Question

the electronic distribution of Mn(25)

Answer 1

The answer is 1s2,2s2,2p6,3s2,3p6,4s2,3d5.
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Answer 2

Answer : The electronic distribution of Mn(25) is, $1s^22s^22p^63s^23p^64s^23d^5$

Explanation :

Electronic distribution : It is defined as the distribution of the electrons of an atom in an atomic orbital.

The given element is, Manganese (Mn)

The atomic number of Mn = 25

Atomic number = Number of protons = Number of electrons = 25

Hence, the electronic configuration of Mn is, $1s^22s^22p^63s^23p^64s^23d^5$