Question

Given the following electronic configurations, identify how many valence electrons each atom has.

A=
B=
C=

Answer 1

Answer : Valence electrons in atom A = 1, B = 6 and C = 2.

Explanation :

Valence electrons means the outermost electrons are the ones which involved in bonding. Electrons are arranged in energy levels or shells.

Valence electrons are found in the highest energy orbitals of an element.

• The electronic configuration of (Na) is $1s^22s^22p^63s^1$. The highest energy orbital is 3s orbital which contains 1 electron. So,

        The number of valence electrons in (Na) = A = 1

Similarly,

• The electronic configuration of (O) is $1s^22s^22p^4$. The highest energy orbital are 2s and 2p orbitals which contains (2 + 4) = 6 electrons. So,

       The number of valence electrons in (O) = B = 6

• The electronic configuration of (Fe) is $1s^22s^22p^63s^23p^64s^23d^6$. The highest energy orbital is 4s orbital which contains 2 electrons. So,

       The number of valence electrons in (Fe) = C = 2

Answer 2

correct answer would be 1,6, and 8 so a=1, 6, c=8.