temperatures and pressures). The most common oxidation state of Manganese is +2. To form that, you'd lose the 4s2 electrons. To form the other states, you must lose one or more of the 3d5 electrons. Even the +1 state probably loses a 3d5 electron so as to keep the 4s2 electrons paired. The states aren't as stable as the +2 or +7 oxidation states, but they're stable. I'm not sure I can tell you all the reasons for this, except that the energy differences aren't all that great.

Thank you so much for your helping... but actually its not really my question... I say Mn in its valence shell has 3d5 that's half full and 4s2 that's full.. and it must be very stable... and when any metal is so stable like that it should not have so much oxidation state... but in inversely Mn is very stable and with that it have 7 oxidation state... I say why its stable and have 7 oxidation state, and when a metal is stable it should not have so much oxidation state because its stable...

Thank you again for your helping..

Well, having the configuration of1s2 2s2 2p6 3s2 3p6 (oxidation state 7+) is even more stable than 1s2 2s2 2p6 3s2 3p6 4s2 3d5 (ground state).

The Mn 2+ ion, 1s2 2s2 2p6 3s2 3p6 3d5 is also stable.

Now, for the rest, I can't say for sure. However, they are not as stable as the 7+ or 2+ oxidation state since they get easily oxidised or reduced to the extreme oxidation states 7+ and 2+.