An atom's configuration based on its number of electrons ends at 3p3. another atom has nine more electrons. starting at 3p, what is the remaining configuration?

3p64s23d4
3p34s23d7
3p63d44s2
3p33d74s2

3p34s23d7

A) 3p64s23d4

Explanation:

First off, it's important to understand the following concepts;

An atom's electron configuration is a numeric representation of its electron orbitals. The S-orbital takes a maximum of two electrons, p-orbital take a maximum of six electrons, d-orbital takes a maximum of 10 electrons, and so fourth.

The electron filling follows this pattern; 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p

An electron configuration can quickly and simply tell a reader how many electron orbitals an atom has as well as the number of electrons populating each of its orbitals.

Now back to the question, since the atom's configuration ends at 3p3 and we need to add nine more electrons,

1) P orbital can only accommodate 6 electrons, so we add three electrons to the 3p3 to make it 3p6 leaving us off with (9-3 = 6 electrons left).

2) The next orbital is the 4s. The s orbital can only take in 2 electrons so we add two electrons leaving us with 4 (6-2) electrons. we have 4s2 here.

3) The next orbital on the order is 3d. Since the d orbital can take in op to 10 electrons, it means we can add the remaining 4 electrons to give us 3d4.

The final electronic configuration from 3p is now given as;

3p64s23d4

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